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Student’s Book
Numerical Functional Analysis
Editor: Stefan Engblom
Fall 2014
Preface
This little book collects some of the student’s output during the course Numerical Functional Analysis, which was given for the first time in the fall 2014 at the Department of
Information technology, Uppsala university.
The course consisted of a total of five lectures, the first three of which went over Metric
spaces, Normed spaces, and Inner product spaces, thus following closely the first three
chapters of Kreyszig’s book Introductory functional analysis with applications. Two more
lectures were devoted to the five ‘Big’ theorems of functional analysis as presented by the
students themselves: the Hahn-Banach theorem, the Uniform boundedness theorem, the
Open mapping theorem, the Closed graph theorem, and the Banach fixed point theorem.
The final part of the course consisted of short essays on various topics, typically connecting Numerical analysis and Functional analysis in one way or the other. The essays
were improved by double open review among the students themselves after which the final
version entered this collection. A selection of student’s solution to book exercises has also
been included: the selection was very loosely based on the difficulty of the exercise itself
and on the solution quality.
Stefan Engblom
Uppsala, January 2015
Contents
I
Student’s essays
3
Dual-consistent approximations of PDEs and super-convergent functional
output
Martin Almquist
4
Lax equivalence theorem
Siyang Wang
15
The Schauder fixed point theorem
Hannes Frenander
20
Existence of a weak solution to the Stokes equations
Hanna Holmgren
24
The Lax-Milgram theorem
Andrea Alessandro Ruggiu
31
A time dependent mapping from Cartesian coordinate systems into curvilinear coordinate systems and the geometric conservation law
Samira Nikkar
37
The Babuˇ
ska inf-sup condition
Simon Sticko
42
Uniformly best wavenumber approximations by spatial central difference
operators
Viktor Linders
47
The Ekeland variational principle with applications
Markus Wahlsten
54
Construction of Sobolev spaces
Cristina La Cognata
59
1
The Arzel´
a -Ascoli theorem
Tomas Lundquist
65
A completeness theorem for non-self-adjoint eigenvalue problems
Saleh RezaeiRavesh
69
The metric Lax and applications
Cheng Gong
77
Fixed-point proof of Lax-Milgram theorem
Fatemeh Ghasemi
82
II
87
Solutions to exercises
1 Metric spaces
88
2 Normed spaces
96
3 Inner product spaces
112
2
Part I
Student’s essays
3
Dual-consistent approximations of PDEs and
super-convergent functional output
Martin Almquist∗
December 22, 2014
Abstract
Finite-difference operators satisfying the summation-by-parts (SBP)
rule can be used to obtain high-order accurate time-stable schemes for
partial differential equations (PDEs), when the boundary and interface conditions are imposed weakly by the simultaneous approximation
term method (SAT). The most common SBP-SAT discretizations are
accurate of order 2p in the interior and order p close to the boundaries,
which yields a global accuracy of p + 1 for first order PDEs. However,
Berg and Nordstr¨om [2012] and Hicken and Zingg [2014] have shown
that any linear functional computed from the time-dependent numerical solution will be accurate of order 2p if the spatial discretization is
dual-consistent.
This text discusses the concepts of dual problem and dual consistency. We walk through some of the definitions and results in Berg
and Nordstr¨om [2012] and Hicken and Zingg [2014] while emphasizing
the underlying functional analysis considerations.
1
Dual problems and dual consistency
We introduce the concepts of dual problem and dual consistency by considering linear partial differential equations with homogeneous boundary conditions. Let L be a linear differential operator of order m on a domain Ω and
consider the problem
Lu − f = 0, x ∈ Ω,
(1.1)
∗
Division of Scientific Computing, Department of Information Technology, Uppsala
university, SE-751 05 Uppsala, Sweden. [email protected]
1
subject to homogeneous boundary conditions. f ∈ L2 (Ω) is independent of
the solution u ∈ H k (Ω) ⊂ L2 (Ω). Here H k (Ω) = W k,2 (Ω) is the Sobolev
space consisting of functions whose weak derivatives up to order k (for some
integer k) exist and are square-integrable on Ω. We assume that m ≤ k.
For v, w ∈ L2 (Ω) we use the notation
Z
(v, w) := vw dx.
(1.2)
Ω
for the standard L2 inner product. We now introduce the formal adjoint of
the operator L.
Definition 1.1. Let u ∈ H k (Ω). The formal adjoint of a linear differential
operator L is the operator L† such that
(φ, Lu) = (L† φ, u)
The word formal in the definition emphasizes that we are not yet specifying the domain D(L† ) of the adjoint operator. The reason for this definition
is that L is not bounded on L2 (Ω), in general. Hence we need to distinguish
between the formal adjoint and the Hilbert-adjoint (which concerns bounded
operators) even though L2 (Ω) is a Hilbert space. Note also that the definition of L† is abstract; an explicit expression for L† φ can be obtained via
(repeated) integration by parts on (φ, Lu).
We now consider a bounded linear functional of the solution u,
J (u) = (z, u)
(1.3)
for some z ∈ L2 (Ω). Note that since L2 (Ω) is a Hilbert space, the representation (1.3) includes all linear functionals, by Riesz’s representation theorem.
To derive the adjoint equation of (1.1) with respect to the functional J , we
take the inner product of (1.1) with a function φ ∈ D(L† ) and express J as
J (u) = (z, u) = (z, u) − (φ, Lu − f ) = (φ, f ) − (L† φ − z, u).
(1.4)
We now obtain the adjoint or dual equation by seeking φ such that J becomes
independent of u, which requires
L† φ − z = 0,
yielding J = (φ, f ).
2
(1.5)
Remark. To simplify the analysis we have disregarded the boundary conditions. In general, the dual problem depends on the type of boundary
conditions in the primal problem, but not on the particular boundary data.
Boundary conditions for the dual problem can be constructed as the minimal
set of homogeneous conditions which, together with the homogeneous primal
boundary conditions, make all boundary terms resulting from the integration
by parts procedure vanish. We outline the procedure for a model problem in
Section 2.3.
Definition 1.2. The continuous dual problem associated with the primal
problem (1.1) and the functional J is
L† φ − z = 0
subject to the dual boundary conditions. The solution φ is called the dual
variable.
Note that the forcing function f and functional representation z in the
primal problem have switched roles in the dual problem.
We now turn to the discrete case. Let
Lh uh − f = 0
(1.6)
be a consistent discretization of (1.1) including boundary conditions, where
uh ∈ Rn may hold basis function coefficients and/or collocation values. Lh is
a bounded linear operator which can be represented by matrix multiplication,
Lh : Rn −→ Rn . Let (·, ·)h denote a discrete inner product on Rn × Rn such
that (Rn , (·, ·)h ) is a Hilbert space. We can then use the Hilbert-adjoint:
Definition 1.3. The Hilbert-adjoint of Lh is the operator L†h : Rn −→ Rn
such that
(φh , Lh uh )h = (L†h φh , uh )h
for all φh , uh ∈ Rn .
Let Jh (uh ) = (z, uh )h be an approximation of the functional J . We
now derive the discrete dual problem by taking the inner product of the
discretization (1.6) with φh ∈ Rn and expressing Jh as
Jh (uh ) = (z, uh )h − (φh , Lh uh − fh )h = (φh , fh )h − (L†h φh − z, uh )h . (1.7)
We obtain the discrete dual problem by seeking φh such that Jh is independent of uh , which requires
L†h φh − z = 0 .
(1.8)
3
Definition 1.4. The discrete dual problem associated with (1.6) and the
functional Jh is
L†h φh − z = 0
Definition 1.5. A discretization (1.6) of a continuous problem (1.1) is called
dual consistent if the discrete dual problem (1.8) is a consistent discretization
of the continuous dual problem (1.5).
1.1
Time-dependent problems
We now extend the concepts of dual problems and dual consistency to timedependent problems. Consider the problem
ut + Lu − f = 0, x ∈ Ω, t > 0
(1.9)
subject to homogeneous boundary and initial conditions. Let
d
u
dt h
+ Lh uh − f = 0, t > 0
(1.10)
be a semi-descretization of (1.9), including the boundary conditions.
Definition 1.6. The semi-discretization (1.10) is called spatially dual-consistent
if
Lh uh − f = 0
is a dual-consistent discretization of the steady adjoint problem, i.e., the dual
problem corresponding to (1.9) with ut = 0.
2
SBP-SAT discretizations
We begin this section by introducing notation and some basic concepts related to SBP-SAT discretizations, and then proceed to consider stable and
dual-consistent SBP-SAT discretizations.
2.1
Definitions
Consider as an example the advection equation on 0 ≤ x ≤ 1,
ut + ux = f
u(0, t) = g(t).
4
(2.1)
We introduce the N + 1 equidistant grid points
xj = jh,
j = 0, 1, . . . , N,
h=
1
N
(2.2)
and denote the semi-discrete solution vector corresponding to grid size h
by uh , where uh = [uh,0 , uh,1 , . . . , uh,N ]T ∈ RN +1 and uh,j (t) approximates
u(xj , t). The restriction [f (x0 ), f (x1 ), . . . , f (xN )]T of f onto the grid will
also be denoted by f , with no risk for confusion. Spatial derivatives will be
approximated by first-derivative SBP operators:
Definition 2.1. A matrix D is called a first-derivative SBP operator if D
can be written as
D = H −1 Q
where H = H T is positive definite, and thus defines a norm, and Q satisfies
Q + QT = diag[−1, 0, . . . , 0, 1].
In this text we will restrict ourselves to diagonal norm matrices H. In
that case, D consists of a 2pth-order accurate centered difference stencil in
the interior and pth-order accurate one-sided stencils near the boundaries.
The global accuracy for first order problems can then be shown to be p + 1
Sv¨ard and Nordstr¨om [2006].
Let u, v ∈ RN +1 . We define the discrete inner product and norm as
kuk2h = (u, u)h .
(u, v)h = uT Hv,
(2.3)
The inner product space (RN +1 , (·, ·)h ) is complete and thus a Hilbert space.
Now consider a discretization matrix Lh . We can compute the Hilbert-adjoint
L†h with respect to the inner product (2.3) explicitly:
Lemma 2.1. Let Lh be an (N + 1) × (N + 1) matrix with real entries. Then
the Hilbert adjoint is
L†h = H −1 LTh H
(2.4)
Proof. Starting from Definition 1.3, we have
(φh , Lh uh )h = (L†h φh , uh )h ∀φh , uh ∈ Rn
φTh HLh uh = (L†h φh )T Huh ∀φh , uh ∈ Rn
φTh HLh uh = φTh (L†h )T Huh ∀φh , uh ∈ Rn
HLh = (L†h )T H
(L†h )T = HLh H −1
L†h = H −1 LTh H
5
⇐⇒
⇐⇒
⇐⇒
⇐⇒
⇐⇒
2.2
Stable SBP-SAT discretizations
The SBP-SAT semi-discretization of (2.1) can be written
d
uh + Duh = f + τ H −1 e0 (eT0 uh − g)
dt
(2.5)
where e0 = [1, 0, . . . , 0]T and τ is a penalty parameter, which will be chosen
such that the scheme (2.5) is stable. Multiplying (2.5) by uTh H from the left
and setting f = 0, g = 0 yields
d
kuh k2h = (1 + 2τ )u2h,0 − u2h,N ,
dt
(2.6)
from which it is clear that τ ≤ − 21 leads to a stable scheme. Thus, the penalty
parameter τ is allowed to vary in a semi-infinite range. This flexibility creates
a possibility to impose additional conditions on the scheme, such as dual
consistency.
2.3
Dual-consistent SBP-SAT discretizations
Consider again the advection equation (2.1) on 0 ≤ x ≤ 1,
ut + ux = f
u(0, t) = 0.
(2.7)
with some initial condition and an associated linear functional
Z 1
J (u) =
zu dx.
(2.8)
0
Note that J is a time-dependent functional. The dual problem is obtained
by setting ut = 0 and finding φ such that J = (φ, f ). Using integration by
parts and inserting the boundary condition in (2.7), we obtain
Z 1
Z 1
J (u) = J (u) −
φ(ux − f ) dx = −φ(1, t)u(1, t) + (z + φx )u dx + (φ, f ),
0
0
and hence the dual problem is
−φx = z
φ(1, t) = 0 .
(2.9)
Note that the sign has changed and the boundary condition is located at the
opposite boundary, compared to the primal problem. We are now ready to
state the main result in this text:
6
Theorem 2.2. Let
d
uh + Lh uh = f
(2.10)
dt
be a stable and spatially dual consistent SBP-SAT approximation of the continuous problem
ut + Lu = f.
(2.11)
A linear functional
J (u) = (z, u)
(2.12)
Jh (uh ) = (z, uh )h
(2.13)
Jh (uh ) = J (u) + O(h2p )
(2.14)
approximated by
satisfies
Proof. The proof can be found in Berg and Nordstr¨om [2012].
We now investigate whether the scheme (2.5) can be made spatially dualconsistent. Assuming homogeneous boundary conditions (g = 0) and neglecting the time-derivative, the scheme can be written as
Lh uh = f
(2.15)
Lh = D − τ H −1 e0 eT0 .
(2.16)
where
By definition the scheme is spatially dual-consistent if the discrete dual problem
L†h φh − z = 0
(2.17)
is a consistent discretization of (2.9). Using Lemma 2.1 and the SBP property, we have
L†h φh −z = H −1 LTh Hφh −z = H −1 −Q + eN eTN − (τ + 1)e0 eT0 φh −z (2.18)
which is a consistent discretization of (2.9) if and only if τ = −1. For τ 6= −1
the scheme imposes a boundary condition at x = 0 which does not exist in the
continuous equation. Note that the requirement τ = −1 does not contradict
the stability requirement τ ≤ − 21 . Thus, we expect convergence of order 2p
for any linear functional of the solution if we choose τ = −1.
7
3
Numerical experiments
For the interested reader, we include a set of numerical experiments which
validate the theoretical results. We solve the advection equation using the
scheme (2.5) and consider the functional
Z 1
cos(4πx)u dx.
(3.1)
J (u) =
0
Using the technique of manufactured solutions, we choose the exact solution
u = cos(4πx) sin(2πt),
(3.2)
which corresponds to the forcing function
f = 2π cos(4πx) cos(2πt) − 4π sin(4πx) sin(2πt).
(3.3)
The analytic solution is used as initial and boundary data. With the solution
(3.2), the exact time-dependent functional becomes
J =
1
sin 2πt.
2
(3.4)
We use an SBP-SAT discretization of internal order 6 (2p = 6) and hence
expect 4th order convergence in the l2 -norm for any τ ≤ − 21 . We expect
the functional error to converge as (at least) h4 for τ ≤ − 12 . For the dualconsistent choice τ = −1, we expect 6th order convergence in the functional.
This is verified by Figures 3.1-3.3, which show the convergence rates for
τ = −0.5, τ = −1 and τ = −1.5, respectively. The l2 -error converges as h4
regardless of τ , whereas the functional error converges as h6 for τ = −1 and
as h4 otherwise.
Figure 3.4 shows both the l2 -error and the functional error as functions
of τ , using N = 256. As expected, the functional error is minimized by the
dual-consistent choice τ = −1. It is interesting to note that also the l2 -error
is minimized by τ ≈ −1 in this case, although none of the theory in this
text leads us to expect that. However, the l2 -error is much less sensitive to
the choice of τ than the functional error, which decreases by approximately
4 magnitudes in the vicinity of τ = −1.
References
J. Berg and J. Nordstr¨om.
Superconvergent functional output for
time-dependent problems using finite differences on summation-by-parts
8
10 0
10 -2
10 -4
10 -6
10
-8
l 2 -error
4th order reference
functional error
6th order reference
10 -10 1
10
10 2
10 3
N
Figure 3.1: Convergence in k·kl2 and functional for τ = −0.5; dualinconsistent discretization.
10 0
10 -5
10 -10
l 2 -error
4th order reference
functional error
6th order reference
10 -15 1
10
10 2
10 3
N
Figure 3.2: Convergence in k·kl2 and functional for τ = −1; dual-consistent
discretization.
9
10 0
10 -2
10 -4
10 -6
10
-8
l 2 -error
4th order reference
functional error
6th order reference
10 -10 1
10
10 2
10 3
N
Figure 3.3: Convergence in k·kl2 and functional for τ = −1.5; dualinconsistent discretization.
10 -6
10
functional error
l 2 -error
10 -4
-5
10 -6
-2
-1.5
-1
10 -8
10 -10
10 -12
-2
-0.5
τ
-1.5
-1
-0.5
τ
(a)
(b)
Figure 3.4: The (a) l2 -error ku−uh kl2 and (b) functional error, as functions of
the penalty parameter τ . Only τ = −1 yields a dual-consistent discretization.
10
form. Journal of Computational Physics, 231(20):6846–6860, Aug. 2012.
ISSN 00219991. doi: 10.1016/j.jcp.2012.06.032. URL http://www.
sciencedirect.com/science/article/pii/S0021999112003476.
J. Hicken and D. Zingg. Dual consistency and functional accuracy: a finitedifference perspective. Journal of Computational Physics, 256:161–182,
Jan. 2014. ISSN 00219991. doi: 10.1016/j.jcp.2013.08.014. URL http://
www.sciencedirect.com/science/article/pii/S0021999113005524.
M. Sv¨ard and J. Nordstr¨om. On the order of accuracy for difference approximations of initial-boundary value problems. J. Comput. Physics, 218:
333–352, October 2006.
11
Lax Equivalence Theorem
Siyang Wang∗
December 26, 2014
Abstract
Convergence of numerical methods for solving differential equations is of great importance, but difficult to prove in a direct way. The
Lax equivalence theorem links consistency, stability and convergence.
This enables us to prove convergence by consistency and stability analysis. In this essay, the Lax equivalence theorem is proved from the
functional analysis point of view. A differential equation example is
also presented.
1
Introduction
The Lax equivalence theorem is originally presented in Lax and Richtmyer
[1956]. It applies to an initial value problem
ut = f, 0 ≤ t ≤ tf ,
u = u0 , t = 0.
(1.1)
(1.2)
The theorem reads
Given a properly posed initial value problem 1.1, 1.2 and a finite difference approximation C(∆t) to it that satisfies the consistency condition,
stability is a necessary and sufficient condition that C(∆t) be a convergent
approximation.1
Remark. The term properly posed is equivalent to wellposed, which is used
more often nowadays. That is, a unique solution U exists, and a small
perturbation of the data leads to a small perturbation of the solution.
∗
Division of Scientific Computing, Department of Information Technology, Uppsala
University, SE-751 05 Uppsala, Sweden. [email protected]
1
This theorem is copied entirely from Lax and Richtmyer [1956].
1
2
The theorem
We prove the Lax equivalence theorem from the functional analysis point of
view. We reformulate the mathematical problem to a more general one.
Find x ∈ X such that
T x = y,
(2.1)
where y ∈ Y , T : X → Y is a linear operator from a normed space X onto
a normed space Y .
We assume that the above problem is wellposed, that is, T −1 : Y → D(T )
is continuous. This ensures that a small perturbation in the data y leads to
a small perturbation in the solution x. Since T is linear, T −1 is also linear.
The spaces X and Y are equipped with appropriate norms.
The mathematical problem 2.1 is solved numerically as
Tn xn = yn ,
(2.2)
where Tn : Xn → Yn , xn ∈ Xn and yn ∈ Yn . The spaces Xn and Yn
are equipped with appropriate norms. Here, n can be considered as the
resolution of 2.2. As n → ∞, we hope for xn → x. In practice (think of 2.1
as a differential equation), the true solution x is a continuous function, while
the numerical solution xn is discrete. In order to compare xn with x to check
convergence, we define the following two operators
RnX : X → Xn ,
RnY : Y → Yn .
These two operators project x and y (in X and Y ) to RnX x and RnY y (in Xn
and Yn ). We make the following definition
• The numerical method 2.2 is consistent with 2.1 if
Tn RnX z → RnY T z,
for any z ∈ D(T ).
• The numerical method 2.2 is stable if
kTn−1 k ≤ K,
for all n, where K is a constant independent of n.
• The numerical method is convergent if
Tn−1 RnY y → RnX T −1 y,
for every y ∈ Y .
2
The Lax equivalence theorem from the functional analysis point of view reads:
If the numerical method is consistent, it is convergent if and only if it is
stable.
Proof.
1. consistent + stable ⇒ convergent:
kTn−1 RnY y − RnX T −1 yk
=kTn−1 RnY T x − RnX xk
=kTn−1 (RnY T x − Tn RnX x)k
≤kTn−1 kkRnY T x − Tn RnX xk
≤K kRnY T x − Tn RnX xk
|
{z
}
→0 by consistency
Tn−1 RnY y
RnX T −1 y.
Therefore,
→
The method is convergent. Note that
this proof does not require deep functional analysis knowledge.
2. consistent + convergent ⇒ stable:
From convergence, we know that the sequence (Tn−1 RnY y) is bounded
[Kreyszig, 1978, Lemma 1.4–2], that is
kTn−1 RnY yk ≤ cy ,
for every y ∈ Y and n = 1, 2, .... By uniform boundedness theorem
[Kreyszig, 1978, Lemma 4.7–3], the sequence of the norms kTn−1 RnY k is
bounded independent of n. That is, there exists a constant β such that
kTn−1 RnY k ≤ β.
In order to proceed, we have to make the following assumption (*):
For every wn ∈ Yn , kwn k ≤ 1, there exists w ∈ Y, kwk ≤ α, such that
RnY w = wn , where α is a constant independent of n. This is a plausible
assumption in most cases in practice, since RnY is just a restriction
operator bringing a continuous function to its corresponding discrete
one. It then follows
kTn−1 RnY wk
kTn−1 wn k
kTn−1 RnY k = sup
= sup
≤β
kwk
kwk
kwk6=0
kwk6=0
The norm of Tn−1 can be written as kTn−1 k = supkzk=1 kTn−1 zk. Choose
wn such that kwn k = 1 and kTn−1 k = kTn−1 wn k. With this wn , we still
have kwk ≤ α. Therefore,
kTn−1 k ≤ αβ.
This proves that the numerical method is stable.
3
Remark. There are interesting examples Sanz-Serna and Palencia [1985] where
the assumption (*) does not hold, so that a consistent and convergent method
is not stable. Also, the uniform boundedness theorem requires that Yn is complete, i.e. a Banach space. This is not assumed or guaranteed, and might be a
strong assumption in practice. Rosinger even claims that the Lax equivalence
theorem is wrong Rosinger [2005].
3
Application
We choose the operator T in 2.1 as a differential operator d/dt. Then we
have a differential equation of the type
xt = y,
0 ≤ t ≤ tf ,
(3.1)
with the initial condition x(0) = c. Comparing with the mathematical problem 2.1, the space X and Y can be chosen as {x ∈ C 2 [0, tf ] | x(0) = c}
equipped with the maximum norm in [0, tf ].
We introduce the time discretization
tj = jk,
j = 0, 1, · · · , N,
where the step size k = tf /N . The corresponding discrete solutions are
x0 , x1 , · · · , xN . We equip the discrete space with the maximum norm. The
finite difference scheme can be written as
xn+1 = P (k)xn ,
n = 0, 1, · · · , N − 1,
where P (k) is the finite difference operator taking the discrete solution from
the current time step to the next time step. The recursion relation can be
written as


  
x1
P (k)x0
1
−P (k)
  x2   0 
1


  

  x3   0 
−P
(k)
1
.

  = 

  ..   .. 
..
..





.
.
.
. 
xN
0
−P (k) 1
{z
}
|
Q(k)
Stability requires that kQ−1 (k)k ≤ cQ , where cQ is independent of k. Due to
4
the special structure of Q(k), its inverse can be computed exactly, yielding


1
 P (k)

1


 2

1
Q−1 (k) =  P (k) P (k)
.


..
.
.
.
.
.
.


.
.
.
.
N −1
2
P
(k) · · · P (k) P (k) 1
Therefore, kQ−1 (k)k =
|P (k)| < 1.
PN −1
i=0
|P i (k)| =
1−|P (k)|N
.
1−|P (k)|
Stability is ensured if
Remark. By taking y = −λx, we get the familiar test equation, which is
often used to analyze the stability properties of numerical methods for time
integration. For forward Euler method, P (k) = 1 − kλ, and stability requires
|1 − kλ| < 1. This is the same result as if we would do the standard stability
analysis.
References
E. Kreyszig. Introductory functional analysis with applications. John Wiley
& Sons, 1978.
P. D. Lax and R. D. Richtmyer. Survey of the stability of linear finite difference equations. Comm. Pure Appl. Math., IX:267–293, 1956.
E. E. Rosinger. What is wrong with the Lax-Richtmyer fundamental theorem
of linear numerical analysis? Technical report, 2005.
J. M. Sanz-Serna and C. Palencia. A general equivalence theorem in the
theory of discretization methods. Math. Comp., 45(171):143–152, 1985.
5
The Schauder fixed point theorem
Hannes Frenander∗
January 5, 2015
Abstract
The Brouwer and Schauder fixed point theorems are presented; the
latter with a complete proof. An application of the Schauder fixed
point theorem regarding solutions to partial diﬀerential equations has
also been investigated.
1
Brouwer fixed point theorem
For completeness, we start by introducing the Brouwer fixed point theorem
(BFPT) since it is essential for understanding the proof of our main theorem: the Schauder fixed point theorem (SFPT). However, we neglect the
proof since it would obscure the main points of this essay. The theorem is
typically used in numerical analysis to prove existence of initial boundary
value problems, which will be demonstrated in the end of this essay. In general, the BFPT states that every continous mapping that maps a convex set
into itself has at least one fixed point. That is:
Theorem 1.1 (Brouwer fixed point theorem). Let X be a compact and convex set and f : X → X a continuous map. Then there is a x0 ∈ X s.t
f (x0 ) = x0 .
To illustrate, lets take a simple example by considering the domain X =
[0, 1] ⊂ R, i.e. the set of all real numbers between zero and one. This set
is compact and convex. BFPT now states that every mapping f (x) that
remains in this domain has at least one fixed point. For example, f (x) = x2
has a fixed point at x0 = 0, 1, f (x) = sin(x) has a fixed at x0 = 0 and so on.
∗
Division of Computational Mathematics, Department of Mathematics, Link¨oping university, SE-58183 Link¨oping, Sweden. [email protected]
1
2
Shauder fixed point theorem
The SFPT was proved 1930 by Juliusz Schauder, but only for special cases.
It remained for Robert Cauty to prove it in its final form in 2001. The
theorem is a generalization of BFPT into vector spaces and claims that every
continuous mapping from a convex vector space into itself has at least one
fixed point. That is:
Theorem 2.1 (Shauder fixed point theorem). Let X be a normed vector
space and K ⊂ X a convex and compact subset of X. Then every continuous
map f : K → K has a fixed point.
Proof. Since K is compact, it has a finite subcover. That is: there is a finite
sequence xi such that the open balls Bϵ (xi ) cover the set K, where ϵ > 0 is
the radius of the balls. Lets define the continuous functions gi (x) as
{
ϵ − ||x − xi || if ||x − xi || ≤ ϵ
gi (x) =
0
if ||x − xi || ≥ ϵ.
∑
Since gi (x) ≥ 0 and
i gi (x) > 0 for all x ∈ K, we can define another
continuous function in K:
∑
gi (x)xi
g(x) = ∑i
.
i gi (x)
1
We can easily∑verify that g is a map from
∑ K into the convex hull K0 of K.
That is, g = i αi xi where αi ≥ 0 and i αi = 1. One can also observe that
||g(x) − x|| ≤ ϵ
for all x ∈ K. The continuous map H = g ◦ f maps K0 into itself. Since K0
is compact, the BFPT tells us that there is a z ∈ K0 such that
H(z) = g(f (z)) = z
and therefore
||f (z) − z|| = ||f (z) − g(f (z))|| ≤ ϵ.
Since ϵ can be arbitrarily small, there is for all m ∈ N a zm ∈ K such that
||f (zm ) − zm || ≤
1
1
.
m
The
∑ convex hull K0∑of K is the smallest convex set that contains K. In this case, the
set { i αi xi |αi > 0,
i αi = 1} is a convex hull of K.
2
Remember that the theorem demands K to be compact, so there is a sub
sequence (zm )k such that f ((zm )k ) → x0 for some x0 ∈ K. Hence, we have
||(zm )k − x0 || = ||(zm )k − f ((zm )k ) + f ((zm )k ) − x0 ||
≤ ||(zm )k − f ((zm )k )|| + ||f ((zm )k ) − x0 ||
1
≤
+ ||f ((zm )k ) − x0 || → 0
mk
which means (zm )k → x0 . Since f is continuous, we also have f ((zm )k ) →
f (x0 ). This means f (x0 ) = x0 , i.e. x0 ∈ K is a fixed point.
3
An application
The Schauder fixed point theorem is commonly used to prove existence of
solutions to partial diﬀerential equations (Pouso [2012],Cao et al. [2014],Agarwal et al. [2013]), and we will in this section consider an example given by
Pouso [2012]. For clarity, we skip some details of the proof and the reader is
referred to Pouso [2012] for the full version.
Consider the problem
xtt = f (t, x)
x(0, t) = x(1, t) = 0
t ∈ I = [0, 1]
(3.1)
under the Carath´eodory’s conditions Pouso [2012]. Hence, in (3.1), the function f (t, x) is allowed to be discontinuous. We now prove that (3.1) has a
solution.
First, we demand that there is a M , such that |f (t, x)| ≤ M (t), and
consider the domain
∫ t
1
K = {x ∈ C (I), x(0) = x(1) = 0, |xt (t) − xt (s)| ≤
M (r)dr}
s
which can be proved to be compact and convex subset of the Banach space
C 1 (I) if we define the norm
||x|| = maxt∈I |x(t)| + maxt∈I |xt (t)|.
Consider now the operator T , defined by
∫ t
T x(t) =
G(t, s)f (s, x(s))ds
0
3
where G(t, s) is the Green’s function to (3.1). T maps K into itself if all the
above conditions are fulfilled. Therefore, according the SFPT, T has a fixed
point:
T x(t) = x(t)
which is then the solution to (3.1).
References
R. Agarwal, S. Arshad, and D. O’regan. A Schauder fixed point theorem in
semilinear spaces and applications. Fixed Point Theory and Applications,
306, 2013.
Z. Cao, C. Yuan, and X. Li. Applications of Schauders fixed point theorem to semipositone singular diﬀerential equations. Journal of Applied
Mathematics, 2014, 2014.
H. Pouso. Schauder’s fixed-point theorem: new applications and a new version for discontinous operators. Boundary Value Problems, 92, 2012.
4
Existence of a weak solution to the Stokes
equations
Application of the Babuška-Brezzi inf-sup theorem
to the Stokes equations
Hanna Holmgren∗
January 5, 2015
Abstract
The dynamics of incompressible viscous fluids are described by the Stokes
equations if stationary laminar flow conditions with low Reynolds numbers
are assumed, see Figure 0.1 for an example. The existence of a unique
solution to the weak formulation of the Stokes equations is proved in this
essay. First the Two Hilbert Spaces H10 (Ω) and L20 (Ω) are introduced and
the weak formulation of the Stokes equations is derived. Then, the BabuškaBrezzi inf-sup theorem is applied to the corresponding variational problem
to show existence and uniqueness of a weak solution.
Figure 0.1: Laminar flow conditions on the River Derwent in North Yorkshire.
1
Division of Scientific Computing, Department of Information Technology, Uppsala university,
SE-751 05 Uppsala, Sweden. [email protected]
∗
1
1
The Stokes Equations
The dynamics of incompressible fluids are generally described by the Navier-Stokes
equations. These equations are derived by considering the conservation of mass,
momentum and energy of a fluid flow. If the flow is “slow and calm” it is called
a laminar flow and an example is illustrated in Figure 0.1. For steady laminar
flows which have a small Reynolds number (intertial forces are small compared with
viscous forces) the Navier-Stokes equations are reduced to the Stokes equations.
In the Stokes equations all time dependent terms as well as the term describing
advection have been neglected. This essay aims at proving the existence of a
unique solution to the weak formulation of the Stokes equations.
Assume an incompressible viscous fluid with kinematic viscosity ν > 0 is filling up
a domain Ω ⊂ R3 with boundary Γ, then the Stokes system of equations is given
by
−ν∆u + ∇p = f, in Ω
∇ · u = 0, in Ω
u = 0, on Γ
(1.1)
(1.2)
(1.3)
where f is a given volume force, u is the unknown vector describing the velocity
of the fluid and p is the unknown pressure.
The Two Hilbert Spaces H10(Ω) and L20(Ω)
2
Before deriving the weak formulation of the Stokes system (1.1)−(1.3) we first
need to introduce two Hilbert spaces related to the two unknowns u and p. We
assume the velocity u to be in the Hilbert space H10 (Ω) (here Ω ⊂ R3 ), which is
the space of all vector fields v = (v1 , v2 , v3 ) with components vi in the space H01
defined by
H01 (Ω) = {v ∈ H 1 (Ω) : v|Γ = 0}.
(2.1)
Remember that the space H 1 (Ω) is equipped with the norm defined by
kvk2H 1 (Ω) = kvk2L2 (Ω) + k∇vk2L2 (Ω) .
The Hilbert space(s) H10 (Ω) is equipped with the inner product and seminorm
defined by
(u,v)
H10 (Ω)
=
Z
Ω
∇u : ∇v dx
|v|2H10 (Ω) = (v,v)H10 (Ω)
1
Picture from http://evidence.environment-agency.gov.uk
2
(2.2)
(2.3)
where
Z
Ω
∇u : ∇v dx =
Z
Ω
∇u1 · ∇v1 dx +
Z
Ω
∇u2 · ∇v2 dx +
Z
Ω
∇u3 · ∇v3 dx.
Further, we assume the pressure p to be in the Hilbert space L20 (Ω) defined by
L20 (Ω)
2
= {q ∈ L (Ω) :
Z
Ω
q dx = 0}
(2.4)
and equipped with the usual inner product and norm of the space L2 (Ω).
3
The Weak Form of Stokes Equations
Now, the weak formulation is derived by first multiplying the momentum equation
(1.1) by a test vector v ∈ H10 (Ω) and then integrating by parts which gives the
following weak formulation of the momentum equation
ν
Z
Ω
∇u : ∇v dx −
Z
Ω
(∇ · v)p dx =
Z
Ω
f · v dx,
where all boundary terms from the integration by parts vanishes since v|Γ = 0.
Further, multiplying the incompressibility constraint (1.2) by a test function
q ∈ L20 (Ω) and integrating yields
Z
Ω
(∇ · u)q dx = 0.
Summarizing, the variational formulation derived from the Stokes system of equations reads:
Find the solution pair (u,p) ∈ H10 (Ω) × L20 (Ω) such that
a(u,v) + b(v,p) = l(v) ∀v ∈ H10 (Ω)
b(u,q) = 0
∀q ∈ L20 (Ω),
(3.1)
(3.2)
where the following linear forms have been introduced:
a(u,v) = ν
Z
b(u,q) = −
l(v) =
Z
ZΩ
Ω
∇u : ∇v dx,
(3.3)
(∇ · u)q dx,
(3.4)
Ω
f · v dx.
3
(3.5)
4
The Babuška-Brezzi inf-sup theorem
For the abstract variational problem:
Find u ∈ V , such that
a(u,v) = l(v),
∀v ∈ V,
where V is a Hilbert space with inner product (·,·), a(u,v) is a coercive continuous
bilinear form on V and l(v) is a continuous linear form on V , the Lax-Milgram
lemma can be used to prove the existence and uniqueness of a solution [1] . Remember that a bilinear form a(·,·) is said to be V -coercive if there is a positive
constant m such that
mkvk2V ≤ a(v,v) ∀v ∈ V.
However, in the case of the weak formulation of the Stokes equations (3.1)-(3.2),
the existence of a unique solution does not follow from the Lax-Milgram lemma
alone, since it is impossible to establish coercivity for the bilinear form b(·, ·) in
(3.4) [2]. In this case we need the Babuška-Brezzi inf-sup theorem [1]:
Theorem 1: The Babuška-Brezzi inf-sup theorem. Let V and M be two
Hilbert spaces, and let a(·,·) : V × V −→ R and b(·,·) : V × M −→ R be two
continuous bilinear forms with the following properties: There exists a positive
constant α such that
a(v,v) ≥ αkvk2V
for all v ∈ U0 = {v ∈ V : b(v,q) = 0 ∀q ∈ M },
i.e. a(·,·) is U0 -coercive, and there exists a positive constant β such that
|b(v,q)|
≥ β,
q∈M,q6=0 v∈V,v6=0 kvkV kqkM
inf
sup
where the latter is called the Babuška-Brezzi inf-sup condition.
Finally, let l : V −→ R and g : M −→ R be two continuous linear forms. Then
the variational problem: Find (u,p) ∈ V × M such that
a(u,v) + b(v,p) = l(v)
b(u,q) = g(q)
has one and only one solution.
4
∀v ∈ V,
∀q ∈ M,
5
Existence and uniqueness of a weak solution
We use the Babuška-Brezzi inf-sup theorem to prove existence and uniqueness of a
solution to the variational problem (3.1)-(3.2) derived from the Stokes equations.
Let
V = H10 (Ω)
and
M = L20 (Ω),
where V and M are the two spaces from the Babuška-Brezzi inf-sup theorem.
First, we prove that a(·,·) is U0 -coercive:
1. a(·,·) is H10 (Ω)-coercive since
a(v,v) = ν|v|2H10 (Ω)
∀v ∈ H10 (Ω).
2. Then a(·,·) is also coercive on U0 = {v ∈ H10 (Ω) : b(v,q) = 0 ∀q ∈ L20 (Ω)}
since the above equality holds for all v ∈ H10 (Ω).
Then, for proving the second part of the Babuška-Brezzi inf-sup theorem (i.e. that
b(·,·) fulfills the Babuška-Brezzi inf-sup condition) we need the following theorem
[1]:
Theorem 2. Let Ω be a domain in RN . Then the injective continuous linear
operator
div : (ker div)⊥ −→ L20 (Ω)
is surjective and has a continuous inverse.
Here “div” is another notation used for the divergence, i.e. divv = ∇ · v for any
v. The space (ker div) is defined by
(ker div) = {v ∈ H10 (Ω) : divv = 0 in L20 (Ω)}.
Now, we prove that b(·,·) fulfills the Babuška-Brezzi inf-sup condition:
1. Since div : (ker div)⊥ −→ L20 (Ω) is surjective (by Theorem 2), there exists
a unique vector field w ∈(ker div)⊥ ⊂ H10 (Ω) such that
divw = q
in L20 (Ω),
and besides, since div : (ker div)⊥ −→ L20 (Ω) has a continuous inverse (also
by Theorem 2), there exists a constant C such that
|w|H10 (Ω) ≤ CkqkL20 (Ω)
5
∀q ∈ L20 (Ω).
2. Now we can prove the Babuška-Brezzi inf-sup condition holds for b(·,·), for
each non-zero q ∈ L20 (Ω):
R
|b(v,q)|
| Ω (∇ · v)q dx|
sup
=
sup
v∈H10 (Ω),v6=0 |v|H10 (Ω) kqkL20 (Ω)
v∈H10 (Ω),v6=0 |v|H10 (Ω) kqkL20 (Ω)
h
∇ · v = divv
i
R
| Ω (divv)q dx|
=
sup
v∈H10 (Ω),v6=0 |v|H10 (Ω) kqkL20 (Ω)
h
i
since w ∈ (ker div)⊥ ⊂ H10 (Ω) we can choose v = w
R
| (divw)q dx|
≥ Ω
|w|H10 (Ω) kqkL20 (Ω)
h
divw = q
i
R
q 2 dx|
=
|w|H10 (Ω) kqkL20 (Ω)
|
Ω
hR
=
q 2 dx = kqk2L2 (Ω)
0
i
kqkL20 (Ω)
|w|H10 (Ω)
h
≥
Ω
|w|H10 (Ω) ≤ CkqkL20 (Ω)
i
1
=β
C
So, by the Babuška-Brezzi inf-sup theorem, there exists a unique solution pair
(u,p) ∈ H10 (Ω) × L20 (Ω) to the variational problem (3.1)-(3.2) derived from the
Stokes equations.
6
References
[1] Philippe G. Ciarlet, Linear and Nonlinear Functional Analysis with Applications. Society for Industrial and Applied Mathematics, Philadelphia, 2013.
[2] Mats G. Larson, Fredrik Bengtzon, The Finite Element Method - Theory, Implementation and Applications. Springer, Umeå, Sweden, 2012.
7
The Lax–Milgram Theorem
Andrea Alessandro Ruggiu∗
January 7, 2015
Abstract
In this essay we state and prove the Lax–Milgram Theorem, a
result which claims well–posedness of variational problems. This theorem is here proven through the classical approach, that is to say in
this work the Banach Fixed Point Theorem is not taken into account.
1
Introduction
Various problems in science can be formulated through PDEs, which necessarly don’t have a classic solution. Solving a problem weakly means find a
solution for the problem, called weak solution, whose derivatives could also
not exist in a classical sense. Not rarely they are the only solutions that
is possible to find. In particular, it’s possible to formulate any differential
problem as a variational problem of the following kind
given F ∈ H ∗ , with H ∗ dual space of H, find u ∈ H such that
a (u, v) = F (v) ,
∀v ∈ H.
(1)
A common example is the Poisson problem
−∆u = f, x ∈ Ω
u
= 0, x ∈ ∂Ω
(2)
that can be formulated as a variational problem through the integration
by parts.
∗
Computational Mathematics, Department of Mathematics, Link¨
oping university, SE581 83 Link¨oping, Sweden. [email protected]
1
Indeed, multiplying the equation by v ∈ H01 (Ω) and integrating by parts
leads to
Z
Z
(3)
∇u · ∇v dx =
f v dx, ∀v ∈ H01 (Ω) .
Ω
Ω
The solution of (3), u ∈
2
H01
(Ω), is said weak solution for (2).
Background theorems
In this section we recall two important theorems that we want to use in
order to prove the Lax–Milgram Theorem. Both of them were studied and
analysed in the Numerical Functional Analysis course. The first one is the
Hilbert Projection Theorem.
Theorem 2.1. Let us consider a Hilbert space H, a non–empty subspace
V ⊂ H closed in H and x ∈ H. Then ∃!v = PV (x) ∈ V such that kx − vk =
inf v∈V kx − vk = d ≥ 0. Moreover
• PV (x) = x
⇐⇒
x∈V,
• QV (x) = x − PV (x) ∈ V ⊥ .
The second theorem that we want to use is the Riesz Representation
Theorem
Theorem 2.2. Let H be a Hilbert space. Then
∀L ∈ H ∗ ∃!u ∈ H s.t. L (v) = (u, v) ∀v ∈ H
and moreover kukH = kLkH ∗ .
3
Lax–Milgram Theorem
In this section we prove the following statement
Theorem 3.1. Let H be a Hilbert space. If a (·, ·) is a bilinear form such
that
• ∃M > 0 such that |a (u, v)| ≤ M kuk kvk,
• ∃α > 0 such that a (u, u) ≥ α kuk2 ,
2
and F ∈ H ∗ is linear and bounded, then ∃!u solution for (1) and
kuk ≤
1
kF k∗ ,
α
(4)
where k·k∗ is the norm associated to H ∗ .
Proof. By Theorem 2.2 ∃!z ∈ H s.t. F (v) = (z, v) , ∀v ∈ H and kF k∗ =
kzk. But a (u, ·) : V 7→ a (u, v) is a linear operator in H ∗ , then, again by
Theorem 2.2 we know that
∃!h ∈ H s.t. a (u, v) = (h, v) ,
∀v ∈ H
with u fixed. Thus we can call h = A (u). Notice that if u is fixed, then h
exists and it is unique.
Therefore (1) can be rewritten as
(A (u) , v) = (z, v) ,
∀v ∈ H,
which implies A (u) = z, A : H → H. In order to show the uniqueness of u
we have to prove that A is invertible. This proof requires 5 steps
• A is linear. As stated before, by Theorem 2.2
(A (u) , v) = a (u, v) ,
∀v ∈ H.
Then we can use the bilinearity of a (·, ·) in order to prove the linearity
of A
(A (λu1 + µu2 ) , v) = a (λu1 + µu2 , v) = λa (u1 , v) + µa (u2 , v) =
= λ (A (u1 ) , v) + µ (A (u2 ) , v) =
= (λA (u1 ) + µA (u2 ) , v) ,
∀v ∈ H. This implies that A (λu1 + µu2 ) = λA (u1 ) + µA (u2 ), that is
to say A is linear. For this reason from now on we write Au instead of
A (u).
• A is bounded. We can write
kAuk2 = (Au, Au) = a (u, Au) ≤ M kuk kAuk .
(5)
If kAuk = 0, then the boundedness follows trivially. Therefore we
assume kAuk =
6 0: dividing by kAuk both the left and the right–hand
side of (5), the boundedness of A follows.
3
• A is injective. This statement is equivalent to nul (A) = {0}, where
nul is the null space of A. Using the hypotheses and Cauchy–Schwarz
inequality in a (u, u) = (Au, u) leads to
α kuk2 ≤ a (u, u) = (Au, u) ≤ kAuk kuk .
This implies that if u 6= 0 (trivially an element of nul (A)) then
kAuk ≥ α kuk, that is to say
kuk ≤
1
kAuk ,
α
∀u ∈ H.
(6)
Therefore if we consider u ∈ nul (A) (kAuk = 0), then u = 0. This
means that the only element in the nullspace of A is 0, then A is
injective.
• The image of A is closed. Let yn ∈ Im (A) an arbitrary convergent
sequence in H, let’s say yn → y. The statement is proven if, and only
if, y ∈ Im (A). But since
yn ∈ Im (A) ⇐⇒ ∃xn s.t. yn = Axn
we can also prove the statement by showing the existence of x s.t.
y = Ax.
As a first step we can prove that xn is convergent. Using (6) and the
linearity of A we can write
0 ≤ kxn − xm k ≤
1
1
1
kA (xn − xm )k = kAxn − Axm k = kyn − ym k .
α
α
α
But since yn is convergent in H, then it is also Cauchy. In particular,
this implies by the above equation that xn is Cauchy in H. The completeness of H leads to the convergence of xn to a certain x ∈ H. By
hypothesis it is known that
Axn = yn → y.
(7)
But, on the other hand, we know by the continuity of A (it is linear
and bounded) that
Axn → Ax.
(8)
By (7), (8) and the uniqueness of the limit we have y = Ax, which
prove the closedness of Im (A) since x is such that y = Ax ∈ Im (A).
4
• A is surjective, that is to say Im (A) = H. We want to prove this
statement by contradiction: let us assume that Im (A) ( H. Then
∃q0 ∈
/ Im (A) and applying Theorem 2.1 (notice that Im (A) is closed
by the previous point) leads to
QIm(A) (q0 ) = q0 − PIm(A) (q0 ) ∈ Im (A)⊥ .
If we define
q0 − PIm(A) (q0 )
q=
q0 − PIm(A) (q0 ) ,
then q ∈ Im (A)⊥ and kqk = 1. Then we can write
0 = (Aq, q) = a (q, q) ≥ α kqk2 = α > 0.
Since we have a contradiction, then is impossible to have Im (A) ( H
and therefore A is surjective.
Finally, we have proven that A is invertible by proving that A is injective
and surjective. In particular Au = z has the unique solution u = A−1 z and
this means that there exists a unique solution for (1).
In order to prove the remaining part of the statement we use (6) and
Theorem 2.2
1
1
1
kuk ≤ kAuk = kzk = kF k∗ .
α
α
α
4
Example of application
Let’s take into account the problem (2) and its weak formulation given by
(3). We can write
Z
Z
a (u, v) ∇u · ∇v dx,
F (v) =
f v dx.
(9)
Ω
Ω
With this notation we have written (3) as a problem of the same kind of
(1) with H ≡ H01 (Ω). It can be shown that a suitable norm for H01 (Ω) is
given by
k·kH 1 (Ω) = k∇·kL2 (Ω)
0
and, equivalently, we can use as inner product in H01 (Ω)
(u, v)H 1 (Ω) = (∇u, ∇v)L2 (Ω) .
0
5
We can easily check that
|a (u, v)| ≤ k∇ukL2 (Ω) k∇vkL2 (Ω) = kukH 1 (Ω) kvkH 1 (Ω)
0
and
0
a (u, u) = k∇uk2L2 (Ω) = kuk2H 1 (Ω) .
0
Moreover F is linear and it is bounded, indeed
Z
f v dx ≤ kf kL2 (Ω) kvkL2 (Ω) ≤ cp kf kL2 (Ω) kvkH 1 (Ω) ,
F (v) =
0
Ω
where we have used the Cauchy–Schwarz and the Poincar´
e inequalities with
constant cp .
Then the Lax–Milgram theorem holds with α = M = 1 and therefore (2)
admits a unique solution u ∈ H01 (Ω) that is such that
kuk ≤ cp kf kL2 (Ω) .
References
[1] Evans L.C., Partial Differential Equations, American Mathematical Society, PP.297-299, 1997.
[2] Salsa S., Equazioni a derivate parziali. Metodi, modelli e applicazioni,
Springer, Chapter 6, 2nd edition, 2010.
6
A time dependent mapping from Cartesian
coordinate systems into curvilinear coordinate
systems and the geometric conservation law
Samira Nikkar∗
January 9, 2015
Abstract
A time-dependent transformation that maps a moving domain
in the physical space (in terms of the Cartesian coordinates) into
a fixed domain in the computational space (in terms of the curvilinear coordinates) is considered. The continuous transformation is
usually non-singular, which means that the Geometric Conservation
Law (GCL) holds exactly. In the numerical set up, the differential operators and the metric formulations must satisfy some certain criteria
in order for the Numerical GCL (NGCL) to be fulfilled. With the use
of the Summation-by-Parts (SBP) operators for the space and time
discretization, we can prove that even for the discrete mapping, the
NGCL is also guaranteed.
1
The continuous problem
Consider a time-dependent transformation from the Cartesian coordinates
into curvilinear coordinates, which results in a fixed spatial domain, as
x = x(τ, ξ, η), y = y(τ, ξ, η), t = τ,
ξ = ξ(t, x, y), η = η(t, x, y), τ = t.
(1.1)
such that 0 ≤ ξ ≤ 1, 0 ≤ η ≤ 1, 0 ≤ τ ≤ T , see Figure 1. The Jacobian
∗
Division of scientific computing, Department of Mathematics, Linkoping university,
Sweden. [email protected]
1
y
η
c
b
c'
d'
d
Ω
n
a
x
ab
bc
cd
da
a'
b'
ξ
a'b' : South (s)
b'c' : East (e)
c'd' : North (n)
d'a' : West (w)
Figure 1.1: A schematic of the moving and fixed domains and boundary
definitions.
matrix of the transformation is


xξ yξ 0
[J] =  xη yη 0 ,
xτ y τ 1
(1.2)
Jξt = xη yτ − xτ yη , Jξx = yη ,
Jξy = −xη
Jηt = yξ xτ − xξ yτ , Jηx = −yξ , Jηy = xξ ,
(1.3)
The relation between [J], and its inverse, which transforms the derivatives
back to the Cartesian coordinates leads to the metric relations
in which J = xξ yη −xη yξ > 0 is the determinant of [J].
If we want to preserve any conservation property while transforming the
physical domain into the computational domain, the transformation and the
metric relations must satisfy the GCL Farhat et al. (2001); Sj¨ogreen et al.
(2014), summarized as the following
Jτ + (Jξt )ξ + (Jηt )η = 0,
(Jξx )ξ + (Jηx )η = 0,
(Jξy )ξ + (Jηy )η = 0.
(1.4)
In practice (1.4) holds for any non-singular transformation. Now let us investigate the requirements for the discrete transformation such that the numerical version of (4) also holds.
2
2
The discrete problem
The spatial computational domain Ω is a square in ξ, η coordinates, see
Figure 1.1, and discretized using N and M nodes in the direction of ξ and
η respectively. In time we use L time levels from 0 to T. The fully-discrete
grid vector is a column vector of size LM N organized as follows

Z0


 ..
 .



Z=
 [Zk ]


 ..
 .


ZL








Z0
Z0


 .. 
 .. 

 . 
 . 





; [Zk ] =  [Zi ] ; [Zi ] =  [Zj ] ,
k 





 .. 
 .. 

 . 
 . 


ZM ki
ZN k


(2.1)
where Zkij = Z(τk , ξi , ηj ) and Z ∈ {x, y}.
The first derivative xξ is approximated by Dξ x, where Dξ is a so-called
SBP operator of the form
Dξ = Pξ−1 Qξ ,
(2.2)
and x = [x0 , x1 , · · · , xN ]T is the x coordinate of the grid points. Pξ is a symmetric positive definite matrix, and Q is an almost skew-symmetric matrix
that satisfies
Qξ + QTξ = E1 −E0 = B = diag(−1, 0, ..., 0, 1).
(2.3)
In (2.3), E0 = diag(1, 0, ..., 0) and E1 = diag(0, ..., 0, 1). The η and τ directions
are discretized in the same way.
A finite difference approximation including the time discretization Nordstr¨om and Lundquist (2013), on SBP-SAT form, is constructed by extending
the one-dimensional SBP operators in a tensor product fashion as
Dτ = Pτ−1 Qτ ⊗ Iξ ⊗ Iη ,
Dξ = Iτ ⊗ Pξ−1 Qξ ⊗ Iη ,
Dη = Iτ ⊗ Iξ ⊗ Pη−1 Qη
(2.4)
where ⊗ represents the Kronecker product Loan (2000). Note that in (2.4) we
have used the same names for the differential operators in multi dimensional
cases as the operators in the one dimensional case. All matrices in the first
position are of size L×L, the second position N×N , the third position M×M .
I denotes the identity matrix with a size consistent with its position in the
Kronecker product.
3
The Kronecker product is bilinear and associative. For square matrices
the following rules exist Nordstr¨om and Berg (2013),
(A⊗B)(C ⊗D) = (AC ⊗BD), (A⊗B)−1 = A−1 ⊗B −1 , (A⊗B)T = AT ⊗B T . (2.5)
For later reference we need
Lemma 2.1. The difference operators in (2.4) commute.
Proof. The properties (2.5) of the Kronecker product lead to
Dτ Dξ = (Pτ−1 Qτ ⊗ Iξ ⊗ Iη )(Iτ ⊗ Pξ−1 Qξ ⊗ Iη )
= Pτ−1 Qτ ⊗ Pξ−1 Qξ ⊗ Iη
= (Iτ ⊗ Pξ−1 Qξ ⊗ Iη )(Pτ−1 Qτ ⊗ Iξ ⊗ Iη ) = Dξ Dτ .
The proof is analogous for the other coordinate combinations.
We show
Lemma 2.2. The Numerical Geometric Conservation Law (NGCL) holds:
Jτ + (Jξt )ξ + (Jηt )η = 0,
(Jξx )ξ + (Jηx )η = 0,
(Jξy )ξ + (Jηy )η = 0.
(2.6)
Proof. Consider the following definitions,
Jτ
(Jξ t )ξ
(Jη t )η
(Jξx )ξ
(Jξy )ξ
(Jηx )η
(Jηy )η
=
=
=
=
=
=
=
diag[Dτ (Dη M (1) − Dξ M (2) )]
diag[Dξ (Dτ M (2) − Dη M (3) )]
diag[Dη (Dξ M (3) − Dτ M (1) )]
diag[Dξ (Dη y)]
diag[−Dξ (Dη x)]
diag[−Dη (Dξ y)]
diag[Dη (Dξ x)]
(2.7)
in which x and y are the discrete Cartesian coordinates defined in (2.1). Also
M (1) = diag(y)(Dξ x), M (2) = diag(y)(Dη x) and M (3) = diag(y)(Dτ x). By
Lemma 2.1 we find that the NGCL holds exactly.
3
Summary and conclusions
We have considered a time-dependent curvilinear coordinates. By using SBP
operators in space and time and a special definitions for the metric coefficients the Geometric Conservation Law is proven to hold numerically.
4
References
C. Farhat, P. Geuzaine, C. Grandmont, The discrete geometric conservation
law and the nonlinear stability of ALE schemes for the solution of flow
problems on moving grids, Journal of Computational Physics 174 (2001)
669–694.
B. Sj¨ogreen, H. C. Yee, M. Vinokur, On high order finite-difference metric
discretizations satisfying GCL on moving and deforming grids, Journal of
Computational Physics 265 (2014) 211–220.
J. Nordstr¨om, T. Lundquist, Summation-by-parts in time, Journal of Computational Physics 251 (2013) 487–499.
C. F. V. Loan, The ubiquitous Kronecker product, Journal of Computational
and Applied Mathematics 123 (2000) 85–100.
J. Nordstr¨om, J. Berg, Conjugate heat transfer for the unsteady compressible
Navier-Stokes equations using a multi-block coupling, Computers & Fluids
72 (2013) 20–29.
5
The Babuˇska Inf-Sup Condition
Simon Sticko∗
January 12, 2015
Abstract
A general weak formulation, typically arising from a linear timeindependent partial differential equation, is discussed. The Babuˇska
inf-sup condition is shown to guarantee that the weak formulation is
a well-posed problem.
1
Introduction
The finite element method applied to a linear time-independent partial differential equation typically leads to a weak formulation of the following form:
find u ∈ U such that
a(u, v) = l(v) ∀v ∈ V.
(1.1)
Here U and V are Hilbert spaces, a(·, ·) : U × V → R a bounded bilinear
form:
|a(u, v)| ≤ αC kukU kvkV , αC ∈ (0, ∞)
and l(v) : V → R is a bounded linear functional. Our goal is to find a
condition for when (1.1) is a well-posed problem. That is, the problem should
fulfill the following requirements:
(E) There exist a solution u.
(U) The solution is unique.
(C) The solution depends continuously on initial data.
∗
Division of Scientific Computing, Department of Information Technology, Uppsala
university, SE-751 05 Uppsala, Sweden. [email protected]
1
2
Theory
In order to understand if the problem in (1.1) is well posed or not we would
like to reformulate it into something which is a bit more familiar. In the
following we denote the scalar product in V and U by h·, ·iV and h·, ·iU respectively. Since U and V are Hilbert spaces a(·, ·) has a Ritz-representation
A : U → V such that:
a(u, v) = hAu, viV ,
where A is a bounded linear operator with the same norm as a(·, ·). Also l(·)
has a Ritz-representations f ∈ V :
l(v) = hf, viV .
Using this we can now express (1.1) as
hAu − f, viV = 0 ∀v ∈ V.
Since this should hold for all v ∈ V it holds in particular for v = Au − f . So
this leads to the equation
Au = f.
(2.1)
This formulation is equivalent to (1.1), but it’s easier to understand what is
required for the problem to be well posed. In order for a solution of (2.1)
to exist for an arbitrary f ∈ V we must require that there is at least one
u ∈ U such that Au = f . Thus A must be a surjective mapping. If we
want uniqueness we can not have two different elements u1 , u2 ∈ U such
that Au1 = f and Au2 = f . Thus we must also require A to be injective.
Finally, consider two different right-hand sides f1 , f2 , such that Au1 = f1
and Au2 = f2 . Since A is a linear operator we obtain
A(u1 − u2 ) = f1 − f2 ,
which leads to
ku1 − u2 kU ≤ A−1 V kf1 − f2 kV .
Thus if we want a small change in right hand side to generate a small change
in the solution we must require that A−1 is a bounded operator. So in
summary we have the following requirements for our problem to be well
posed:
2
(E) A is surjective.
(U) A is injective.
(C) A−1 is a bounded operator.
Our goal is to show that A fulfills these three requirements. This was shown
by Babuˇska [1971] and can be summarized as:
Theorem 2.1 (Babuˇska-Lax-Milgram). The weak formulation in (1.1) is
well posed if there exists αV , αU > 0 such that a(·, ·) fulfills:
αV kukU ≤ sup
v∈V
αU kvkV ≤ sup
u∈U
|a(u, v)|
∀u ∈ U
kvkV
|a(u, v)|
∀v ∈ V.
kukU
(2.2)
(2.3)
Proof. The three requirements (E), (U ) and (C) are proved separately in
Lemma 2.2, 2.3 and 2.4.
Remark. The conditions (2.2) and (2.3) are frequently referred to as inf-sup
conditions. The reason for this is that they can be formulated as: there exist
αU , αV > 0 such that
|a(u, v)|
≥ αV
inf sup
u∈U v∈V kuk kvk
U
V
inf sup
v∈V u∈U
|a(u, v)|
≥ αU .
kukU kvkV
Lemma 2.2 ((U ) Injectivity). If a(·, ·) satisfies (2.2) its Ritz-representation
is injective.
Proof. By the definition of the Ritz-representation and the Cauchy-Schwartz
inequality we have
| hAu, viV |
kAukV kvkV
|a(u, v)|
=
≤
= kAukV .
kvkV
kvkV
kvkV
By taking the supremum and utilizing (2.2) we obtain:
αV kukU ≤ sup
v∈V
|a(u, v)|
≤ kAukV .
kvkV
3
(2.4)
This gives us that
Au = 0 =⇒ u = 0
which implies injectivity since
Au1 − Au2 = 0 =⇒ A(u1 − u2 ) = 0 =⇒ u1 − u2 = 0.
Lemma 2.3 ((E) Surjectivity). If a(·, ·) satisfies (2.2) and (2.3) its Ritzrepresentation is surjective.
Proof. Consider the range of A: R(A), this is a subspace of V . In order to
show that this is a complete subspace let {yj }∞
j=1 be a Cauchy sequence in
R(A). For every > 0 there exists N such that
kyi − yj kV ≤ ∀i, j ≥ N.
But since yi ∈ R(A) we have yi = Axi with xi ∈ U so we get
kAxi − Axj kV ≤ ∀i, j ≥ N,
which by linearity and (2.4) gives us:
αV kxi − xj kU ≤ kA (xi − xj )kV ≤ ∀i, j ≥ N.
This shows us that {xj }∞
j=1 is a Cauchy sequence, which converges since U is
a Hilbert space. Denote the limit by x and let y = Ax. Since we originally
assumed that A is a bounded operator we now see that yj converges to y:
ky − yj kV = kAx − Axj kV ≤ αC kx − xj kU → 0 j → ∞.
Thus the conclusion is that R(A) is a complete subspace of V . This can only
happen if R(A) is also closed. Since R(A) is a closed subspace of V we can
decomposed V as a direct product space:
M
V = R(A)
R(A)⊥ .
(2.5)
But consider now a point v ∈ R(A)⊥ . From (2.3) we get
αU kvkV ≤ sup
u∈U
| hAu, vi |
= 0,
kukU
since Au ∈ R(A) and v ∈ R(A)⊥ . So this implies v = 0, which means that
R(A)⊥ = {0} since v was arbitrary. By (2.5) we must then have R(A) = V ,
so that A is surjective.
4
Lemma 2.4 ((C) Bounded Ritz-Inverse). Given that a(·, ·) fulfills (2.2), the
inverse of the Ritz-projection fulfills
−1 A ≤ 1 ,
U
αV
and thus is a bounded operator.
Proof. By rewriting (2.4) we obtain
kukU ≤
1
kAukV .
αV
Since we now know that A is a bijection we knot that for each u ∈ U there
exist an unique f ∈ V such that u = A−1 f . Inserting this gives us
−1 A f ≤ 1 kf k .
V
U
αV
Since this holds for all f ∈ V we obtain
−1
−1 A = sup kA f kU ≤ 1 .
V
kf kV
αV
f ∈V
References
I. Babuˇska. Error-bounds for finite element method. Numerische Mathematik, 16:322–333, 1971. ISSN 0029599X. doi: 10.1007/BF02165003.
5
Uniformly best wavenumber approximations
by spatial central difference operators
Viktor Linders∗
January 12, 2015
Abstract
We show that the problem of finding uniformly best wavenumber approximations by central difference schemes is equivalent to approximating a continuous function from a finite dimensional subspace. A
characterisation theorem for best approximations is proven.
1
Introduction and motivation
Consider a central difference approximation of a first derivative ux of some
function u(x, t) at the point x = xi ,
∂u
∂x
i
p
+ O(∆x2p ) =
1 X (p)
c (ui+k − ui−k ).
∆x k=1 k
(1.1)
(p)
Here ∆x is the spatial step size of the discretisation and ck are the coefficients of the 2pth order classical central difference scheme. It is well known
that the numerical dispersion relation of this scheme is
ξ¯c = 2
p
X
(p)
ck sin (kξ).
(1.2)
k=1
where ξ = κ∆x is the normalised wavenumber. The subscript c serves as a reminder that (1.2) corresponds to a classical stencil. Here we have introduced
an overbar to signify that the wavenumber is a numerical approximation.
∗
Division of Computational Mathematics, Department of Mathematics, Link¨
oping University, SE-581 83 Link¨oping, Sweden. [email protected]
1
Noting that the smallest wavelength the scheme can resolve is λmin = 2∆x,
the largest wavenumber is κmax = 2π/λmin = π/∆x so that we have |ξ| ≤ π.
Typically, waves with high frequencies and wavenumbers require small spatial increments, ∆x, in order to be properly resolved. Over sizeable intervals,
¯ may come to dominate the error in the
the so called dispersion error, |ξ − ξ|,
approximation, which gravely restricts the ∆x that may be used. Problems of this type are commonly encountered in computational fluid dynamics, aeroacoustics, electromagnetism, elasticity, seismology, and other fields
where energy is propagated. Wave properties encoded within the dispersion
relation include phase velocity, group velocity and dissipation. It is therefore
of interest to develop difference schemes that preserve the analytic dispersion relation of the governing equations for a wide range of spatial increments.
2
Problem formulation
Let us perturb the stencil (1.1) by adding n new coefficients, without increasing the accuracy. Thus, the scheme we consider has the form
p+n
1 X
∂u
2p
ak (ui+k − ui−k ).
(2.1)
+ O(∆x ) =
∂x i
∆x k=1
Such a scheme uses (p + n) points on either side of xi to approximate the
derivative, however the formal order of accuracy remains O(∆x2p ).
In view of (1.2) the numerical wavenumber corresponding to this scheme is
ξ¯ = 2
p+n
X
ak sin (kξ).
(2.2)
k=1
Our goal may thus be formulated in terms of the following problem:
Choose ak , k = 1, . . . , p + n, such that the error function
¯∞
kEk∞ := kξ − ξk
is minimised, ensuring that the stencil has formal
accuracy O(∆x2p ).
(p)
(2.3)
Note that if ap+j = 0, ∀j ≥ 1, we have ak = ck . By linearity we may thus
write
n
X
(p)
(j)
ak = c k +
αk ap+j , k = 1, . . . , p
(2.4)
j=1
2
(j)
where the parameters αk are independent of ap+j . We summarise in the
following proposition:
Proposition 2.1. The numerical wavenumber of the 2pth order central difference stencil (2.1) can be written as
ξ¯ = ξ¯c +
n
X
ap+j φj (ξ)
j=1
where ξ¯c is the wavenumber apprpximation of the classical central difference
stencil and
p
X
(j)
φj (ξ) = 2 sin ((p + j)ξ) + 2
αk sin (kξ).
k=1
(j)
The coefficients αk are independent of ap+j for all k = 1, . . . , p and j =
1, . . . , n.
Proof. Inserting (2.4) into (2.2) and collecting terms multiplying ap+j immediately gives the desired result.
The set {φj } is linearly independent and thus span some n-dimensional vector space, which we will denote by Ξn . We may thus consider each φj as a
basis function of Ξn .
P
Let ψ(ξ, a) = nj=1 ap+j φj (ξ). We see from Proposition 2.1 that the problem
(2.3) may equivalently be written
find a ∈ Rn that minimises kE(ξ, a)k∞ = kEc (ξ) − ψ(ξ, a)k∞ ,
(2.5)
where Ec (ξ) = ξ − ξ¯c is the error of the classical stencil and ψ(ξ, a) ∈ Ξn .
3
Approximation theory
For convenience we use the following definition:
Definition 3.1. If a vector a solves problem (2.5), then ψ(ξ, a) is called a
best approximation.
Finding and characterising best approximations are central problems in the
field of approximation theory. It is well-known that a solution to problem
(2.5) exists Riesz [1918]. It is in place to introduce some relevant concepts:
3
Definition 3.2. A set of functions, {θi }, i = 1, . . . , n forms a Chebyshev
set on the interval [0, ξmax ] if any non-trivial linear combination has at most
(n − 1) zeros in [0, ξmax ].
We will also need the following definition:
Definition 3.3. Let M (a) denote the set of points {ξ1 , ξ2 , . . . } at which
|E(ξ, a)| = kE(ξ, a)k∞ . Then E(ξ, a) is said to alternate n times on [0, ξmax ]
if M (a) contains (n + 1) points 0 ≤ ξ1 < ξ2 < · · · < ξn+1 ≤ ξmax such that
E(ξi , a) = −E(ξi+1 , a),
i = 1, 2, . . . , n.
The set {ξi } is referred to as an alternating set.
The usefulness of the above definitions becomes apparent when we consider
the following well-known extension of the Chebyshev alternation theorem:
Theorem 3.1. Let {θi (ξ)}, i = 1, . . . , n be a Chebyshev set on [0, ξmax ].
Then a solves problem (2.5) if and only if there is an alternating set of
(n + 1) points in [0, ξmax ]. Such a solution exists and is unique.
For a proof, see e.g. Watson [1980].
Theorem 3.1 identifies best approximations as those whose error oscillates
between a positive and a negative fixed value. The main task left is thus to
show that the set {φj } constitutes a Chebyshev set.
4
Characterisation of best approximations
The following observation is useful. The proof is omitted for brevity.
Lemma 4.1. Consider the function
f
(p)
(x, a) = 1 − 2
p+n
X
ak kTk (x)
k=1
where ak is defined as before and Tk (x) is the k th order Chebyshev polynomial
of the first kind, uniquely defined through the relation Tk (cos (φ)) = cos (kφ).
Then f (p) has a root at x = 1 of multiplicity p.
We are in position to prove our main result:
Theorem 4.2. The set {φj }, j = 1, . . . , n defined in Proposition 2.1 forms
a Chebyshev set on the semi-open interval (0, ξmax ].
4
Proof. Recall the form of the basis functions,
φj (ξ) = 2 sin ((p + j)ξ) + 2
p
X
(j)
αk sin (kξ).
k=1
Note first that φj (ξ) is a trigonometric polynomial. Thus the number of zeros
of any non-trivial combination, ψ(ξ, a) is finite in a bounded interval.
The function f (p) (x, a) is a polynomial in x of degree p + n with a root of
multiplicity p at x = 1, i.e.
f (p) (x, a) = (1 − x)p Pn (x, a),
where Pn (x, a) is some polynomial of degree n that depends linearly on a.
Let x = cos (ξ). We have
f (p) (cos (ξ), a) = 1 − 2
p+n
X
k=1
ak kTk (cos (ξ)) = 1 − 2
p+n
X
ak k cos (kξ) =
k=1
dE
.
dξ
Thus, E(ξ, a) has an extreme point at ξ = 0 and at most n other extreme
points ξ1 ≤ ξ2 ≤ · · · ≤ ξn in [0, ξmax ]. If ξ1 = 0 the stencil would be of higher
order than O(∆x2p ) so we may assume that 0 < ξ1 .
Consider now another error function, E(ξ, b). Linearity gives
∂ (E(ξ, a) − E(ξ, b))
= (1 − cos (ξ))p [Pn (cos (ξ), a) − Pn (cos (ξ), b)]
∂ξ
= (1 − cos (ξ))p Pn (cos (ξ), a − b)
but also
∂ (E(ξ, a) − E(ξ, b))
∂
=
[(Ec (ξ) − ψ(ξ, a)) − (Ec (ξ) − ψ(ξ, b))]
∂ξ
∂ξ
∂
=
ψ(ξ, b − a),
∂ξ
where, as before, ψ is a linear combination of the basis functions, {φj }. The
vectors a and b are general so we may write c = b − a for any vector c ∈ Rn .
Thus ψ(ξ, c) has at most n extrema in the open interval (0, ξmax ]. Obviously
there is at most one zero between any two consecutive extreme points. Thus
ψ(ξ, c) has at most n − 1 zeros in (0, ξmax ]. This completes the proof.
5
Corollary 4.3. Let {φj (ξ)}, j = 1, . . . , n be defined as in Proposition 2.1.
Then a solves problem (2.5) if and only if there exists an alternating set of
(n + 1) points in (0, ξmax ]. Such a solution exists and is unique.
Proof. Theorem 4.2 shows that for any δ satisfying 0 < δ < ξmax , {φj } is a
Chebyshev set on [δ, ξmax ]. Thus Theorem 3.1 applies. Letting δ → 0 and
noting that at ξ = 0 the approximation is exact yields the desired result.
5
Example
We conclude with a brief example: Figure 5.1 shows the error of a stencil
optimised with respect to the max-norm. The stencil is second order accurate
and is 31 points wide. Here we have set ξmax = π/2. In this region the
dispersion error is bounded from above by kEk∞ ∼ O(10−12 ). Clearly the
alternation property holds as expected from Theorem 3.1.
1.5
kEk∞ = 1.337520e − 12
×10 -12
1
E
0.5
0
-0.5
-1
-1.5
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
ξ
Figure 5.1: Optimised stencil for the case p = 1 and n = 14, i.e. the stencil
has formal accuracy O(∆x2 ) and is 31 points wide.
6
References
¨
F. Riesz. Uber
lineare functionalgleichungen. Acta Mathematica, 41:71–98,
1918.
G. A. Watson. Approximation Theory and Numerical Methods. John Wiley
& Sons Ltd., 1980.
7
The Ekeland Variational Principle With
Applications
Markus Wahlsten
∗
January 12, 2015
Abstract
We state and prove the Ekeland Variational Principle and exemplify its usefulness with some applications.
1
Introduction
The Ekeland Variational Principle is an important tool in areas such as nonlinear analysis and optimization theory. The principle is often used to establish existence of solutions for optimization problems where the Weierstrass’
theorem does not apply. Further, it’s proven to be a valuable tool when
studying partial differential equations Ekeland [1974].
2
The Ekeland Variational Principle
Before we state The Ekeland Variational Principle we recall the Cantor’s
intersection theorem and some definitions.
Theorem 2.1. (Cantor’s intersection theorem) Let {Sk } be a sequence of
non-empty, closed and bounded sets satisfying
S0 ⊇ S1 ⊇ · · · Sk ⊇ Sk+1 · · · ,
then
∞
\
k
Sk
!
∗
6= ∅.
(2.1)
(2.2)
Department of Mathematics, Computational Mathematics, Link¨
oping University, SE581 83 Link¨oping, Sweden [email protected]
1
Definition 2.1. (Upper and lower semicontinuity) Let f be a function and
x0 a point such that f : R → R and x0 ∈ R. The functiom f is said to be
upper semicontinuous at the point x0 if
f (x0 ) ≥ lim sup f (x)
(2.3)
f (x0 ) ≤ lim inf f (x)
(2.4)
x→x0
or lower semicontinuous if
x→x0
Theorem 2.2. (The Ekeland Variational Principle) Let X be a complete
metric space and φ : X → R an upper semi-continuous map that is bounded
from above. Then, for any > 0 there exists a y ∈ X such that,
φ(y) > φ(x) − d(x, y),
∀x ∈ X\{y}
(2.5)
Remark. The theorem can be modified by replacing the words ”upper” with
”lower” when also ”above” is replaced by ”below” and finally in (2.5) the
”−” and ”>” is replaced by ”+” and ”<” respectively.
Proof. By the fact that sup φ(X) < ∞, ∃z0 ∈ X such that,
φ(z0 ) > sup φ(X) − .
(2.6)
∞
If we now construct two sequences {zi }∞
i=0 and {Si }i=0 inductively in the
following manner
1. Assume zi is given
2. Define Si := {x ∈ X : φ(zi ) ≤ φ(x) − d(x, zi )}
3. Pick any element zi+1 ∈ Si such that
1
φ(zi+1 ) ≥ sup φ(Si ) − i+1
(2.7)
Since φ is upper semi-continuous, the map φ(·) − d(·, zi ) is also upper semicontinuous, which leads to the fact that Si is closed.
Next, we need to prove that Si+1 ⊆ Si for i = 0, 1, . . . . First, we pick an
arbitrary i ∈ Z+ and let x ∈ Si+1 . Then, by using the triangle inequality,
and the facts that x ∈ Si+1 and zi+1 ∈ Si , we obtain the relation
φ(x) − d(x, zi ) ≥ φ(x) − d(x, zi+1 ) − d(zi+1 , zi )
≥ φ(zi+1 ) − d(zi+1 , zi )
≥ φ(zi )
which means x ∈ Si and since x was arbitrarily chosen Si+1 ⊆ Si .
2
(2.8)
Further, in order to apply the Cantor Frechet Intersection Theorem we
also need to prove that
δ(Si ) = sup d(x, y) ≤
x,y∈Si
2
.
i
(2.9)
By picking an i ∈ Z+ , and an x ∈ Si+1 , which also is x ∈ Si which we know
from the proof above. By using this fact and the inequality in (2.7) we get,
d(x, zi ) ≤ φ(x) − φ(zi )
≤ sup φ(Si−1 ) − φ(zi )
≤ 1i
(2.10)
which proves the desired result. By applyingT
the Cantor Frechet Intersection
Theorem we conclude that the intersection +∞
i=0 Si = {y} for some y ∈ X.
To conclude our proof we need to show that the element y satisfies (2.7).
This is done by assuming the contrary, let x ∈ X\{y}, hence we have
φ(x) − d(x, y) ≥ φ(y),
(2.11)
φ(x) − d(x, y) ≥ φ(y) ≥ φ(zi ) + d(x, zi )
(2.12)
then
since (2.12) is satisfied for all i = 0, 1, . . . we have
φ(x) ≥ φ(zi ) + (d(x, y) + d(y, zi ))
≥ φ(zi ) + d(x, zi )
(2.13) is true for all i = 0, 1, . . . , hence x ∈
hence a contradiction.
3
T+∞
i=0
(2.13)
Si , which means x = y,
Applications
In this section several useful applications of the Ekeland Variational Principle
is presented.
Theorem 3.1. Caristi’s Fixed Point Theorem Let ψ be a self-map on a
complete metric space X. If
d(x, ψ(x)) ≤ φ(x) − φ(ψ(x)),
∀x ∈ X
(3.1)
for some lower semi-continuous φ ∈ RX that is bounded from below, then ψ
has a fixed point in X.
3
Proof. From the Ekeland Variational Principle, we know that there exists
some y ∈ X such that,
φ(y) < φ(x) + d(x, y),
∀x ∈ X\{y}.
(3.2)
However, by (3.1) we have that
φ(y) ≥ φ(ψ(y)) + d(y, ψ(y)).
(3.3)
So ψ(y) ∈
/ X\{y}, that is ψ(y) = y.
Definition 3.1. (Palais-Smale condition) We say that a C 1 functional φ :
X → R satisfies the Palais-Smale condition if every sequence {un }∞
n=1 ∈ X
which satisfies
kψ(un )k ≤ const,
and ψ(un ) → 0 in X
possesses a convergent subsequence.
(3.4)
Theorem 3.2. Let ψ : X → R be a C 1 functional where X is a Hilbert
space. Let S be a closed and convex subset of X. Suppose that
• K ≡ I − ψ 0 which maps S into S
• ψ is bounded from below in S
• ψ satisfies definition 3.1 in S
Then there exists a u0 ∈ S such that ψ 0 (u0 ) = 0 and inf ψ(u) = ψ(u0 )
u∈S
Proof. We start by applying the Ekeland Variational Principle to ψ. Given
there exists a u ∈ C such that ψ(u ) ≤ inf ψ(u) + and
u∈S
ψ(u ) ≤ ψ(u) + ku − u k ,
∀u ∈ S
(3.5)
If we now let u = (1 + t)u + tKu , where 0 < t < 1. Next, we use Taylors
formula to expand ψ(u + t(Ku − u )) about u and obtain,
2
t kψ 0 (u )k ≤ kψ 0 (u )k + O(t),
(3.6)
which implies kψ 0 (u )k < . Finally we use the property (3.1).
4
Summary
The Ekeland Variational Principle has been stated and proved in section 2
followed by some useful applications of the theorem in section 3.
4
References
I. Ekeland. On the variational principle. Journal of Mathematical Analysis
and applications, 47:324–353, 1974.
5
Construction of Sobolev spaces
Cristina La Cognata∗
January 15, 2015
Abstract
Sobolev spaces are Banach spaces that are characterized by the definition of ”weak” derivatives. The functions in Sobolev spaces and their
weak derivatives lie in Lp spaces.
Introduction
This essay is an introduction to the theory of “Sobolev” spaces. They are
powerful tools from Functional Analysis for defining the solutions of Partial
Differential Equations (PDE). In particular, they are used in those classes of
problems in which smoothness and continuity are too strong requirements.
In a certain sense, functions that belong to Sobolev spaces represent a good
compromise as they have some, but not all, smoothness properties [Evans,
2002]. We start with the definition of the concept of weak derivatives and
then we will apply it to the construction of the Sobolev spaces. We conclude
with an overview of some of the most interesting properties concerning these
spaces.
1
Weak derivatives
A kth-order PDE can be symbolically represented by the following expression
F (Dk u(x), Dk−1 u(x), ..., Du(x), u(x), x) = 0,
x ∈ U,
(1.1)
where k ≥ 1 is a fixed integer, U is an open set in Rn and u : U → R is
the unknown. The equation (1.1) involves an unknown function u : U → R,
∗
Division of Computational Mathematics, MAI, Link¨
oping University, SE-581 83
Link¨oping, Sweden. [email protected]
1
of variable x ∈ U , and some of its partial derivatives. By solving a PDE
of type (1.1) we mean (ideally) finding a simple (in the best case), explicit
expression for such a u, or at least guaranteeing its existence together with
some properties. Let us ignore for the moment the strict concept of a wellposed PDE problem. What do we require for u to be a “solution” of (1.1)?
It is natural to desire having a solution to be at least k times continuously
differentiable (u ∈ C k ). A solution of this type is called “classical” solution.
Unfortunately there exist certain PDEs (or most of them) which cannot be
solved in a classical sense. For instance, in general the conservation law of
the form
ut + F (u)x = 0,
x ∈ U,
(1.2)
has no classical solutions Evans [2002],R.LeVeque [1992]. Then we need to
allow a more general or weaker class of functions to be our solutions.
Let us denote by Cc∞ (U ) the spaces of infinitely many times differentiable
functions φ : U → R, with compact support on U . We will refer to a function
in Cc∞ (U ) as a test function. Furthermore, we will widely use Lp spaces and
their norms. For a a reader which is not familiar with these type of spaces
we remand to Evans [2002], R.LeVeque [1992].
Definition. Suppose that u, v ∈ L1loc (U )1 and α is a multi-index. We say
that v is the αth-weak derivative of u if
Z
Z
α
|α|
uD φ dx = (−1)
vφ dx
(1.3)
U
U
for all test functions φ ∈ Cc∞ (U ).
In other words, a weak derivative is a generalization of the concept of the
derivative of a function (strong derivative) for functions not assumed differentiable, but only integrable in the Lebesgue sense. Note that, since u is not
necessarily a differentiable function, the locally summability assumption is
needed to ensure the existence of the right-hand side of (1.3).
The notion of weak derivatives originated with the works of J. Leray (19061998) and later from S. Sobolev (1908-1989) when he came to deal with
non-continuous functions with almost everywhere existing derivatives.
Lemma 1.1. The αth-weak derivative of u is uniquely defined almost everywhere2
1
2
see Appendix
see Appendix
2
Proof. Let us assume that v, w ∈ L1loc (U ) both satisfy
Z
Z
Z
α
|α|
|α|
uD φ dx = (−1)
vφ dx = (−1)
wφ dx,
U
for all test functions φ ∈
U
Cc∞ (U ).
Z
U
U
Then
(v − w)φ dx = 0,
for all test functions φ ∈ Cc∞ (U ). Hence, (v − w) = 0 a.e.
Example. Let n = 1, U = (a, b) and consider u, v defined as follows
x
a < x ≤ (a + b)/2
u(x) =
1
(a + b)/2 ≤ x < b
1
a < x ≤ (a + b)/2
v(x) =
0
(a + b)/2 ≤ x < b.
We can show that v is the weak derivative of u by choosing any test function
φ ∈ Cc∞ (U ) and calculate
Rb 0
R (a+b)/2 0
Rb
uφ dx = a
xφ dx + (a+b)/2 φ0 dx
a
R (a+b)/2
= − a
φ dx + [φ((a + b)/2) − φ((a + b)/2)] ((a + b)/2) =
Rb
= − a vφ dx.
as required.
2
Sobolev spaces
We now define a function space which contains elements with weak derivatives
of various orders from Lp spaces. Let us fix 1 ≤ p ≤ ∞ and let k be a nonnegative integer.
Definition (Sobolev spaces). The Sobolev space W k,p (U ) consists of all locally summable functions u : U → R such that for each multi-index α with
|α| ≤ k, the weak derivative Dα u exists and lie in Lp (U ).
Definition (Sobolev norms). W k,p (U ) is a normed space equipped with the
norm defined as
 1/p
R
 P
α p
(1 ≤ p ≤ ∞)
|α|≤k U |D u| dx
kukW k,p (U ) =
,
(2.1)
P

α
ess
sup
|D
u|
(p
=
∞)
U
|α|≤k
where u ∈ W k,p (U ). For the definition of ess sup f see 3
3
We finally introduce the following notation for convergence:
k,p
Definition (Sobolev convergence). We say that {um }∞
(U )
m=1 , um ∈ W
k,p
converge to u ∈ W (U ) provided that
lim kum − ukW k,p (U ) = 0.
m→∞
Next we list some elementary properties Sobolev functions. The proofs can
be found in Evans [2002]. Note that the following properties are trivially
true for functions in C k (U ) spaces. For Sobolev functions, instead, we must
think about these rules as in the weak definition.
Theorem 2.1. Assume u, v ∈ W k,p (U ), |α| ≤ k. Then
1. Dβ (Dα u) = Dα (Dβ u) = Dα+β u for all |α| + |β| ≤ k.
2. For each λ, µ ∈ R, λu + µv ∈ W k,p (U ) and λDα u + µDα , |α| ≤ k.
3. if V is an open set in U , then U ∈ W k,p (V ).
Some of these properties are needed to prove that the Sobolev spaces are
Banach spaces.
Theorem 2.2 (Sobolev spaces as a function spaces). For each k = 1, ..., and
1 ≤ p ≤ ∞, the Sobolev space W k,p (U ) is a Banach space.
Proof. First we check that (2.1) is a norm. It is easy to check from the
definition that
kλukW k,p (U ) = |λ|kukW k,p (U )
kukW k,p (U ) = 0
and
iff
u=0
a.e.
(2.2)
Next we assume that u, v ∈ W k,p (U ) with 1 ≤ p < ∞. Then
ku + vkW k,p (U ) =
P
|α|≤k kD
u+D
α
vkpLp (U )
1/p
p 1/p
α
α
p (U ) + kD vkLp (U )
kD
uk
L
|α|≤k
P
1/p P
1/p
p
p
α
α
≤
kD
uk
+
kD
vk
p
p
|α|≤k
|α|≤k
L (U )
L (U )
= kukW k,p (U ) + kvkW k,p (U ) .
≤
P
α
Note that in the second inequality we have used Minkowski’s relation defined
in 3. Next we prove that W k,p (U ) is complete by showing that any Cauchy
sequence um ∈ W k,p (U ) converges.
4
1. For any |α| ≤ k, Dα um is a Cauchy sequence in Lp (U ). Since Lp (U ) is
complete, the existence of u and uα , ∀α, such that
um → u and Dα um → uα
in Lp (U ),
(2.3)
is guaranteed.
2. Consider any test function φ ∈ Cc∞ (U ). Then we have
R
R
α
uD
φ
dx
=
lim
u Dα φ dx
m→∞
U
U m R
= limm→∞R(−1)|α| U Dα um φ dx
= (−1)|α| U uα φ dx.
Thus, uα = Dα u, |α| ≤ k.
3. From the two previous steps we find that um → u in Lp (U ) and the uα
are the weak derivatives of u. Therefore um → u in W k,p (U ).
Thus W k,p (U ) is a Banach space.
We conclude by noting that in the special case p = 2, the Sobolev space has
a more interesting structure.
Definition (Hilbert-Sobolev spaces). We define the Hilbert-Sobolev space
as H k (U ) = W k,2 (U ). The space H k (U ) is a Hilbert space endowed with the
inner product
XZ
(u, v)H k (U ) =
Dα uDα v dx.
|α|≤k
U
Note that H 0 (U ) = L2 (U ).
For the curious reader, more details and results can be found in Brezis
[1983].
3
Appendix
Definition (Almost everywhere). If a property holds everywhere in Rn , except for a measurable set of Lebesgue measure zero, we say that the property
is true almost everywhere (a.e.).
Definition (ess sup). If a real function f is measurable, the essential supremum of f is defined as
ess sup f = inf {µ ∈ R s.t. |f > µ| = 0} .
5
Definition (L1loc functions).
L1loc = {u : U → R s.t. u ∈ Lp (V ) for each V ⊂⊂ U } .
Definition (Minkowski’s inequality). Assume 1 ≤ p ≤ ∞ and u, v ∈ Lp (U ).
Then
ku + vkLp (U ) ≤ kukLp (U ) + kvkLp (U ) .
The Minkowski inequality is the triangle inequality in Lp (U ).
References
H. Brezis. Analyse Fonctionnelle. Masson, 1983.
L. Evans. Partial Differential Equations. AMS, 2002.
R.LeVeque. Numerical Methods for Conservation Laws. Birkh¨auser, 1992.
6
The Arzel´a-Ascoli theorem
Tomas Lundquist∗
January 15, 2015
Abstract
We discuss the Arzel´
a-Ascola theorem from a functional analysis
point of view, and give examples of applications within the field, in
particular relating to the theory of compact embeddings in function
spaces.
1
Introduction
The Arzel´a-Ascola theorem is of fundamental importance in both real analysis and functional analysis, as well as in topology. In functional analysis, it is
particularly useful in function space theory, providing a constructive criterion
for compactness of subsets in C(X), the set of complex valued continuous
functions on a metric space X.
In its original form, the theorem was formulated for sequences of continuous functions on compact intervals of the real line, giving a criterion for
the existence of uniformly bounded subsequences. Later it was extended to
compact metric spaces then further to compact Hausdorﬀ topological spaces.
Here we will focus on the functional analysis version of the theorem. In section 2 we present the theorem and discuss some of its implications. Section
3 is a brief introduction to the theory of compact embeddings in continuous
function spaces, to which the theorem is closely related.
2
Statement of the theorem
The Arzel´a-Ascoli theorem gives a necessary and suﬃcient condition for compactness of continuous function spaces. Before we proceed with stating the
theorem, we need the following definition.
∗
Division of Computational Mathematics, Department of Mathematics, Link¨oping university, SE-581 83 Link¨oping, Sweden. [email protected]
1
Definition 2.1. A set F of functions between two metric spaces X and
Y is said to be equicontinuous if for each ϵ > 0, ∃ a δ > 0 such that
d(f (x0 ), f (x1 )) < ϵ for all f ∈ F , and x0 , x1 ∈ X with d(x0 , x1 ) < δ. The
parameter δ may depend on x0 but not on f .
Note the contrast between equicontinuity and uniform continuity. In the
former case, ϵ is not allowed to depend on f , whereas in the latter case, ϵ is
not allowed to depend on x0 .
The metric space version of the Arzel´a-Ascoli theorem now follows below.
Theorem 2.1. Let X be a compact metric space, and denote with C(X)
the space of real (or complex) valued continuous functions on X with respect
to the supremum norm. Then a subset M of C(X) is relatively compact in
C(X) if and only if M is equicontinuous and bounded.
Proof. A proof can be found e.g. in Nagy [2006].
¯ of
We recall that relatively compact in this case means that the closure M
¯
M is compact in C(X); that is, each sequence in M has a convergent subsequence. The Arzela-Ascola theorem is used in the proof of many important
theoretical results. One example is Peano’s theorem of local existence for
continuous ordinary diﬀerential equations. Another is to prove that an operator between Banach spaces is compact if and only if the adjoint operator
is compact (see e.g. Garrett [2012]).
3
Application: compact embeddings of continuous functions
The concept of compact embeddings between function spaces is important
not the least in the study of numerical methods for partial diﬀerential equations. For example, the famous Sobolev embedding theorems are examples
of results relating to compact embeddings. These results are important when
applying Sobolev spaces to partial diﬀerential equations. For a discussion of
this, see e.g. Dlotko [2014].
An embedding E : X ← Y from a subset X ⊂ Y into Y is simply the identity operator. As an example, consider the space C 1 (Ω) of
C 1 −diﬀerentiable functions on the domain Ω ∈ Cn , equipped with the norm
∥f ∥C 1 = sup|f | + sup|∇f |. With this definition, the embedding E(C 1 (Ω))
into C(Ω) is the space formed by the set C 1 (Ω) and standard supremum
norm. As we shall later see, this embedding is indeed compact.
2
Definition 3.1. Let (X, ∥·∥X ) and (Y, ∥·∥Y ) be two normed spaces, where X
is a subset of Y . We say that X is compactly embedded in Y if the following
holds:
1. ∃ c > 0 : ∥x∥Y ≤ c∥x∥X ∀x ∈ X.
(uniform continuity)
2. The embedding of X into Y is a compact operator. (compactness)
We recall that an operator is compact if it maps bounded sets to relatively
compact sets.
The Arzel´a-Ascoli theorem provides a constructive way to show compactness of embeddings. Too see this, we continue with the example above. Thus,
consider a bounded subset M of the space C 1 (Ω). The boundedness of sup|f |
and sup|∇f | then implies equicontinuity of the set (this part is left as an exercise). From the Arzel´a-Ascoli theorem, it also follows that M is relatively
compact in C(Ω). i.e. the embedding E(M) into C(Ω) is compact. Moreover, uniform continuity follows directly from the definitions of the norms
involved:
∥f ∥C(Ω) = sup|f | ≤ sup|f | + sup|∇f | = ∥f ∥C 1 (Ω) .
Thus the space C 1 (Ω) is compactly embedded in C(Ω) with respect to the
norm given above.
Analogous results to the one given above can be derived using the Arzol´aAscoli theorem for a variety of continuous function spaces. In particular
the Sobolev embedding theorems, of fundamental importance for Sobolev
space theory, can be proven using similar arguments. The most famous of
these result states that the embedding H m+1 (Ω) in H m (Ω) is compact, where
H m (Ω) is the Sobolev space with square integrable weak derivatives up to
order m. Note that H 0 (Ω) = L2 (Ω), so that all Sobolev spaces are compactly
embedded in the space of square integrable functions. A full proof of this,
generalized to Lp (Ω), is given e.g. in Adams [1975].
4
Summary
We have reviewed the metric space formulation of the Arzol´a-Ascoli theorem,
and shown its usefulness as a constructive criterion for compactness of continuous function spaces. An example of application is given by the famous
Sobolev embedding theorems.
References
R. Adams. Sobolev spaces. Academic Press, New York, 1975.
3
T. Dlotko. Sobolev spaces and embedding theorems.
www.icmc.usp.br/~ andcarva/sobolew.pdf, 2014.
P.
Technical report,
Garrett.
Compact
operators
on
banach
spaces:
Fredholm-riesz.
Technical
report,
http://www.math.umn.edu/~ garrett/m/fun/fredholm-riesz.pdf,
2012.
G. Nagy. A functional analysis point of view of the arzela-ascoli theorem.
Real Analysis Exchange, 32:583–586, 2006.
4
A Completeness Theorem for Non-Self-Adjoint
Eigenvalue Problems
Saleh RezaeiRavesh∗
January 16, 2015
Abstract
An important completeness theorem for a general class of nonself-adjoint eigenvalue problems is stated. The corresponding proof is
reproduced in connection with basic material in spectral theory and
functional analysis.
1
Introduction
In many applications in hydrodynamic stability analysis we are confronted
with eigenvalue problems in the form
(L − λM )ϕ = 0
(1.1)
where L and M are differential operators and λ ∈ σ(L, M ). The associated
boundary conditions for (1.1) are ϕ(0) = Dϕ(0) = ϕ(1) = Dϕ(1) = 0, where,
D := d/dy is the differentiation operator. One of the famous examples is the
Orr-Sommerfeld equation which is derived in linearized stability analysis of
parallel flows (e.g., cf. Drazin and Reid [1981]). The general idea for studying
the spectrum of (1.1) is to write L := Ls +B where Ls and M are self-adjoint
operators and B is called perturbation.
In Section 2, some of the definitions and theorems required in this document
are expressed. Then, theorem 3.1 shows how we can find a self-adjoint extension operator G for M −1 Ls in such a way that two problems Gϕ = λϕ
and Ls ϕ = λM ϕ become equivalent. Subsequently, theorem 3.2 proves the
completeness of the generalized eigenvectors of (1.1). Finally, an application
of the completeness theorem is briefly explained.
∗
Division of Scientific Computing, Department of Information Technology, Uppsala
university, SE-751 05 Uppsala, Sweden. [email protected]
1
2
Spectral Theory
Definition 2.1 (Resolvent Set & Spectrum of T ) Let X 6= {0} be a
complex normed space and T : D(T ) → X 1 a linear operator with D(T ) ⊂ X.
Then, (a) The resolvent set ρ(T ) of T is the set of all regular values λ of
T . If T : X → X is a bounded linear operator and X is a Banach space,
then, ρ(T ) = {λ ∈ |(T − λI) is one-to-one and onto} Evans [2010] (Any
λ ∈ σ(T ) is called a spectral value of T (cf. 7.2-1 Kreyszig [1989])), (b) The
spectrum of T is the complement of ρ(T ); i.e., σ(T ) = − ρ(T ) (cf. 7.2-1
Kreyszig [1989]).
Definition 2.2 (Resolvent Operator) If Tλ := (T − λI) has an inverse,
denoted by Rλ (T ) := Tλ−1 = (T − λI)−1 , we call it the resolvent operator of
T (cf. 7.2 Kreyszig [1989]).
Definition 2.3 (Point, Continuous, Residual Spectra) The point or
discrete spectrum σp (T ) is the set such that Rλ (T ) does not exist. λ ∈ σp (T )
is called an eigenvalue of T .
The continuous spectrum σc (T ) is the set such that Rλ (T ), exists and has a
domain dense in X, but is unbounded.
The residual spectrum σr (T ) is the set such that Rλ (T ), (a) exists (may be
bounded or not) but its domain is not dense in X (cf. 7.2-1 Kreyszig [1989]).
C
C
Theorem 2.1. (Distance from σ(T )) If T ∈ B(X, X), where X is complex Banach space and λ ∈ ρ(T ), then (cf. 7.5-3 Kreyszig [1989])
kRλ (T )k ≥
1
,
d(λ, σ(T ))
d(λ, σ(T )) := inf |λ − s|
s∈σ(T )
(2.1)
Definition 2.4 (Eigenvalue & Eigenvector) We say that λ ∈ σ(T ) is
an eigenvalue of T provided N (T − λI) 6= {0}. We write σp (T ) to denote
the collection of eigenvalues of T ; clearly, σp (T ) is a point spectrum. If λ
is an eigenvalue and ϕ 6= 0 satisfies T ϕ = λϕ, we say ϕ is an associated
eigenvector (cf. D.3 Evans [2010]).
Definition 2.5 (Generalized Eigenvector of (L, M )) Let L and M
be linear operators with D(L) ⊂ D(M ) dense in a Hilbert space H. We say
that ψ ∈ H is a generalized eigenvector of (L, M ) if and only if for some λ
and some integer p > 1, there exist p non-zero vectors ψ1 , ψ2 , · · · , ψp = ψ
such that Lψi = λM ψi + M ψi−1 , i = 1, · · · , p, where ψ0 = 0. By allowing
p = 1, eigenvectors become included in generalized eigenvectors (DiPrima
and Habetler [1969]).
1
In this manuscript, D(T ), R(T ), and N (T ) respectively denote domain, range and
null space of operator T .
2
Definition 2.6 (Completeness) The generalized eigenvectors of a linear
operator with domain dense in H are said to span H if the following is true.
Let ϕk1 , ϕk2 , · · · , ϕknk be the eigenvectors (and possibly generalized eigenvectors) corresponding to the eigenvalues in rk−1 < |λ| < rk for k > 1 and
|λ| < r1 for k = 1; then each ϕ ∈ H has a representation
ϕ=
ni
∞ X
X
aij ϕij
(2.2)
i=1 j=1
with convergence in the norm of H (DiPrima and Habetler [1969]).
Theorem 2.2. (Friedrichs Extension) A positive-bounded below (hence
positive definite) operator with domain dense in a Hilbert space can be extended to a self-adjoint operator which possesses an inverse defined on the
entire space (cf. p 11 Mikhlin [1965]).
Theorem 2.3. (Naimark’s Theorem) If T is a closed linear operator
on a dense subspace of a Hilbert space H with norm k.k whose resolvent
(T − λI)−1 is compact for some λ = λ0 , and if there exists a sequence of
circles Ck , |λ| = rk such that (a) there are no eigenvalues of T on any Ck ,
(b) limk→∞ rk = ∞, and (c) limk→∞ supλ∈Ck k(T − λI)−1 k = 0, then the
generalized eigenvectors of T span H in the sense of (2.2) (e.g., see DiPrima
and Habetler [1969]).
3
Completeness Theorem
Theorem 3.1. (Equivalent EVP’s) Let Ls and M be linear operators
with D(Ls ) ⊂ D(M ) dense in a Hilbert space H and such that,
1. M and Ls are positive-bounded below with R(M ) = R(Ls ),
2. G−1 is compact and R(G−1 ) ⊂ D(M ) where G−1 is the inverse of the
self-adjoint extension of M −1 Ls in HM .2
Then, the two problems Gϕ = λϕ and Ls ϕ = λM ϕ are equivalent (DiPrima
and Habetler [1969]).
Proof: I Step 1. (Construct G−1 ): M is psoitive-bounded below, i.e.,
∀x ∈ D(M ) ∃α > 0 : hM x, xi ≥ αkxk2
2
A new Hilbert space HM which is embedded in H can be constructed by completing
the pre-Hilbert space HM (D(M ), [., .]), where [f, g] = hf, M gi for f, g ∈ D(M ) is the
1
corresponding inner product and the norm is kf kHM = [f, f ] 2 (cf. 3 Mikhlin [1965]).
3
Combining this with the Schwarz inequality results in,
αkxk2 ≤ hM x, xi ≤ kM xkkxk
Note that y = M x for x and y in D(M ) and R(M ) of M , respectively; hence,
kM −1 yk ≤
1
kyk
α
Therefore, M −1 exists and is bounded. Furthermore, M −1 Ls is positivebounded below in HM ; since ∀x ∈ D(Ls ) ⊂ D(M ),
[x, M −1 Ls x] = hx, M M −1 Ls xi = hx, Ls xi ≥ βkxk2 ,
β>0
where the last inequality holds since Ls is positive-bounded below in H.
According to Theorem 2.2, M −1 Ls has a self-adjoint extension (call it G)
that has an inverse G−1 as mentioned in hypothesis 2.
I Step 2. (Find the Spectrum of G): Since G−1 is compact, then it has a
point spectrum, σp (G) (cf. 8.3-1, 8.3-3 Kreyszig [1989]). Furthermore, the
set of eigenvalues G is countable (may be finite and even empty) and λ = 0 is
the only possible point of accumulation of that set. In general, on D(G) ⊂ X
which is a complex normed space two cases can be considered; if X is finite
dimensional, then G has a matrix representation and clearly 0 may or may
not belong to σ(G) = σp (G); that is, if dimX < ∞ we may have 0 ∈
/ σ(G);
then 0 ∈ ρ(G). However, if dimX = ∞, then we must have 0 ∈ σ(G) (cf.
pp. 420 and 433 Kreyszig [1989]).
I Step 3. (Equivalent Spectra): We show that σ(G) = σp (G) = σp (Ls , M ).
3.a) Suppose λ ∈ σ(G); then, there exists ϕ ∈ D(G) such that Gϕ = λϕ.
Since from hypothesis 2, D(G) = R(G−1 ) ⊂ D(M ), then we have ϕ ∈ D(M )
and hence Gϕ ∈ D(M ). But, since R(M ) = R(Ls ), M −1 Ls is onto D(M ).
This means that there exists ψ ∈ D(Ls ) such that M −1 Ls ψ = Gϕ, and since
G = M −1 Ls on D(Ls ) and G−1 exists, we have ψ = ϕ. As a result,
M −1 Ls ψ = M −1 Ls ϕ = λϕ ⇒ Ls ϕ = λM ϕ ⇒ λ ∈ σp (Ls , M )
3.b) Conversely, consider λ ∈ σp (Ls , M ) and correspondingly, Ls ϕ = λM ϕ,
ϕ 6= 0. Then immediately we have M −1 Ls ϕ = λϕ an hence Gϕ = λϕ and
λ ∈ σ(G). Consequently, σp (Ls , M ) ∈ σ(G) and an eigenvector of G is
an eigenvector of (Ls , M ). This completes the proof of equivalency of the
Gϕ = λϕ and Ls ϕ = λM ϕ .
Theorem 3.2. (Completeness) Let L = Ls + B, and let Ls , B, and M
be linear operators with D(Ls ) = D(B) ⊂ D(M ) dense in a Hilbert space
H, such that,
4
1. Ls and M satisfy conditions of theorem 3.1,
2. M −1 B is bounded in HM and D(Ls ) is dense in HM ,
3. for some λ, R(Ls + B + λM ) = R(M ) and (G + M −1 B + λI) has an
inverse,
4. there exists a sequence of concentric circles {Ck } with the radii {rk }
such that, (a) limk→∞ rk = ∞, (b) d(Ck , σp (Ls , M )) > 0 for all k, (c)
limk→∞ d(Ck , σp (Ls , M )) = ∞.
Then,
1. the eigenvalues of (L, M ) lie within circles of radii kM −1 BkHM about
those of (Ls , M ), and,
2. the generalized eigenvectors of (L, M ) span HM in the sense of (2.2)
(DiPrima and Habetler [1969]).
Proof: I Step 1. (Define T and Show it is Closed) Consider T :=
G + M −1 B where M −1 B is a unique bounded extension from D(B) to D(G).
Now, we show that T satisfies the condition of Naimark’s theorem 2.3. According to the proof of theorem 3.1, G is self-adjoint and therefore closed.
In addition by hypothesis 2., M −1 B is bounded and when we add it to the
closed operator G, we end up with closed operator T .
I Step 2. (Show that ∃λ0 : Rλ (T ) = (T − λ0 I)−1 is Compact): Let’s choose
λ0 such that d(λ0 , σ(G)) > kM −1 BkHM . This is always possible; since according to hypothesis 4.(c), it is enough to choose λ0 on Ck for sufficiently
large k. G−1 is compact and G is closed, so (G − λI)−1 is compact for all
λ ∈ ρ(G) (cf. 5.7 Engel and Nagel [2006]), and hence for λ = λ0 . Now, we
can show that (T − λ0 I)−1 is compact. we have,
(T − λ0 I)−1 = (G + M −1 B − λ0 I)−1 = [I + (G − λ0 I)−1 M −1 B]−1 (G − λ0 I)−1
By using theorem 2.1, we can show that [I +(G−λ0 I)−1 M −1 B]−1 is bounded:
kI+(G−λ0 I)−1 M −1 BkHM ≥ 1−k(G−λ0 I)−1 kHM kM −1 BkHM ≥ 1−
kM −1 BkHM
>0
d(λ0 , σ(G))
Therefore, (T − λ0 I)−1 which is the product of a bounded operator and a
compact operator, is compact (cf. 8.3-2 Kreyszig [1989]).
I Step 3. (Obtain Conclusion 1.): 3.a) Since (T − λ0 I)−1 is compact and
T is closed, then (T − λI)−1 is compact for all λ ∈ ρ(T ) (see 5.7 Engel
and Nagel [2006]). Therefore, σ(T ) is a point spectrum such that each λ ∈
σ(T ) = σp (T ) has finite multiplicity (cf. 8.3-1 Kreyszig [1989]). On the other
hand, if λ is such that d(λ, σ(G)) > kM −1 BkHM then necessarily λ ∈ ρ(T ).
Consequently, for λ ∈ σ(T ), we have d(λ, σ(G)) < kM −1 BkHM . In other
5
words, σ(T ) lies within the union of a set of circles of radii kM −1 BkHM and
center σ(G) = σp (Ls , M ).
3.b) By a similar discussion made in Step 3. of the proof of theorem 3.1,
it can be shown that σ(T ) = σp (M −1 L). For this, replace Ls by L + λM ;
similarly, existence of G−1 is replaced by existence of (G + M −1 B + λI)−1 for
λ mentioned in hypothesis 3. Since it is possible that G + M −1 B does not
have an inverse, it is required that such λ exists. Hence, σp (M −1 L) lies within
the union of a set of circles of radii kM −1 BkHM centered about σp (M −1 Ls ).
I Step 4. Obtain Conclusion 2.: According to hypothesis 4.(c) we can choose
a k0 such that for all integers k ≥ k0 , d(Ck , σp (Ls , M )) = d(Ck , σp (G)) >
kM −1 BkHM . Then for such a k and λ ∈ Ck , we have
k(T − λI)ϕkHM = k(G − λI)ϕ + M −1 BϕkHM ≥ k(G − λI)ϕkHM − kM −1 BϕkHM
≥ {d(λ, σp (Ls , M )) − kM −1 BkHM }kϕkHM
Furthermore, k(T − λI)ϕkHM ≤ k(T − λI)kHM kϕkHM . Combining this with
the latest inequality and then taking supremum of both sides results in,
sup k(T − λI)−1 kHM ≤
λ∈Ck
1
d(λ, σp (Ls , M )) − kM −1 BkHM
Clearly, by hypothesis 4.(c), limk→∞ supλ∈Ck k(T − λI)−1 kHM = 0. Therefore
all the conditions of Naimark’s theorem are satisfied by T for circles Ck with
k ≥ k0 . Consequently, the generalized eigenvectors of T , and hence those of
(L, M ) span HM in the sense of (2.2) .
4
Application
To show an application of theorem 3.2, consider the Orr-Sommerfeld equation
rewritten in the form of equation (1.1); then
λ = iαcRe
Ls = (D − α ) , B = −iαRe[U (D2 − α2 ) − U 00 ]
M = −(D2 − α2 )
2
2 2
(4.1)
(4.2)
(4.3)
Here, U (y) is the base flow profile and U 00 = d2 U/dy 2 . Moreover, α and Re are
wavenumber and Reynolds number, respectively. In hydrodynamic temporal
stability analysis we are interested in finding the phase speed c appearing in
λ for a set of input parameters. In association with the completeness theorem
mentioned above, consider
H = L2 (Ω) = H 0 (Ω),
HM = H01 (Ω),
6
where Ω = [0, 1]
Then, through a detailed process, one can show that differential operators
in the Orr-Sommerfeld equation satisfy all hypotheses of theorems 3.1 and
3.2. Here as a step of that process, the construction of concentric circles is
presented. Consider σp (Ls , M ) = {µn } ∪ {ϑn } where the µn are the ordered
positive eigenvalues of the problem
Ls ϕ1 = µM ϕ1 ,
ϕ1 (0) = ϕ01 (0) = ϕ1 (1) = ϕ01 (1) = 0
(repeated eigenvalues, if any occurs, are listed once) and the numbers νn are
the ordered positive eigenvalues of the problem
M ϕ2 = ϑϕ2 ,
ϕ2 (0) = ϕ2 (1) = 0
For the operators in the Orr-Sommerfeld equation, it can be shown that as
n → ∞, µn+1 − µn → ∞ and ϑn+1 − ϑn → ∞. Thus, the sequence {Ck }∞
k=1
is a set of concentric circles with center at the origin and with radii rk as
follows. For n ≥ 1 consider the interval (µn , µn+1 ). Then,
(a) if there is no ϑl ∈ (µn , µn+1 ), take rn = 21 (µn + µn+1 );
(b) if there is only one ϑl ∈ (µn , µn+1 ), then when ϑl ∈ (µn , 21 (µn +µn+1 )) take
3
1
rn = 14 µn + 34 µn+1 and when ϑl ∈ ( 21 (µn + µn+1 ), µn+1 ) take rP
n = 4 µn + 4 µn+1 ;
m
1
(c) if there are several (say m) ϑl in (µn , µn+1 ), take rn = m i=1 ϑi (DiPrima
and Habetler [1969]).
As a result of theorem 3.2, for a bounded flow, the spectrum of the OrrSommerfeld equation consists of an infinite number of discrete eigenvalues
and the associated eigenvectors span HM = H01 (Ω) in the sense of (2.2). By
this completeness, an arbitrary initial disturbance which satisfies the physical boundary conditions can be expanded in terms of these eigenfunctions.
This technique is very useful in studying different hydrodynamic instability
mechanisms.
References
R. C. DiPrima and G. J. Habetler. The completeness theorem for nonselfadjoint eigenvalue problems in hydrodynamic stability. Archieve for Rational
Mechanics and Analysis, 34(3):218–227, 1969.
P. G. Drazin and W. H. Reid. Hydrodynamic Stability. Cambridge Univ.
Press, 1981.
K.-J. Engel and R. Nagel. A Short Course on Operator Semigroups. Springer,
2006.
7
L. C. Evans. Partial Differential Equations. American Mathematical Society,
2nd edition, 2010.
E. Kreyszig. Indtroductory Functional Analysis with Applications. John Wiley & Sons, 1989.
S. G. Mikhlin. The Problem of the Minimum of a Quadrature Functional.
Holden-Day INC., 1965.
8
The Metric Lax and Applications∗
Cheng Gong†
January 16, 2015
Abstract
We define consistency, convergence, and stability of numerical methods in the setting of metric spaces. The Lax equivalence theorem is
formulated and a proof is given in this setting. We also indicate its
use in some applications.
1
Introduction
The Lax (or Lax-Richtmyer) equivalent theorem Lax and Richtmyer [1956] is
the fundamental theorem in numerical analysis. It was established by Peter
Lax and Robert D. Richtmyer in 1956. The theorem states that “a consistent finite difference method for a well-posed linear initial value problem is
convergent if and only if it is stable”.
The convergence of a numerical method is a measurement of the differences between the numerical solution and the analytic solution from the
mathematical problem. In general, the analytic solutions are usually not accessible and the convergence is difficult to be observed directly. However, the
consistency and stability can be evaluated from the properties of the numerical method, e.g. error estimate, Fourier analysis and Von Neumann analysis
Gustafsson et al. [2013]. Thus, the Lax equivalent theorem provides a practical way to prove the convergence of a numerical method from stability. This
principle can be formulated and proved in metric spaces.
∗
This essay is based on the notes from Stefan Engblom. [email protected]
Division of Scientific Computing, Department of Information Technology, Uppsala
university, SE-751 05 Uppsala, Sweden. [email protected]
†
1
2
Definition
˜ be two metric spaces and we define the
Let X = (X, d) and Y = (Y, d)
mathematical problem as: find x ∈ X, such that
T x = y,
(2.1)
where T : X −→ Y is an operator defined by the problem and y ∈ Y is
given.
Definition 2.1. The mathematical problem in (2.1) is well-posed if T −1
exists and is continuous in some neighborhood containing y.
Then, the solution of a well-posed problem with the same form as in (2.1)
is given by the inverse operator of T as
x = T −1 y.
(2.2)
The problem in (2.1) is approximated by a sequence of numerical problems:
find xn ∈ X such that
Tn xn = y,
(2.3)
where Tn : X −→ Y is an operator of the corresponding numerical method
and y ∈ Y is the same right-hand-side as in (2.1). The family of equations in
(2.3) is called the numerical approximations for the problem in (2.1) [Hunter
and Nachtergaele, 2001] and n refers to the indices of the sequence such that
Tn −→ T as n −→ ∞.
Definition 2.2. Let D(T ) be the domain of T . The numerical method in
(2.3) is consistent, if ∀x ∈ D(T ), where D(T ) ⊂ X is the domain of T in
X, then Tn x −→ T x as n −→ ∞.
Definition 2.3. The numerical method in (2.3) is stable if ∀n ∈ N, Tn−1
exists and is continuous in some neighborhood containing y.
Definition 2.4. The numerical method in (2.3) is convergent if xn −→ x
as n −→ ∞.
For a stable numerical method, the solution is given as xn = Tn−1 y by the
inverse operator of Tn−1 . If the numerical method is convergent, the numerical
solution will converge to the analytic solution of the mathematical problem.
2
3
The Lax Equivalent Theorem
A general form of Lax equivalent theorem is given in Theorem 3.1.
Theorem 3.1. A consistent method applied to a well-posed problem is convergent if and only if it is stable.
By introducing the definitions from Section 2, the Lax equivalent theorem
is formulated in Theorem 3.2.
Theorem 3.2. Define a mathematical problem as (2.1) and the numerical
approximation as (2.3). Suppose T −1 is continuous and Tn x −→ T x for any
x in the domain of T . Then, Tn−1 y −→ x as n −→ ∞ if and only if Tn−1 is
continuous for all n.
Proof. [Engblom, 2014]
‘=⇒’
By the consistency of Tn , for any given δ, ∃N ∈ N such that, for all n > N
˜ n x, T x) < δ.
d(T
(3.1)
From the assumption in (2.1) and (2.3),
T x = y = Tn xn ,
(3.2)
together with the well-posedness of the problem,
˜ n x, Tn xn ) < δ.
d(T
(3.3)
By the stability of the numerical method, Tn−1 is continuous at y for all n.
Denote y1 = Tn x, then ∀ > 0 and for all y1 satisfying (3.3), ∃δ > 0, such
that
d(Tn−1 y1 , Tn−1 y) < ,
(3.4)
which is equivalent to
d(x, xn ) < .
(3.5)
Therefore, xn −→ x as n −→ ∞ and the numerical method is convergent.
‘⇐=’
If the numerical method in (2.3) is not stable, then ∃ > 0, for ∀δ > 0,
˜ 1 , y) < δ, we have
∀y1 ∈ Y satisfying d(y
d(Tn−1 y1 , Tn−1 y) > .
(3.6)
From the consistency, there exists an N ∈ N, ∀n > N ,
˜ n x, T x) < δ.
d(T
3
(3.7)
If we take y1 = Tn x, then (3.6) is written as
d(Tn−1 Tn x, Tn−1 Tn xn ) > ,
(3.8)
d(x, xn ) > ,
(3.9)
by (3.2)
and consider the well-posedness of the problem that T −1 exists in some neighborhood of y,
(3.10)
d(T −1 y, Tn−1 y) > ,
which indicates that the numerical method is not convergent. Therefore, a
consistent and convergent numerical method is stable.
4
Applications
In a metric space, the metric can be used to define convergence, continuity
and compactness. In section 3, the metric form of Lax equivalent theorem
is shown in Theorem 3.2. Usually, this theorem is applied to the problems
defined in normed spaces, or even in Banach spaces, to prove the numerical
method is convergent by using the stability. There are many linear and
nonlinear examples for this forward direction of Lax equivalent theorem, e.g.
[Sanz-Serna and Palencia, 1985, Palencia and Sanz-Serna, 1984].
On the other hand, to get stability from convergence is not very practical in some finite difference method since the convergence always relates to
the analytic solution of the problem which is not accessible in many cases.
However, this direction of the Lax equivalent theorem can be use to show
a consistent numerical method does not converge from the condition that it
is unstable. An interesting example is to show that Newton-Cotes formulas for numerical integration do not converge [Engels, 1980] by knowing that
Newton-Cotes formulas are numerically unstable [Krommer and Ueberhuber,
1998].
The importance of the Lax equivalent theorem is that the convergence of a
numerical method is difficult to be established while it is usually desired. The
consistency is straightforward to be verified and the stability which concerns
the ‘local’ properties of numerical method is much easier to show than the
convergence. Moreover, the metric Lax equivalent theorem only requires the
operator for the numerical method to be defined on a metric space which may
provide a wider use than the classical one for numerical analysis. However,
the operators Tn and T are defined on the same metric space which could
cause some troubles for the applications in the future.
4
References
S. Engblom. Notes of lax equivalent theorem, 2014.
H. Engels. Numerical quadrature and cubature. 1980.
B. Gustafsson, H.-O. Kreiss, and J. Oliger. Time-dependent problems and
difference methods, volume 121. John Wiley & Sons, 2013.
J. K. Hunter and B. Nachtergaele. Applied analysis. World Scientific, 2001.
A. R. Krommer and C. W. Ueberhuber. Computational integration, volume 53. Siam, 1998.
P. D. Lax and R. D. Richtmyer. Survey of the stability of linear finite difference equations. Communications on Pure and Applied Mathematics, 9(2):
267–293, 1956.
C. Palencia and J. Sanz-Serna. An extension of the lax-richtmyer theory.
Numerische Mathematik, 44(2):279–283, 1984.
J. Sanz-Serna and C. Palencia. A general equivalence theorem in the theory
of discretization methods. Mathematics of computation, 45(171):143–152,
1985.
5
Fixed-point proof of Lax-Milgram theorem
Fatemeh Ghasemi∗
January 16, 2015
Abstract
Lax-Milgram theorem gives conditions under which a bilinear form
can be inverted to show existence and uniqueness of weak solution of a
given partial differential equation (PDE). There are different ways for
proving this theorem. Here we use the well-known Banach fixed-point
theorem.
1
Introduction
Riesz representation theorem is of the fundamental theorem about Hilbert
space that is the basis for the existence and uniqueness for (symmetric) variational problems. For using this theorem, the variational problems needs to
be symmetric. But a wide variety of variational problems are not symmetric and this means that this theorem is not applicable. Fortunately, there
is an analogous result, the Lax-Milgram theorem, that does apply to nonsymmetric problems.
The Lax-Milgram lemma has played a critical role over the last half century by establishing existence and uniqueness of weak solutions of operator
equations Au = f where A is a continuous and coercive linear operator from a
Hilbert space H into its topological dual H 0 . The result has had far-reaching
impact on the analytic and numerical study of elliptic and parabolic partial
differential equations; boundedness and coercivity of the weak operators are
all that need be verified.
While in 1954 the paper (1) of Lax and Milgram provided a nonconstructive proof of the result, constructive proofs appeared over the next several
∗
Division of Computational Mathematics, Department of Mathematics, Link¨
oping University, SE-581 83 Link¨oping, Sweden . [email protected]
1
years. In 1960, Zaran- tanello provided a generalization of Lax-Milgram to
strongly monotone nonlinear operators via Banach fixed points. A similar
fixed point proof in the linear case was given by Lions and Stampacchia in
a 1967 paper (2) on variational inequalities. In both cases, the mapping
shown to be contractive involves applying the Riesz map for H to a residual
Av − f ∈ H for some v ∈ H. Also in 1967, Petryshyn gave a constructive
proof of the result based on so-called upper and lower semi-orthogonal bases
for subspaces of H.
In this essay, after introducing some preliminaries in the first section, we
prove Lax-Milgram theorem by using Banach fixed-point theorem.
2
Preliminaries
Let H denote a real Hilbert space equipped with inner product (., .)H and
0
0
associated norm k.k. H denotes the topological dual with h., .i the H × H
duality pairing. In this essay we use Banach fixed-point theorem and Riesz
Representation theorem in order to prove Lax-Milgram Theorem. In the
following we review these two theorems.
Theorem 2.1. (Banach fixed-point theorem)
Let (X, d) be a non-empty complete metric space, and T : X → X a map
such that d(T x, T y) ≤ kd(x, y) for some real positive k < 1. Then T has
a unique fixed point, i.e. there exists a unique x ∈ X such that T x = x;
furthermore for any initial x0 , the iterates T n x0 converge to the fixed point
x.
Theorem 2.2. (Riesz Representation theorem) For every bounded linear
functional f ∈ H there exists a unique element u ∈ H such that
f (v) = (u, v),
3
∀v ∈ H.
Lax-Milgram Theorem
Theorem 3.1. Suppose that B : H × H → R is a bilinear form that is
1. bounded, i.e. there exists an α > 0 such that
|B(u, v)| ≤ αkukkvk,
and
2
∀u, v ∈ H,
2. coercive, i.e. there exists a β such that
B(u, u) ≥ βkuk2 ,
∀u ∈ H.
0
Then for any f ∈ H there exists a unique uf ∈ H such that
B(uf , v) = f (v),
∀v ∈ H.
(3.1)
Furthermore
kuf k ≤ β −1 kf kH 0 ,
(3.2)
where
kf kH 0 =
sup
v∈H,kvk=1
|hf, vi|.
in particular uf depends continuously on f , i.e.
kuf − ug k ≤ β −1 kf − gk.
Proof. The uniqueness is straightforward. Suppose there exist two solutions u, u¯ then
B(u, v) = B(¯
u, v) = f (v), ∀v ∈ H.
Since B is bilinear, we have B(u − u¯, v) = 0 for all v ∈ H. Let v = u − u¯.
Using the coercivity of B leads to
βku − u¯k2 ≤ B(u − u¯, u − u¯) = 0,
i.e. u = u¯. Let v = uf in (3.1) to get
B(uf , uf ) = f (uf )
By definition of k.kH 0 we know that f (uf ) ≤ kf kH 0 . By using this inequality
and coercivity of B, the bound (3.2) is derived. The continuity result follows
by considering
B(uf , v) − B(ug , v) = B(uf − ug , v) = (f − g, v),
and setting v = uf − ug . So only existence requires any work.
Fix u ∈ H and consider the map v → B(u, v). We claim that this defines a
bounded linear functional on H. Since B is bilinear and bounded, it is clearly
linear and bounded. It follows from the Riesz Representation Theorem that
there exists a w ∈ H such that
(w, u) = B(u, v),
3
∀v ∈ H.
We define Au = w. By this definition we have
(Au, u) = B(u, v),
∀v ∈ H.
and claim that this definition yields a bounded linear operator from H into
itself.
Indeed, for every v ∈ H
(A(α1 u1 + α2 u2 ), v) =B(α1 u1 + α2 u2 , v)
=α1 B(u1 , v) + α2 B(u2 , v)
=α1 (Au1 , v) + α2 (Au2 , v)
=(α1 Au1 + α2 Au2 , v)
Since this holds for every v ∈ H, it follows that
A(α1 u1 + α2 u2 ) = α1 Au1 + α2 Au2 ,
and this implies that A is linear. To show that A is bounded, note that
kAuk2 = (Au, Au) = B(u, Au) ≤ αkukkAuk,
hence kAuk ≤ αkuk and A is bounded.
Using the Riesz Representation, we know that there exists a ϕ ∈ H such that
(ϕ, v) = f (v),
∀v ∈ H.
We can rewrite our equation in the following way
(Au, v) = (ϕ, v),
∀v ∈ H.
This means that u satisfies (3.1) if and only if Au = ϕ. Now we have to
show that this equation has a solution. In this step of proof we use Banach
fixed-point theorem.
Clearly, for any λ > 0
Au = ϕ ⇔ u = u − (Au − ϕ).
Let T u = u − λ(Au − ϕ). So we are investigating a fixed point for T. Since
H is a Hilbert space, it is enough to show that T is a contraction. We have
kT u − T vk2 =k(u − v) − λA(u − v)k2
=ku − vk2 − 2λ(A(u − v), u − v) + λ2 kA(u − v)k2
=ku − vk2 − 2λB(u − v, u − v) + λ2 kA(u − v)k2
≤ku − vk2 − 2λβku − vk2 + λ2 α2 ku − vk2
=(1 − 2λβ + λ2 α2 )ku − vk2 ,
where we have used the coercivity of B and the fact that A is bounded. It
follows that if we choose sufficiently small λ, T is a contraction. It therefore
has a unique fixed point, which provides our solution.
4
References
[1] P. D. Lax and A. N. Milgram, Parabolic equations , in Contributions
to the theory of partial differential equations, Annals of Mathematics
Studies, no. 33, Princeton University Press, Princeton, N. J., 1954, pp.
167-190.
[2] J.-L. Lions and G. Stampacchia, Variational inequalities ,Comm. Pure
Appl. Math. Comm. Pure Appl. Math., 20 (1967), pp. 493-519.
5
Part II
Solutions to exercises
87
Chapter 1
Metric spaces
88
Ex. 1.2.4
Consider the sequence
1
, n ≥ 2.
log (n)
Clearly this sequence converges to zero. However, log (n) grows slower than any polynomial
sequence so for each 1 ≤ p < ∞ there exists an Np ∈ N such that ξnp ≥ 1/n for all n ≥ Np .
Therefore, since the harmonic series diverges, the sequence ξn is not in any space lp .
ξn =
Ex. 1.3.8
Show that the closure B(x0 ; r) of an open all B(x0 ; r) in a metric space can differ from the
˜ 0 ; r).
closed ball B(x
Solution: Let x0 = (0, 0) and r = 1. Define a metric space (X, d), where X = {(x1 , x2 ) |
x21 + x22 ≤ 1, x1 and x2 are real and nonnegative} ∪ {(x1 , x2 ) = (cos(1.25π), sin(1.25π))},
and d is the Euclidean distance. The points in set X are shown in Figure 1.1. According
1.5
1
0.5
0
−0.5
−1
−1.5
−1.5
−1
−0.5
0
0.5
1
1.5
Figure 1.1: Points in X
to the definitions,
B(x0 ; r) = {(x1 , x2 ) ∈ X | x21 + x22 < 1},
˜ 0 ; r) = {x ∈ X | d(x, x0 ) ≤ 1} = X,
B(x
B(x0 ; r) = {(x1 , x2 ) ∈ X | (x1 , x2 ) 6= (cos(1.25π), sin(1.25π))}.
89
˜ 0 ; r).
Therefore, B(x0 ; r) can differ from B(x
Ex. 1.3.8
Let X be a set with at least two elements and let the metric space (X, d) have the discrete
metric
(
0 x=y
d(x, y) =
.
1 x 6= y
˜ 0 , 1) be the closed ball around x0 ∈ X
Let B(x
˜ 0 , 1) = {x ∈ X : d(x, x0 ) ≤ 1}.
B(x
By the definition of the metric this means
˜ 0 , 1) = X.
B(x
Let now B(x0 , 1) = {x ∈ X : d(x, x0 ) < 1} be the open ball around x0 . Which means
B(x0 , 1) = {x0 }.
Let now x˜ ∈ X be an accumulation point of B(x0 , 1). This means that
d(˜
x, x0 ) < , ∀ > 0
which implies x˜ = x0 thus the closure of the open ball is
B(x0 , 1) = {x0 },
which is not equal to the closed ball.
Ex. 1.3.8
Consider the discrete metric on R2 , X = (R2 , d) where B(x0 ; 1) = X and B(x0 ; 1) =
{x0 }, clearly differing.
Ex. 1.3.12
Let us consider the family of characteristic functions
X = x (t) = χ(a,c) (t) , c ∈ (a, b)
90
that is non–countable because (a, b) is a non–countable set. Notice that
d (x, y) = 1,
x 6= y, x, y ∈ X.
Now, let us assume that there exists
a dense set in B [a, b], call it G: then there exist as
1
many non–empty balls B x0 , 2 , x0 ∈ X as the elements of X. On the other hand, we
know that
1
1
0
0
B x0 ,
∩ B x0 ,
= ∅, x0 6= x0
2
2
and therefore G is non–countable as well. Thus, it does not exist a countable dense set in
B [a, b], since we have proved that every dense set in X is non–countable.
Ex. 1.3.12
Consider the subset M ⊂ B[a, b] consisting of the functions δx = 1 at x and zero everywhere
else on [a, b]. Then M is uncountable, and
sup |δx − δy | = 1
whenever x 6= y. Consider now all open balls B(δx ; 1/2). These are disjoint, and so if
a subset S ⊂ B[a, b] is dense, then it must intersect each ball in a different point. This
renders S uncountable. Since S is general this shows that B[a, b] is not separable.
Ex. 1.4.2
If xnk → x, then for every ε > 0 there is an N1 = N1 (ε) such that
d (xnk , x) <
ε
2
∀nk > N1
and, since xn is Cauchy we know that there is an N2 = N2 (ε) such that
d (xn , xm ) <
ε
2
∀n, m > N2 .
Hence by the triangular inequality we obtain for n, nk > max {N1 , N2 }
d (xn , x) ≤ d (xn , xnk ) + d (xnk , x) <
and the claim follows.
91
ε ε
+ =ε
2 2
Ex. 1.4.2
For the subsequence we have
d(xnk , x) −→ 0,
and since (xn ) is Cauchy,
Using the triangle inequality we get
d(xn , xnk ) −→ 0.
d(xn , x) ≤ d(xn , xnk ) + d(xnk , x) −→ 0
which proves that (xn ) is convergent with the limit x.
Ex. 1.4.8
If d1 and d2 are metrics on the same set X and there are positive numbers a and b such
that for all x, y ∈ X,
ad1 (x, y) ≤ d2 (x, y) ≤ bd1 (x, y),
(1.0.1)
show that the Cauchy sequences in (X, d1 ) and (X, d2 ) are the same.
Solution: We use the relation (1.0.1) to show
(xn ) is Cauchy in (X, d1 ) ⇐⇒ (xn ) is Cauchy in (X, d2 ) .
First assume that (xn ) is Cauchy in (X, d1 ). That is, for any ε > 0 there is an N such that
d1 (xn , xm ) <
ε
b
∀m, n > N.
Using (1.0.1) we obtain
d2 (xn , xm ) ≤ bd1 (xn , xm ) < ε ∀m, n > N.
which proves that (xn ) is Cauchy also in (X, d2 ).
Now assume that (xn ) is Cauchy in (X, d2 ). That is, for any ε > 0 there is an N such that
d2 (xn , xm ) < aε ∀m, n > N.
Using (1.0.1) we obtain
1
d1 (xn , xm ) ≤ d2 (xn , xm ) < ε ∀m, n > N.
a
which proves that (xn ) is Cauchy also in (X, d1 ). Thus, (xn ) is Cauchy in (X, d1 ) iff (xn )
is Cauchy in (X, d2 ), i.e., the Cauchy sequences in (X, d1 ) and (X, d2 ) are the same.
92
Ex. 1.5.6
Show that the set of all real numbers constitutes an incomplete metric space if we choose
d(x, y) = |arctan x − arctan y|.
Solution: Let X = R. We show that (X, d) is incomplete by constructing a Cauchy
sequence in X which does not converge.
Consider the sequence (xn ), where xn = n. Clearly, (xn ) does not converge. Since
π
,
2
lim arctan x =
x→∞
for any ε > 0 we can find N = N (ε) such that
π ε
arctan
n
−
<
2
2
∀n > N.
Hence,
π
π d(xn , xm ) = |arctan xn − arctan xm | = arctan n − − arctan m + 2
2
π π ε ε
≤ arctan n − + arctan m − < + = ε ∀n, m > N
2
2
2 2
which shows that (xn ) is Cauchy.
Ex. 1.5.8
Let y1 , y2 ∈ Y . Since Y is a subspace it is enough to show that Y is closed. Let {yn }∞
n=1
be a sequence in Y with yn → y ∈ C[a, b], this means
∀ > 0 ∃N : sup |y(t) − yn (t)| < |.
t∈[a,b]
But by definition
|y(a) − yn (a)| ≤ sup |y(t) − yn (t)|
t∈[a,b]
so we get
and by the same argument also
yn (a) → y(a),
yn (b) → y(b).
But since each yn ∈ Y we have yn (a) = yn (b) so we get
|y(a) − y(b)| ≤ |y(a) − yn (a)| + |y(b) − yn (b)| → 0
which shows that y(a) = y(b). This means that y ∈ Y and thus is a closed subspace and
complete.
93
Ex. 1.5.9
Assume: that (xm ) is the sequence of continuous functions on [a, b] that uniformly converges on [a, b];
Prove: that the limit function x is continuous on [a, b].
Solution: Since (xm (t)) is converging to x,
∀ > 0 ∃N (),
∀t ∈ [a, b],
|xm (t) − x(t)| <
3
(1.0.2)
On the other hand, (xm (t)) is continuous at any to ∈ [a, b]; so, we have,
∀ > 0 ∃δ > 0, |t − to | < δ
⇒ |xm (t) − xm (to )| <
3
(1.0.3)
Now, we want to show that x(t) is continuous on [a, b]; therefore,
∀ > 0 ∃δ1 > 0, |t − to | < δ1
⇒ |x(t) − x(to )| < (1.0.4)
For this purpose, we can find the following relation, using (1) and (2):
|x(t) − x(to )| = |x(t) ± xm (t) ± xm (to )| =
= |(x(t) − xm (t)) + (xm (t) − xm (to )) + (xm (to ) − x(to ))|
≤ |x(t) − xm (t)| + |xm (t) − xm (to )| + |xm (to ) − x(to )|
+ + =
≤
3 3 3
which is equivalent to (3).
Ex. 1.6.14
The only question here is whether d(x, y) = 0 ⇒ x = y or not.
i) Let d(x, y) = 0 and let h(t) = x(t) − y(t) which implies
Z
a
b
|h(t)|dt = 0.
Assume that there exist t0 ∈ [a, b] such that |h(t0 )| =
6 0. Since x and y are continuous h is
continuous. So let 0 < = |h(t0 )| and let δ be so small that if
|t − t0 | < δ
we have
|h(t) − h(t0 )| < .
94
The reverse triangle inequality gives us:
|h(t0 )| − |h(t)| ≤ |h(t) − h(t0 )| < .
Since = |h(t0 )| this now implies
|h(t)| > 0, ∀t ∈ [t0 − δ, t0 + δ].
But since:
Z
a
this implies
b
|h(t)|dt ≥
Z
a
Z
t0 +δ
t0 −δ
|h(t)|dt
b
|h(t)|dt > 0
which is a contradiction. So no t0 with |h(t0 )| =
6 0 can exist, thus x(t) = y(t).
ii) For Riemann-integrable functions d(x, y) = 0 does not imply x = y. An example is:
let tm = (a + b)/2 and
(
0 t < tm
x(t) =
1 t ≥ tm
(
0 t ≤ tm
y(t) =
1 t > tm
We have x(tm ) 6= y(tm ) but d(x, y) = 0.
95
Chapter 2
Normed spaces
96
Ex. 2.3.10
Suppose that the Banach space (X, k · k) has a Schauder basis {en }∞
n=1 so that for any
x ∈ X there is a unique sequence of scalars, α1 , α2 , . . . such that
x=
∞
X
αk ek .
k=1
We will for simplicity assume that αk ∈ R, though generalisation to complex numbers is
straight forward.
Given any > 0 there is an N such that
kx −
N
X
αk ek k < .
2
k=1
Now, recall that Q is dense in R, meaning that for any αk ∈ R we can find a βk ∈ Q such
that |αk − βk | < /2k+1 . By the triangle inequality we have
N
X
N
X
N
X
N
X
N
X kx −
βk ek k ≤ kx −
αk e k k + k
αk ek −
βk ek k < +
< .
2 k=1 2k+1
k=1
k=1
k=1
k=1
Thus the Schauder basis constitutes a basis for a dense subset of X, which is countable
since βk ∈ Q for all k. Consequently, X is separable.
Ex. 2.3.10
Let Y be the set,
N
X
Y ={
αj ej : αj ∈ R, N ∈ N+ }.
j=1
Let x ∈ X, since we have a Schauder-basis there exist {αj }∞
j=1 such that
kx − xn k → 0
where
xn :=
n
X
αj ej .
j=0
Thus for every x ∈ X we have a sequence in Y converging to x. This shows that Y¯ = X.
97
Let now instead xn =
set
Pn
j=1
αj ej ∈ Y be an arbitrary element in Y and let M be the
N
X
M ={
βj ej : bj ∈ Q, N ∈ N+ },
j=1
k
which is countable. Since Q is dense in R there exist a sequence {βjk }∞
k=1 with βj ∈ Q such
that
|αj − βjk | ≤
, ∀k ≥ K,
Cn
where
C = max kej k .
j∈{1,...,n}]
Let xkn =
Pn
j=1
βjk ej ∈ M , we get:
n
n
X
X
xn − xkn = (αj − βjk )ej |αj − βjk | kej k ,
≤
j=1
j=1
so for k ≥ K we get
n
n
X
X
xn − xkn ≤
kej k ≤
1 ≤ .
Cn
n
j=1
j=1
Showing that xkn → xn when k → ∞. Thus for every xn ∈ Y we can find a sequence in M
¯ = Y and in total
converging to xn . So M
¯) = M
¯
X = Y¯ = (M
showing that M is dense in X so that X is separable.
Ex. 2.4.4
Show that equivalent norms on a vector space X induce the same topology for X.
Solution: Assume that k·k1 and k·k2 are equivalent norms on X. Let T1 be the topology
for X induced by k·k1 and let T2 be the topology for X induced by k·k2 . In other words, T1
and T2 are the collections of all open subsets of X in the sense of k·k1 and k·k2 , respectively.
We show that T1 = T2 by showing that any subset of X is open in the sense of k·k1 iff it
is open in the sense of k·k2 .
Since k·k1 and k·k2 are equivalent, we have
a k·k1 ≤ k·k2 ≤ b k·k1
(2.0.1)
for some a, b > 0. Let Bi (x; r) denote the open ball of radius r around x in the sense of
k·ki . Now assume that M is an open subset of X in the sense of k·k1 . By the definition of
an open set, M contains an open ball around each of its points, i.e., for any x ∈ M ∃ε > 0
98
such that B1 (x; ε) ⊂ M . Because of (2.0.1) we have B2 (x; aε) ⊂ B1 (x; ε), which means
that M contains an open ball in the sense of k·k2 around each of its points. This proves
that any M ⊂ X is open in the sense of k·k2 if it is open in the sense of k·k1 . By reversing
the above argument, we can show that any M ⊂ X is open in the sense of k·k1 if it is open
in the sense of k·k2 .
Ex. 2.4.4
Suppose that the norms kk0 and kk1 are equivalent. Then there are constants c1 and c2
with 0 ≤ c1 ≤ c2 such that c1 k x k0 ≤k x k1 ≤ c2 k x k0 for all x ∈ X. Let U be a subset of
X which is open with respect to the topology on X induced by the norm kk1 . For u ∈ U
there is an such that {x ∈ X k u − x k1 < } ⊂ U . But since the norms are equivalent
{x ∈ X k u − x k1 < } ⊂ {x ∈ X k u − x k0 < /c2 }. This means that U is open with
respect to the topology induced by norm kk0 . By the same argument, we can show that
any open subset of X with respect to the topology induced by the norm kk0 is open with
respect to the topology induced by kk1 .
Ex. 2.5.4
Show that for an infinite subset M in the space s to be compact, it is necessary that there
are numbers γ1 , γ2 , . . . such that for all x = (ξk (x)) ∈ M we have |ξk (x)| ≤ γk .
Solution: The metric on the space s is
d(x, y) =
∞
X
1 |ξk − ηk |
2k 1 + |ξk − ηk |
k=1
where x = (ξj ), y = (ηj ). Assume that not all (ξk ) in M are bounded by |ξk | ≤ γk . Then
there is at least one k0 such that
|ξk0| is unbounded. Now consider a sequence of elements
(m)
(m)
in M , x1 , x2 , . . ., where xm = ξk . We choose the sequence such that ξk0 = m. We
then have, for m 6= n,
∞
(m)
(n)
X
1 |m − n|
1 1
1 |ξk − ξk |
≥ k0
≥ k0 .
|xm − xn | =
(m)
(n)
k
2 1 + |ξk − ξk |
2 1 + |m − n|
2 2
k=1
Clearly (xm ) can not have a convergent subsequence. Thus, M is not compact.
Ex. 2.5.10
Let X and Y be metric spaces, X compact, and T : X → Y bijective and continuous.
Show that T is a homeomorphism.
99
In the proof, I use the result from Prob. 14 in Section 1.3: a mapping T : X → Y is
continuous if and only if the inverse image of any closed set M ⊂ Y is a closed set in X.
This result is equivalent to a mapping T −1 : Y → X is continuous if and only if for any
closed set M ⊂ X, T (M ) is a closed set in Y . (*)
Let M be a closed subset in X. Since X is a compact metric space, M is compact
(Prob. 2.5-9). Since T is continuous, according to Theorem 2.5-6 we have that T (M ) is
compact. By using Lemma 2.5-2, we have that T (M ) is closed and bounded. Since M
is arbitrary and T is bijective, by (*) we have that T −1 is continuous. Therefore, T is a
homeomorphism.
Ex. 2.5.10
Let X and Y be metric spaces, X be compact, T : X → Y bijective and continuous and
F = T −1 .
• Let V ⊆ X and assume that V is closed and bounded so that it is compact.
• Since T is continuous, it follows that T (V ) is compact.
• This implies that T (V ) is closed and bounded.
• But T (V ) = F −1 (V ) so F −1 (V ) is closed and bounded.
Hence F is continuous, and thus T is a homeomorphism.
Ex. 2.5.10
We study the continuous and bijective operator T : X → Y where X is compact. By
theorem 2.5-6, Y is compact. Lets assume that T −1 is discontinuous, that is there is a
y ∈ Y and x ∈ X such that
dy (yn , y) → 0
dx (xn , x) > δ > 0
(2.0.2)
(2.0.3)
where the sequences xn , yn satisfy T xn = yn and T x = y. However, X is compact, so there
is a subsequence of xn such that xk → x0 6= x. We have
dy (T x0 , T x) ≤ dy (T xk , T x) + dy (T x0 , T xk ) = dy (yk , y) + dy (y0 , yk ).
Since T is continuous, we have dy (y0 , yk ) = dy (T x0 , T xk ) → 0 as k → ∞. Therefore, we
have
dy (T x0 , T x) ≤ dy (yk , y) + dy (y0 , yk ) → 0
100
as k → ∞. However, according to (8) we should have for bijective T dy (T x, T x0 ) > > 0,
which is a contradiction. T −1 must therefore be continuous and T is then a homeomorphism.
Ex. 2.5.10
A Homeomorphism is a continuous bijective mapping whose inverse is continuous. Since
we have a continuous and bijective mapping T : X −→ Y (X and Y metric spaces, X
compact), we need to show that the inverse T −1 is continuous.
Generally, a mapping T : X −→ Y is continuous if and only if, for any closed set
K ⊂ Y , the inverse image T −1 (K) is a closed set in X (cf. Prob. 14, Sec. 1.3).
So here, to show that T −1 : Y −→ X is continuous, we want to show that T (M ) is a
closed set in Y , for any closed set M ⊂ X.
1. Let M be a closed subset in X-compact, then from Prob. 9, Sec. 2.5, M is also
compact.
2. Since T is continuous, and M is compact ⇒ T (M ) is compact
(Theorem 2.5-6)
3. Since T (M ) is a compact subspace of the metric space Y , T (M ) is closed and bounded
(Lemma 2.5-2)
Ex. 2.7.2
• Consider T bounded. If there exists M > 0 such that kxk ≤ M , then
kT xk ≤ c kxk = cM
and therefore T maps bounded sets in X into bounded sets in Y .
• Let us consider T which maps bounded sets in X into bounded sets in Y . If x = 0,
then T is trivially bounded since
T (0) = T (0R z) = 0R T (z) = 0 ∀z ∈ X ⇒ kT 0k = 0.
Let us consider x 6= 0. We know that exists C > 0 such that kT xk ≤ C, ∀x 6= 0 such
that kxk = 1. Therefore
x T
≤C
kxk and for the homogeneity of the norm we have kT xk ≤ C kxk.
101
Ex. 2.7.2
Let X and Y be normed spaces. A linear operator T : X → Y is bounded iff T maps
bounded sets in X into bounded sets in Y .
If T is bounded:
We want to show that for any bounded set M ⊂ D(T ) s.t. for any x, y ∈ M we have
that kx − yk ≤ r < ∞ than also kT x − T yk < ∞. Since D(T ) is a vector space then
x, y ∈ M ⊂ D(T ) → x − y = z ∈ D(T ). Furthermore kT zk ≤ ckzk since T is bounded.
Hence kT x − T yk = kT (x − y)k = kT zk ≤ ckzk < ∞.
If T maps bounded sets in X into bounded sets in Y :
We know that for any x, y ∈ M bounded than kT x − T yk < ∞. Then we call
δ(M ) = sup kx − yk and δ(T (M )) =
x,y∈M
sup
T x,T y∈T (M )
kT x − T yk.
For any zD(T ), z 6= 0 we define
x=y+
δ(M )
z
kzk
⇒ kx − yk = δ(M ).
Then
kT x − T yk = kT (x − y)k = kT
which means that
kT zk = kT x − T yk
where c =
δ(M )
δ(M )
z k=
kT zk
kzk
kzk
δ(T (M ))
kzk
≤
kzk = ckzk.
δ(M )
δ(M )
δ(T (M ))
.
δ(M )
Ex. 2.7.2
A linear operator T : D(T ) −→ Y , (where D(T ) ⊂ X and X and Y are normed spaces),
is bounded if there is a real number c such that for all x ∈ D(T ),
kT xk ≤ ckxk.
(2.0.4)
Further, a subset K in a normed space X is bounded if and only if there is a positive
number a such that
kxk ≤ a
(2.0.5)
for every x ∈ K.
Let M be a bounded set in X.
102
1. Show that if T is bounded ⇒ T (M ) is a bounded set in Y :
Let x ∈ M and T x ∈ T (M ), then from (2.0.4) and (2.0.5) we have
kT xk ≤ ckxk ≤ ca
(2.0.6)
where ca is a positive number. Since x ∈ M is arbitrary this proves T (M ) is a
bounded set in Y (according to the definition (2.0.5) of a bounded subset).
2. Show that if T (M ) is a bounded set in Y ⇒ T is bounded:
Let x ∈ M and T x ∈ T (M ), then using the definition for the operator norm we have
kT kkxk =
kT xk
kT xk
kxk ≥
kxk = kT xk,
kxk
x∈M,x6=0 kxk
sup
x 6= 0.
(2.0.7)
Since kT k is a real number, call it c, we have
kT xk ≤ ckxk,
x 6= 0.
(2.0.8)
For the case when x = 0, since for linear operators T 0 = 0, we have
kT xk = kT 0k = 0 ≤ ckxk,
x = 0.
(2.0.9)
Since inequality (2.0.4) holds for all x ∈ M , T is bounded.
Ex. 2.7.7
Given: T : X → Y is bounded and linear.Furthermore,
∃b > 0 3 ∀x ∈ X, kT xk ≥ bkxk ⇒ kxk ≤
kT xk
b
(2.0.10)
We want to show that T −1 : Y → X exists and is bounded. According to [Theorem
2.6-10(a)], T −1 exists if and only if T x = 0 implies x = 0. In our case,
T x = 0 ⇒ 0 ≥ bkxk
Since b > 0 the only way for the latter inequality to be valid is that, x = 0. Hence, T −1
exists. Also, based on the definition of mapping we have,
∀x ∈ X ∃y ∈ Y 3 y = T x ∧ x = T −1 y
kT xk
kyk
⇒ kT −1 yk = kxk ≤(1)
=
b
b
1
⇒ kT −1 yk ≤ kyk
b
therefore, according to [Definition 2.7-1], T −1 is bounded.
103
(2.0.11)
(2.0.12)
(2.0.13)
Ex. 2.7.8
If we consider T : l∞ → l∞ , x 7→ y defined by
x = {αj } ,
y = {βj } , s.t. βj =
αj
j
the bounded set y = {1} ⊂ R (T ) is mapped by T −1 into x = {j}j∈N , that is not bounded.
Ex. 2.7.8
Let X be a normed space with points x = {ξk } being sequences of real or complex numbers
that has only finitely many non-zero terms. Further let kxk = supk |ξk |. Now define the
operator T : X → X as
T x = (ξ1 , ξ2 /2, . . . ) = {ξk /k}.
If y = {ηk } and α and β are numbers then
T (αx + βy) = {(αξk + βηk )/k} = α{ξk } + β{ηk } = αT (x) + βT (y)
so T is a linear operator. Also note that
kT xk = sup |ξk |/k ≤ kxk sup 1/k ≤ kxk
k
k
so T is bounded. Now, note that T −1 x = {kξk }. Consider the sequence x = {δk,n } ∈ X.
Then kxk = 1 and kT −1 xk = n. Since n is an arbitrary integer and x and infinite sequence,
we may let n → ∞, which shows that T −1 is not bounded.
Ex. 2.8.6
The space C 1 [a, b] is the normed space of all continuously differentiable functions on J =
[a, b] with norm defined by
kxk = max |x(t)| + max |x0 (t)|.
t∈J
(2.0.14)
t∈J
(N1) Since |x(t)| ≥ 0 and |x0 (t)| ≥ 0, kxk ≥ 0.
(N2) ”⇒”: kxk = 0 ⇔ max |x(t)| = − max |x0 (t)|⇔ x(t) = 0 and x0 (t) = 0.
t∈J
t∈J
”⇐”: x = 0 ⇒ kxk = max |0| + max |0| = 0
t∈J
t∈J
(N3) kαxk = max |αx(t)| + max |αx0 (t)| = |α| max |x(t)| + |α| max |x0 (t)| = |αk|xk
t∈J
t∈J
t∈J
t∈J
(N4) kx + yk = {differentiation is linear} = max |x(t) + y(t)| + max |x0 (t) + y 0 (t)|
t∈J
t∈J
≤ max |x(t)| + max |y(t)| + max |x0 (t)| + max |y 0 (t)| = kxk + kyk.
t∈J
t∈J
t∈J
t∈J
104
Define f (x) = x0 (c), c = (a + b)/2. This functional on C 1 [a, b] is bounded since
a + b a + b a + b 0
0
kxk = max |x(t)| + max |x0 (t)| ≥ x
+ x
≥ x
t∈J
t∈J
2
2
2
⇔
|f (x)| ≤ kxk.
(2.0.15)
Further, f is linear since
f (αx + βy) = {differentiation is linear} = αx0
= αf (x) + βf (y).
a + b
2
+ βx0
a + b
2
(2.0.16)
If f is now considered as a functional on the subspace of C[a, b] which consists of all
continuously differentiable functions, we want to show that f is not bounded. We consider
the sequence of continuous and differentiable functions (xε ) = arctan( εt ), where ε is a real
number. If we let ε −→ 0, then for c = 0,
lim x0ε (c) = ∞,
ε−→0
see Figure 2.1.
So, given a constant C, there exists an ε such that
|f (xε )| > Ckxε k,
which proves that f is not bounded (since kxε k = max | arctan( εt )| ≤
t∈J
ε).
π
2
independent of
Ex. 2.8.6
The space C 1 [a, b] is the normed space of all continuously differentiable functions on J =
[a, b] with norm defined by
kxk = max |x(t)| + max |x0 (t)|.
t∈J
t∈J
(2.0.17)
Show that the axioms of a norm are satisfied. Show that f (x) = x0 (c), c = (a + b)/2,
defines a bounded linear functional on C 1 [a, b]. Show that f is not bounded, considered
as a functional on the subspace of C[a, b] which consists of all continuously differentiable
functions.
Solution: First, we show that the four axioms of a norm are satisfied in this problem.
(N1). By the definition of maximum and absolute function, kxk ≥ 0.
(N2). If kxk = 0, as both max|x(t)| and max|x0 (t)| are nonnegative, |x(t)| = |x0 (t)| = 0.
t∈J
t∈J
If x(t) = 0 is a constant function, then |x0 (t)| = 0. Therefore, kxk = 0.
105
2
1.5
arctan(x/eps)
1
0.5
0
−0.5
−1
−1.5
−2
−100
−80
−60
−40
−20
0
20
40
60
80
100
x
Figure 2.1: The function arctan( xε ). Smaller and smaller values of ε leads to a steeper and
steeper function.
(N3).
kαxk = max |αx(t)| + max |αx0 (t)|
t∈J
t∈J
= |α|(max |x(t)| + max |x0 (t)|)
t∈J
= |α|kxk.
106
t∈J
(2.0.18)
(N4). The triangle inequality is satisfied
kx + yk = max |x(t) + y(t)| + max |x0 (t) + y 0 (t)|.
t∈J
t∈J
≤ max |x(t)| + max |y(t)| + max |x0 (t)| + max |y 0 (t)|.
t∈J
t∈J
t∈J
t∈J
= kxk + kyk.
(2.0.19)
Then, we show that functional f (x) is linear and bounded on C 1 [a, b]. As the space
C 1 [a, b] is a normed space, it defines the algebraic operations. For x, y ∈ C 1 [a, b] and any
scalar α, the operation f satisfies
f (x + y) = (x(c) + y(c))0 = f (x) + f (y),
(2.0.20)
f (αx) = (x(c) + y(c))0 = f (x) + f (y).
(2.0.21)
and
This shows that f is linear. Consider the norm in C 1 [a, b],
|f (x)| = |x0 (c)| ≤ max |x0 (t)| ≤ max |x(t)| + max |x0 (t)| = kxk,
t∈J
t∈J
t∈J
(2.0.22)
followeding the definition in 2.8-2, f is bounded on C 1 [a, b].
However, on the space C[a, b], where the norm is defined by
kxk = max |x(t)|,
t∈J
(2.0.23)
let xn (t) = sin(ntπ/c), where n ∈ N. Then kxn k ≤ 1 and
nπ
f (xn ) = x0n (c) =
cos(nπ).
(2.0.24)
c
So kf (x)k = nπ/c. As n ∈ N is arbitrary, this shows that there is no fixed number α such
that kf (x)k ≤ αkxk, so f is unbounded.
Ex. 2.10.4
Let X and Y be normed spaces and Tn : X → Y , n = 1, 2, 3, .., be bounded linear operators.
Show that convergence Tn → T implies that for every > 0 there is an N such that for all
n > N and all x in any given closed ball we have kTn x − T xk < .
˜ 0 ; r) ⊂ X. Then, for all x ∈ B(x
˜ 0 ; r),
Solution: Denote the given closed ball by B(x
there exists a positive number C such that kxk ≤ C. From the convergence of Tn , for a
given > 0, there exists N ∈ N such that for all n > N ,
(2.0.25)
kTn − T k ≤ .
C
˜ 0 ; r)
Therefore, for all x ∈ B(x
kTn x − T xk ≤ kTn − T kkxk ≤ .
107
(2.0.26)
Ex. 2.10.8
∞
Since {ek }∞
k=1 with ek = (δkj )j=1 is a Schauder basis for c0 , for every x ∈ c0 we can write
∞
X
x=
λk e k ,
k=1
where x = (λk ) and limn→∞ λk = 0 and we can approximate x by
sn =
n
X
λk ek ,
k=1
then limn→∞ k sn − x k∞ = 0. Let f ∈ c00 , which is the dual space of c0 . This means that
f : c0 → R is a bounded linear functional, so f (sn ) → f (x), as n → ∞. So
f (x) =
∞
X
λk f (ek ).
k=1
n)
for f (en ) 6= 0 and γn = 0
Now we show that f (ek ) is in l1 . Define the sequence γn = |ff (e
(en )|
for f (en ) = 0. Consider the sequence xn = (γ1 , γ2 , · · · , γn , 0, 0, · · · ) = γ1 e1 + · · · + γn en . It
is obvious that k xn k∞ ≤ 1, so
|f (xn )| =
So
Pn
k=1
n
X
k=1
|f (ek )| ≤k xn k∞ k f k≤k f k .
|f (ek )| ≤k f k for all n. Since this inequality holds for all n, we have
∞
X
k=1
|f (ek )| ≤k f k .
(2.0.27)
This means that (f (ek )) ∈ l1 . Now we want to show that for every y = (λk ) ∈ l1 , we can
obtain a corresponding bounded linear operator g : c0 → R. Let
g(x) =
∞
X
λk µk , x = (µk ).
k=1
Clearly g is linear and
|g(x)| = |
∞
X
k=1
λk µk | ≤ sup |µj |
j
∞
X
k=1
|λk | ≤k x k∞
∞
X
k=1
|λk |.
So g is bounded. Taking the supremum over all x with norm 1, we get
|g(x)| ≤
∞
X
k=1
108
|λk |.
This inequality and the inequality (2.0.27) gives
∞
X
k=1
|f (ek )| =k f k .
So the mapping f : c00 → l1 with f → (f (ek )) is bijective and preserves norms and hence
isomorphism.
Ex. 2.10.8
We consider f a bounded linear functional on c0 . Since a Schauder basis for c0 is {ek }k∈N ,
with ek = δkj , then every x = {αk } ∈ c0 has a unique representation
x=
∞
X
αk ek
k=1
and, from the linearity of f , we know also that
f (x) =
∞
X
αk f (ek ) .
k=1
We can write
|f (x)|
=
∞
X
αk f (ek )
≤
k=1
∞
X
k=1
|αk | |f (ek )|
≤
kxk∞
∞
X
k=1
|f (ek )|
=
c kxk∞ ,
where c = |f (ek )| < ∞ because f (x) is bounded. This implies that {f (ek )} ⊂ l1 and
0
therefore c0 ⊆ l1 .
On the other hand, for every b = {βk }k∈N ∈ l1 , we can obtain a bounded linear functional
g in c0 . In fact, we may define g on c0 by
f (x) =
∞
X
αk βk .
k=1
Then g is
• Linear
g (λx + µy) =
∞
X
(λαk + µγk ) βk =
k=1
∞
X
=λ
k=1
αk βk + µ
∞
X
k=1
where y = {γk }k∈N ∈ c0 .
109
γk βk = λg (x) + µg (y) ,
λ, µ ∈ R
• Bounded
∞
∞
∞
X
X
X
|g (x)| = α k βk ≤
|αk βk | ≤ kxk∞
βk = kxk∞ kbk1 .
k=1
0
k=1
k=1
0
Then g ∈ c0 and therefore l1 ⊆ c0 .
P
We finally prove that kgk = kbk1 . Indeed, since ∞
k=1 |βk | converges, then
∞
X
∀ε > 0, ∃N = N (ε) s.t.
k=n+1
|βk | < ε,
∀n > N.
Let us consider x s.t.
αk =
x ∈ c0 , and
(
|βk |
,
βk
0,
k < N,
k≥N
.
∞
∞
∞
X
X
X
|g (x) − kbk1 | = αk βk −
|βk | = |βk | < ε.
k=1
k=1
k=N +1
0
This shows that the bijective linear mapping on l1 onto c0 defined by g is an isomorphism.
Ex. 2.10.8
Show that the dual space of the space c0 is l1 .
c0 is the space of all sequences of scalars converging to zero. The norm on c0 is k · k∞ .
0
0
Let c0 be the dual space of c0 . We want to show that c0 is isomorphic with l1 . The norm
on l1 is k · k1 . We prove it step by step.
Step 1: Let x ∈ c0 , then x has a unique representation
x=
∞
X
ξk ek ,
k=1
where (ek ) = (δkj ) is a Schauder basis for c0 , (δkj ) has one in the k th place
P and zeros
otherwise, and ξk → 0 as k → ∞. Since f is linear, we have f (x) = ∞
k=1 ξk f (ek ).
Since f is bounded on c0 , we have
kf k = sup
x∈c0
110
|f (x)|
< ∞.
kxk∞
We construct s sequence (¯
xk ) such that
(
x¯k =
|f (ek )|
f (ek )
0
k ≤ n and f (ek ) 6= 0
k > n or f (ek ) = 0
|f (¯
x)|
Obviously, (¯
xk ) ∈ c0 . Therefore, k¯
< ∞. Since the elements of (¯
xk ) can only take
xk∞
values 0 , 1, or -1, we have k¯
xk∞ = 1. We notice that
f (¯
x) =
∞
X
x¯k f (ek ) =
k=1
n
X
k=1
|f (ek )|,
f (¯
x) ≤ kf kk¯
xk∞ = kf k
Since n is arbitrarily large, we have
∞
X
k=1
|f (ek )| ≤ kf k.
(2.0.28)
Step 2: Take any x ∈ c0 , we have
|f (x)| = |
∞
X
ξk f (ek )|
k=1
≤ kxk∞
∞
X
k=1
|f (ek )|.
Taking all x such that kxk∞ = 1 gives
kf k ≤
∞
X
k=1
|f (ek )|.
(2.0.29)
P
1
By (2.0.28) and (2.0.29), we get that kf k = ∞
k=1 |f (ek )|. Therefore, f is in l . This also
1
shows that given
P∞ a sequence (ηk ) in l , there exists a linear bounded operator g on c0 such
that g(x) = k=1 ξk ηk , where x = ξk ∈ c0 . g is linear by construction. Boundedness can
be proved similar to (2.0.29).
0
We conclude that c0 is isomorphic with l1 .
111
Chapter 3
Inner product spaces
112
Ex. 3.3.4
A X is a Hilbert space (complete) and M ⊂ X is closed. A closed subset of a complete one
is also complete, so M is complete. M is also a vector space, which is convex by definition.
Hence, M has all the properties needed for Theorem 3.3-1. B Appolonius identity is
1
1
||x − yn ||2 + ||x − ym ||2 = ||yn − ym ||2 + 2||x − (yn + ym )||2 .
2
2
(3.0.1)
Since M is convex, we have
1
2||x − (yn + ym )||2 ≥ 4δ 2
2
where δ = ||x − y|| the minimum distance. Putting vn = yn − x and using this in (3.0.1),
one obtains
1
||vn ||2 + ||vm ||2 ≥ ||vn − vm ||2 + 2δ 2 .
(3.0.2)
2
Using the parallelogram equality on the left-hand side of (3.0.2), one can conclude that
||vn − vm ||2 ≥ 4δ 2
which is the first part of the proof. The rest of the proof can then proceed as usual.
Ex. 3.3.4
(a) Show that the conclusion of Theorem 3.3-1 also holds if X is a Hilbert space and
M ⊂ X is a closed subspace. (b) How could we use Appolonius’ identity (Sec. 3.1, Prob.
5) in the proof of Theorem 3.3-1?
Solution The problem states that: Let X be a Hilbert space and a closed subspace
M ⊂ X. For every given x ∈ X there exists a unique y ∈ M , such that δ = inf kx − y˜k =
y˜∈M
kx − yk.
1) Existence. From theorem 3.2-4, we know that M is complete.
From theorem 3.3-4, the Hilbert space X can be represented as X = M ⊕ M ⊥ . Then
for any given x ∈ X there exists y ∈ M and z ∈ M ⊥ such that x = y + z. Then we prove
such a y satisfies kx − yk ≤ kx − y˜k for any y˜ ∈ M . Since y − y˜ ∈ M , the norm
kx − y˜k2 =
=
=
=
≥
kx − y + y − y˜k2
kz + y − y˜k2
kzk2 + < z, y − y˜ > + < y − y˜, z > +ky − y˜k2
kzk2 + ky − y˜k2
kzk2 .
113
2) Uniqueness.
Suppose there are two different elements in M satisfy the minimizing condition, denote
as y1 and y2 such that δ = inf kx − y˜k = kx − y1 k = kx − y2 k. By using the Appolonius’
y˜∈M
identity,
ky1 − y2 k2 = 2kx − y1 k2 + 2kx − y2 k2 − 4kx −
= 2δ 2 + 2δ 2 − 4kx −
y1 + y2 2
k.
2
y1 + y2 2
k
2
(3.0.3)
2 2
2
As y1 6= y2 , we have kx − y1 +y
k < δ 2 . Then, y = y1 +y
is a better approximation for
2
2
the norm kx − yk in M ., contradiction. Therefore, there is a unique y ∈ M satisfies the
minimizing condition.
Ex. 3.3.4
(a) Show that the conclusion of Theorem 3.3-1 (minimizing vector) holds also if X is a
Hilbert space and M ⊂ X is a closed subspace. (b) How could we use Appolonius’ identity
in the proof of Theorem 3.3-1?
Solution: (a) The assumptions in Theorem 3.3-1 are that X is an inner product space
and M 6= ∅ is a convex subset which is complete. We now assume that X is a Hilbert space
and M is a closed subspace. Since M is closed and X is a Hilbert space, M is complete.
Since M is a subspace, M is also convex. Thus, all the assumptions in Theorem 3.3-1 are
fulfilled and the conclusions of the Theorem follow.
(b) Appolonius’ identity reads
kc − ak2 + kc − bk2 =
1
1
ka − bk2 + 2kc − (a + b)k2
2
2
or, equivalently,
1
ka − bk2 = 2 kc − ak2 + 2 kc − bk2 − 4kc − (a + b)k2 .
2
(3.0.4)
To prove uniqueness of the minimizing vector in Theorem 3.3-1, assume that both y ∈ M
and y0 ∈ M satisfy ky − xk = δ and ky0 − xk = δ. By Appolonius’ identity (3.0.4) with
a = y, b = y0 and c = x, we obtain
1
ky − y0 k2 = 2 kx − yk2 + 2 kx − y0 k2 − 4kx − (y + y0 )k2
2
1
= 2δ 2 + 2δ 2 − 4kx − (y + y0 )k2 .
2
Because M is convex, 21 (y + y0 ) ∈ M , so that kx − 12 (y + y0 )k ≥ δ, which implies
ky − y0 k2 ≤ 2δ 2 + 2δ 2 − 4δ 2 = 0.
114
But since ky − y0 k2 ≥ 0, we must have ky − y0 k = 0 and thus y = y0 , which proves that
the minimizing vector is unique.
Ex. 3.3.10
We want to show that if Y ⊂ H is a closed subspace such that M ⊂ Y implies that
M ⊥⊥ ⊂ Y.
As a first step let y⊥ ∈ Y ⊥ . This implies hy⊥ , mi = 0 ∀m ∈ M since M ⊂ Y , which shows
us that y⊥ ∈ M ⊥ . This means that
Y ⊥ ⊂ M⊥
(3.0.5)
since y⊥ was arbitrary.
L ⊥
Let now x ∈ M ⊥⊥ . Since Y is a closed subspace we have that X = Y
Y . So x can
⊥
be decomposed as x = y + y⊥ with y ∈ Y and y⊥ ∈ Y . But from (3.0.5) we have that
hx, y⊥ i = 0 so we get
0 = x, y ⊥ = hy + y⊥ , y⊥ i = ky⊥ k2 ,
which means that y⊥ = 0. Thus x = y ∈ Y , and since x was arbitrary we get
M ⊥⊥ ⊂ Y.
Ex. 3.3.10
If M ⊂ H, show that M ⊥⊥ is contained in any closed subspace Y ⊂ H such that M ⊂ Y .
There are two cases: Case I) if M is closed; then, M = M ⊥⊥ by Theorem (3.3-6). Since
M ⊂ H, then M ⊥⊥ is the smallest subset of H that contains M .
Case II) if M is not closed; we know that M ⊥⊥ contains M ; i.e.,
∀x ∈ M, x⊥M ⊥ ⇒ x ∈ M ⊥⊥ ⇒ M ⊂ M ⊥⊥
If there exists a closed subspace Y ⊂ H, then,
M ⊂ Y ⇒ Y ⊥ ⊂ M ⊥ ⇒ M ⊥⊥ ⊂ Y ⊥⊥ = Y
Y was arbitrary, so M ⊥⊥ is the smallest closed subset of H which contains M .
115
Ex. 3.4.6
(Minimum property of Fourier coefficients) Let {e1 , . . . , en } be an orthonormal set in an
inner product space X, where n is fixed. Let x ∈ X be any fixed element and y =
β1 e1 + . . . + βn en . Then kx − yk depends on β1 , . . . , βn . Show by direct calculation that
kx − yk is minimum if and only if βj = hx, ej i, where j = 1, . . . , n.
Solution:
kx − yk2 = hx − y, x − yi = hx, xi − hx, yi − hy, xi + hy, yi
= kxk2 + kyk2 − hx, yi − hx, yi = kxk2 + kyk2 − 2Re {hx, yi}
≥ kxk2 + kyk2 − 2|hx, yi|
n
n
X
X
= kxk2 +
|βk |2 − 2|hx,
βk ek i|
k=1
≥ kxk2 +
2
= kxk +
By choosing βk = hx, ek i,
n
X
k=1
n
X
k=1
k=1
|βk |2 − 2
n
X
k=1
|βk ||hx, ek i|
2
2
2
(|βk | − |hx, ek i|) − |hx, ek i| ≥ kxk −
n
X
k=1
|hx, ek i|2
k = 1, . . . , n, we obtain
kx − yk2 = kxk2 + kyk2 − 2Re {hx, yi}
)
(
n
n
X
X
βk e k i
= kxk2 +
|βk |2 − 2Re hx,
( n k=1
)
X
|βk |2 − 2Re
βk hx, ek i
k=1
= kxk2 +
n
X
= kxk2 −
n
X
k=1
k=1
)
( n
n
X
X
|hx, ek i|2
= kxk2 +
|hx, ek i|2 − 2Re
k=1
k=1
k=1
|hx, ek i|2
which shows that the choice βk = hx, ek i,
k = 1, . . . , n minimizes kx − yk
116
Ex. 3.4.6
Since X is an inner product space and x, y ∈ X, x − y is also in X, and we can use the
Bessel inequality for x − y:
2
kx − yk ≥
=
n
X
k=1
n
X
k=1
=
=
n
X
k=1
n
X
k=1
|hx − y, ek i|2
|hx, ek i − hy, ek i|2
|hx, ek i −
n
X
j=1
βj hej , ek i|2
|hx, ek i − βk |2 .
Thus, the minimun of kx − yk is obtained if and only if βk = hx, ek i.
Ex. 3.4.6
Let {e1 , ..., en } be an orthonormal set in an inner product space X, where n is fixed. Let
x ∈ X be any fixed element and y = β1 e1 + ... + βn en . We want to show that kx − yk is
minimum if and only if βi = hx, ei i.
The set {e1 , ..., en } is a basis of a closed finite vectorial subspace Y ⊂ X, i.e.
Y = span(e1 , ..., en ), dim(Y ) = n.
which is a convex and complete (see Theorem 2.4-2) subspace of X (see Exercise 3.3.4).
Then we can say that y = β1 e1 + ... + βn en ∈ Y for any list of n scalars (β1 , ..., βn ). From
Theorem 3.3-1, we know that for every given x ∈ X there exists a unique y ∈ Y such that
kx − yk = inf kx − y˜k,
y˜∈Y
(3.0.6)
which in our context we can rephrase as follows: there exists a unique list of n scalars
(β1 , ..., βn ) such that y = β1 e1 + ... + βn en minizes kx − yk. From Lemma 3.3-2 we also
know that a such y must be the orthogonal projection of x on Y , i.e. z = x−y ∈ Y ⊥ . Then
we need to show that if we choose the coeffincients of y to
Pbe βi = hx, ei i, then z = x − y
is orthogonal to any other element of Y . We take v ∈ Y , nk=1 αk ek and we multiply it by
117
z with respect to the inner product of X:
hz, vi = hx − y, vi
= hx, vi − hy, vi
= hx,
=
=
=
=
Pn
k=1
αk ek i − h
Pn
αk hx, ek i −
Pn
αk hx, ek i −
k=1
Pn
k=1
k=1
Pn
k=1
αk hx, ek i −
Pn
k=1
Pn
k=1
Pn
k=1
Pn
k=1
βk ek ,
Pn
βk hek ,
βk
Pn
k=1
Pn
j=1
αk ek i
k=1
αk ek i
αj hek , ej i
βk α k
αk [hx, ek i − βk ]
which prove
Pnthat hz, vi = 0 iff βi = hx, ei i . Note that in the second last equality we have
used that j=1 αj hek , ej i = αk since hek , ej i = 1 only when k = j and zero otherwise.
3.4.8
From Bessel’s inequality, we have
∞
X
|hx, ei i|2
j=1
||x||2
≤ 1.
Assume we have nm inner products hx, ei i satisfying |hx, ei i| >
nm
X
|hx, ek i|2
k=1
||x||2
>
nm
X
k=1
(3.0.7)
1
.
m
We have
1
nm
=
.
m2 ||x||2
m2 ||x||2
We must then (at least) have
nm < m2 ||x||2 ,
since (3.0.7) cannot hold otherwise. This proves the statement.
Ex. 3.5.4
If (xj ) is a sequence in an inner product space X such that the series
kx1 k + kx2 k + kx3 k + . . .
converges, show that Sn is a Cauchy sequence, where Sn = x1 + . . . + xn
118
(3.0.8)
Solution: We let
S n = x1 + x2 + x3 + . . . + xn
(3.0.9)
Then
kSn − Sm k = kxm + xm+1 + · · · + xn k ≤ kxm k+kxm+1 k+. . .+kxn k → 0
as n, m → ∞
(3.0.10)
Since the series (3.0.8) converges.
Ex. 3.5.6
Given: x =
P∞
i=1
αi ei and y =
hx, yi = h
=
∞
X
αi ei ,
i=1
∞
∞
XX
P∞
j=1
∞
X
j=1
βj ej where, (ei ) is an orthonormal sequence. Then,
∞ X
∞
∞ X
∞
X
X
βj ej i =
hαi ei , βj ej i =
αi βj hei , ej i
αi βj δij =
i=1 j=1
i=1 j=1
∞
X
i=1 j=1
αi βi
i=1
It must be added that the series of x and y are absolutely convergent. For example, for x:
∞
∞
∞
∞
X
X
X
X
kxk = hx, xi = h
αi e i ,
αj e j i =
αi αi =
|αi |2
2
i=1
i=1
i=1
i=1
Ex. 3.6.4
We study
hx + y, x + yi = hx, xi + hx, yi + hy, xi + hy, yi
which can be re-written by using Parsevals identity
X
X
X
hx + y, ek ihx + y, ek i =
hx, ek ihx, ek i +
hy, ek ihy, ek i + hx, yi + hy, xi. (3.0.11)
By using the linear properties of the inner product and rearranging, (3.0.11) can be written
X
X
hx, yi + hy, xi =
hy, ek ihx, ek i +
hx, ek ihy, ek i.
k
k
P
P
P
Since hx, yi+hy, xi = 2Re(hx, yi) and k hy, ek ihx, ek i+ k hx, ek ihy, ek i = 2Re( k hy, ek ihx, ek i)
we can conclude that
X
Re(hx, yi) = Re( hx, ek ihy, ek i).
k
119
As for the imaginary part, we have
hx + iy, x + iyi = hx, xi − ihx, yi + ihy, xi + hy, yi,
where i is now the usual imaginary number. Repeating the same procedure on this, one
arrives at
X
X
i(−hx, yi + hy, xi) = i( hy, ek ihx, ek i −
hx, ek ihy, ek i).
k
k
P
P
By using i(−hx, yi + hy, xi) = 2Im(hx, yi) and i( k hy, ek ihx, ek i − k hx, ek ihy, ek i) =
P
2Im( k hy, ek ihx, ek i), one can conclude
X
Im(hx, yi) = Im( hx, ek ihy, ek i).
k
Since both the real and imaginary parts are equal, we must have
X
hx, ek ihy, ek i,
hx, yi =
k
which completes the proof.
Ex. 3.6.4
Let M be a total orthonormal set in a Hilbert space H. Then ∀ x, y ∈ H we know that
X
X
x=
hx, em i em
and
y=
hy, en i en
(3.0.12)
m
n
where en and em are the elements of M which generate a countable set of Fourier coeffincients (Lemma 3.5-3) for x and y respectively. In the next step we consider (ek ) = (el ) =
(em ) ∪ (en ) which is a countable set since it is the countable union of countable sets.
P
P
hx, yi = h k hx, ek i ek , l hy, el i el i
=
=
=
P
k
hx, ek i hek ,
P P
k
P
k
P
l
hy, el i el i
l hx, ek i hy, el i hek , el i
hx, ek i hy, ek i
which is another version of the Parseval relation.
120
(3.0.13)
Ex. 3.6.4
P
P
We have Parseval’s relation k |hx, ek i|2 = kxk2 and wish to show that hx, yi = k hx, ek ihy, ek i.
Consider kx − yk. We have
kx − yk2 = hx − y, x − yi = kxk2 + kyk2 − 2<[hx, yi]
but also from Parseval’s relation that
X
X
kx − yk2 =
|hx − y, ek i|2 =
|hx, ek i − hy, ek i|2
X
X
X
=
|hx, ek i|2 +
|hy, ek i|2 − 2<[ hx, ek ihy, ek i]
X
= kxk2 + kyk2 − 2<[ hx, ek ihy, ek i],
where we again have used Parseval’s relation in the final equality. Comparing the two
expressions tells us that
X
<[hx, yi] = <[ hx, ek ihy, ek i].
k
Repeating the exercise using kx + yk yields the analogous result for the imaginary parts,
X
=[hx, yi] = =[ hx, ek ihy, ek i].
k
Consequently the desired result follows.
Ex. 3.6.4
Let x ∈ X, M = {ek } ⊂ X be a total orthonormal set and
X
x˜ =
hx, ej i ej .
j
Since
hx − x˜, ej i = 0 ∀i
we have
x − x˜ ⊥ M.
This implies x = x˜ by Theorem 3.6.2 a) since M is total in X. Thus we can write any
element y ∈ X as
X
y=
hy, ek i ek .
k
This now gives
hx, yi =
=
*
X
j
X
k
hx, ej i ej ,
X
k
hy, ek i ek
+
hx, ek i hy, ek i .
121
=
XX
j
k
hx, ej i hy, ek i hek , ej i =
Ex. 3.6.10
Let M be a subset of a Hilbert space H, and let u, v ∈ H. Suppose that hv, xi = hw, xi
for all x ∈ M implies v = w. If this holds for all v, w ∈ H, show that M is total in H.
Solution: By the linearity of the inner product
hv, xi = hw, xi ⇐⇒ hv − w, xi = 0.
Thus,
hv, xi = hw, xi for all x ∈ M implies v = w, for all v, w ∈ H ⇐⇒
hv − w, xi = 0 for all x ∈ M implies v = w, for all v, w ∈ H ⇐⇒
hu, xi = 0 for all x ∈ M implies u = 0 ⇐⇒
u⊥M implies u = 0 ⇐⇒
M is total in H.
In the last step we used Theorem 3.6-2. Here it is important that H is complete.
Ex. 3.6.10
Let us consider M subset of H Hilbert space and v, w ∈ H. Moreover we admit that
hv, xi = hw, xi , ∀x ∈ M ⇒ v = w
holds ∀v, w ∈ H.
We want to prove that M is total: in order to do this we consider z ⊥ M , that satisfies for
construction hz, xi = 0, ∀x ∈ M .
But since also h0, xi = 0, ∀x ∈ M , then we have
hz, xi = h0, xi
and this implies that z = 0. Thus, by Theorem 3.6.2 the claim follows.
Ex. 3.8.10
Sesquilinearity follows by
h(x1 + x2 , y) = hx1 + x2 , yi = hx1 , yi + hx2 , yi = h(x1 , y) + h(x2 , y)
h(x, y1 + y2 ) = hx, y1 + y2 i = hx, y1 i + hx, y2 i = h(x, y1 ) + h(x, y2 )
h(αx, y) = hαx, yi = α hx, yi = αh(x, y)
¯
h(x, βy) = hx, βyi = β¯ hx, yi = βh(x,
y).
122
From the Cauchy-Schwartz inequality we get:
khk = sup sup
x6=0 y6=0
| hx, yi |
kxk kyk
≤ sup sup
= 1.
kxk kyk
x6=0 y6=0 kxk kyk
But note that for x = y we have
h(x, x)
kxk2
=
=1
kxk2
kxk2
so we get khk = 1.
Ex. 3.8.14
Schwarz inequality. Let X be a vector space and h a Hermitian form on X × X. This form
is said to be positive semidefinite if h(x, x) ≥ 0 for all x ∈ X. Show that then h satisfies
the Schwarz inequality |h(x, y)|2 ≤ h(x, x)h(y, y).
Since h is positive semidefinite, we have
h(x − λy, x − λy) ≥ 0.
By using the fact that h is a Hermitian sesquilinear form, we obtain
¯
¯
h(x, x) − λh(x,
y) − λh(y, x) + λλh(y,
y) ≥ 0.
Choose
λ=
(3.0.14)
h(x, y)
.
h(y, y)
Thus,
¯ = h(x, y) = h(y, x) .
λ
h(y, y)
h(y, y)
Substituting this into inequality (3.0.14), we obtain
h(x, y)
h(x, y)h(y, x)
h(y, x)
h(x, y) −
h(y, x) +
h(y, y)
h(y, y)
h(y, y)
h(y, x)
= h(x, x) −
h(x, y)
h(y, y)
|h(x, y)|2
= h(x, x) −
.
h(y, y)
0 ≤ h(x, x) −
Therefore, |h(x, y)|2 ≤ h(x, x)h(y, y).
Remark: It may not be straightforward to get the correct λ value. The idea is that the
equality in the Schwarz inequality is taken when x and y are linear dependent. Therefore,
we use λ to project y so that λy and x are linear dependent.
123