1. No category

2014 Fall Mathematical Analysis III
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Assignment 10
Problems 1–3 are for systems of differential equations.
1. Consider the Cauchy problem for the third order equation:
y 000 = f (x, y, y 0 ), y(x0 ) = y0 , y 0 (x0 ) = y1 , y 00 (x0 ) = y3 .
Express it as the Cauchy problem for a system of three first order equations.
Solution. Let u1 (x) = y(x), u2 (x) = y 0 (x). Then u(x) = (u1 (x), u2 (x)) solves the system
of first order equations
u01 = u2 , u02 = f (x, u1 , u2 ), (u1 , u2 )(x0 ) = (y0 , y1 ).
Similarly one can express the Cauchy problem for a third order equation as the Cauchy
problem for a system of three first order equations.
2. Optimal. Consider (2.3) for a system of differential equations. Let f ∈ C(G) satisfies the
Lipschitz condition on every compact subset of the open set G. Show that there exists a
solution y ∗ of (2.3) defined on some (α, β) with the properties: (a) whenever y is a solution
of (2.3) on some I, I ⊂ (α, β); and (b) when β is finite, for each compact K in G, there
exists some ρ > 0 such that (x, y ∗ (x)) ∈ G \ K for x ∈ [β − ρ, β).
Solution. Omitted. It is similar to the scalar case.
3. Consider the Cauchy problem (2.3) for a system of differential equations. Assuming that
f (x, y) ∈ C(R2 ) satisfying the Lipschitz condition and the growth condition
f (x, y) ≤ C(1 + |y|), ∀(x, y) ∈ R2 .
Show that (2.3) admits a global solution, that is, aPsolution defined on (−∞, ∞). Hint:
Study the differential inequality satisfied by |y|2 = nj=1 yj2 .
1
Solution. Let E(x) = |y(x)|2 . Then
2
|E 0 (x)| = |y(x) · y 0 (x)| = |y(x) · f (x, y)| ≤ C|y|(1 + |y|) ≤ C0 (1 + |y|2 ) = C0 (1 + |E(x)|).
Hence it follows from the comparison principle that E(x) ≤ (E(0) + 1)eC0 x − 1. Hence
(2.3) admits a global solution.
4. Let In = [−n, n] and kf kn = sup{|f (x)| : x ∈ [−n, n]}. For f, g ∈ C(R), define
∞
X
1 kf − gkn
d(f, g) =
.
2n 1 + kf − gkn
n=1
(a) Show that this is a complete metric on C(R). Hint: Use Problem 8, Ex 4.
(b) Show that {fj } converges to f in this metric implies {fj } converges to f uniformly
on every bounded interval.
Solution. (a). kf −Pgk/(1 + kf − gk) is a metric by a previous exercise. As the infinite
series is bounded by n 2n ≤ 1, it is convergent so d(f, g) is a well-defined metric on C(R).
(b). Well, when {fj } is a Cauchy sequence in this metric, it is also a Cauchy sequence in
the metric kf − gkn /(1 + kf − gkn ) for every n. But this metric is equivalent to kf − gkn .
2014 Fall Mathematical Analysis III
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By the completeness of kf − gkn there exists some f n such that {fj } converges to f n
uniformly on In . From In ⊂ In+1 and the uniqueness of limit we know that f n+1 = f n on
In , so the function f (x) = f n (x), x ∈ In is a continuous function
P on R.−nWe claim that
d(fj , f ) → 0 as j → ∞. For, given ε > 0, fix a large N such that ∞
< ε/2. Then
n=N 2
∞
∞
X
X
1 kfj − f kn
ε
1
≤
< .
2n 1 + kfj − f kn
2n
2
n=N
n=N
On the other hand, we can find a large j0 such that
kfj − f k1 , · · · , kfj − f kN −1 <
ε
,
2N
∀j ≥ j0 .
It follows that d(fj , f ) < ε, for all j ≥ j0 .
(c). From (b) we see that {fj } converges to f on every In . As every bounded set is
contained in some In for large n, our desired conclusion is drawn.
5. Optional. Inspired by the previous problem, introduce a metric on C ∞ (R), the vector
space of all smooth functions on the real line, so that {fn } converges to f in this metric
implies {dk fn /dxk } converges to dk f /dxk uniformly on each bounded interval for each
k ≥ 0.
Solution. Try the metric
∞
X
1 kf − gk∞ + · · · + kf (n) − g (n) kn
.
ρ(f, g) =
2n 1 + kf − gkn + · · · + kf (n) − g (n) kn
n=1
6. Let E be a bounded, convex set in Rn . Show that a family of equicontinuous functions is
bounded in E if it is bounded at a single point, that is, if there is some constant M such
that |f (x0 )| ≤ M for all f in this family.
Solution. By equicontinuity, for ε = 1, there is some δ0 such that |f (x) − f (y)| ≤ 1
whenever |x − y| ≤ δ0 . Let BR (x0 ) a ball containing E. Then |x − x0 | ≤ R for all x ∈ E.
We can find x0 , · · · , xn = x where nδ0 ≤ R ≤ (n + 1)δ0 so that |xn+1 − xn | ≤ δ0 . It follows
that
n−1
X
R
|f (x) − f (x0 )| ≤
|f (xj+1 − f (xj )| ≤ n ≤ .
δ0
j=0
Therefore,
|f (x)| ≤ |f (x0 )| + n + 1 ≤ M +
R
δ0
∀x ∈ E, ∀f ∈ F.
7. Let {fn } be a sequence in C(G) where G is open in Rn . Suppose that on every compact
subset of G, it is equicontinuous and bounded. Show that there is a subsequence {fnj }
converging to some f ∈ C(G) uniformly on each compact subset of G.
Solution. Let Kn be an ascending family of compact sets in G satisfying G = ∪n Kn .
Applying Ascoli-Arezela theorem to {fn } on each Kn step by step and then take a Cantor’s
diagonal sequence.
8. Let {fn } be a sequence of bounded functions in [0, 1] and let Fn be
Z x
Fn (x) =
fn (t)dt.
0
2014 Fall Mathematical Analysis III
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(a) Show that the sequence {Fn } has the Bolzano-Weierstrass property provided there is
some M such that kfn k∞ ≤ M, for all n.
(b) Show that the conclusion in (a) holds when boundedness is replaced by the weaker
condition: There is some K such that
Z 1
|fn |2 ≤ K, ∀n.
0
Solution.
Rx
Rx
(a) Since |Fn | ≤ 0 |fn (t)|dt ≤ M , and |Fn (x) − Fn (y)| ≤ y |fn (t)|dt ≤ |x − y|M , {Fn } is
uniformly bounded and equicontinuous. Then it follows from Arzela-Ascoli theorem
that {Fn } has the Bolzano-Weierstrass property.
(b) It follows from the Cauchy-Schwarz inequality that
Z
x
x
Z
|fn (t)|dt ≤
|Fn (x) − Fn (y)| ≤
y
1/2 Z
1 dt
2
y
x
1/2
√ p
|fn (t)| dt
≤ K |x − y|.
2
y
Similarly one can show that {Fn } is uniformly bounded. Then apply Arzela-Ascoli
theorem.
9. Let K ∈ C([a, b] × [a, b]) and define the operator T by
Z
(T f )(x) =
b
K(x, y)f (y)dy.
a
(a) Show that T maps C[a, b] to itself.
(b) Show that whenever {fn } is a bounded sequence in C[a, b], {T fn } contains a convergent subsequence.
Solution.
(a) Since K ∈ C([a, b] × [a, b]), given ε > 0, there exists δ > 0 such that |K(x, y) −
K(x0 , y)| < ε, whenever |x − x0 | < δ. Then for x, x0 ∈ [a, b], |x − x0 | < δ, one has
0
Z
|(T f )(x) − (T f )(x )| ≤
b
|K(x, y) − K(x0 , y)||f (y)|dy ≤ |a − b|kf k∞ ε.
a
Hence T f ∈ C[a, b].
(b) Suppose supn kfn k∞ ≤ M < ∞. It follows from the proof of (a) that δ can be taken
independent of n. Hence {fn } is equicontinuous. Furthermore, since |(T fn )(x)| ≤
Rb
a |K(x, y)||fn (y)|dy ≤ M (b − a)kKk∞ , {fn } is uniformly bounded. Then it follows
from Arzela-Ascoli theorem that {T fn } contains a convergent subsequence.
10. Let f be a bounded, uniformly continuous function on R. Let fa (x) = f (x + a). Show
that for each l > 0, there exists a sequence of intervals In = [an , an + l], an → ∞, such
that {fan } converges uniformly on [0, l].
Solution. Let {an } be a sequence with an → ∞. Since f is bounded and uniformly
continuous on R, it follows that {fan } is uniformly bounded and equivcontinuous on [0, l].
Applying Ascoli-Arezela theorem to obtain a subsequence converging uniformly on [0, l].
2014 Fall Mathematical Analysis III
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11. Optional. Let {hn } be a sequence of analytic functions in the unit disc satisfying |hn (z)| ≤
M, ∀z, |z| < 1. Show that there exist an analytic function h in the unit disc and a
subsequence {hnj } which converges to h uniformly on each smaller disc {z : |z| ≤ r}, r ∈
(0, 1). Suggestion: Use a suitable Cauchy integral formula.
Solution. Let D = {|z| < 1} be the unit disc. Let rn % 1 be a strictly-increasing positive
sequence and let Dn = {|z| < rn }. For each n ∈ N, since hj is analytic in D, it follows
from the Cauchy integral formula that
Z
hj (ζ)
1
hj (z) =
dζ, for |z| < rn .
2πi |ζ|=rn+1 ζ − z
Hence
|h0j (z)|
1
=|
2πi
Z
|ζ|=rn+1
f (ζ)
M
dζ| ≤
,
(ζ − z)2
2π|rn+1 − rn |2
for |z| < rn .
Since |h0j (z)| is uniformly bounded on Dn , it follows that {hj } is equicontinuous on each
Dn . Applying Ascoli-Arezela theorem to {hj } on each Dn step by step and then taking a
Cantor’s diagonal sequence, one obtains a {hnj } which converges to h uniformly on each
smaller disc {z : |z| ≤ r}, r ∈ (0, 1). It follows from uniform convergence that
Z
h(ζ)
1
dζ, for |z| < rn .
h(z) =
2πi |ζ|=rn+1 ζ − z
Hence h is analytic in D.