1. No category

Math 140B - Winter 2015 - Final Exam
Problem 1.
Consider the sequence fn : [0, 1] → R given by
(
nx for 0 ≤ x ≤ n1
fn (x) =
1
for n1 < x ≤ 1.
(i) Find the pointwise limit f of the sequence fn .
(ii) Calculate the distance between fn and f in the supremum norm
d(fn , f ) = sup |fn (x) − f (x)|.
x∈[0,1]
Deduce that the sequence fn does not converge uniformly to f .
(iii) Show that the sequence fn converges in L2 to the function f .
Solution:
(i) We keep x fixed and make n → ∞. When x = 0, we have fn (0) = 0, hence the limit is
f (0) = 0. For x 6= 0, we have n1 < x for n sufficiently large. Thus fn (x) = 1 for n large,
hence the limit when n → ∞ is f (x) = 1. Thus, the pointwise limit is
(
0 if x = 0
f (x) =
.
1 if 0 < x ≤ 1
(ii) We have


0
fn (x) − f (x) = nx − 1


0
if x = 0
if 0 < x ≤ n1 .
if n1 < x ≤ 1
We have
d(fn , f ) = sup |fn (x) − f (x)| = sup |nx − 1| = sup 1 − nx = 1.
1
x∈(0, n
]
x∈[0,1]
1
x∈(0, n
]
Since d(fn , f ) 6→ 0, it follows that fn does not converge uniformly to f .
(iii) We show that the (square of the) L2 -distance between fn and f converges to zero:
Z 1
|fn (x) − f (x)|2 dx → 0.
0
We compute
Z
1
2
Z
|fn (x) − f (x)| dx =
0
0
1
n
2
Z
|nx − 1| dx =
0
1
n
n2 x2 − 2nx + 1 dx =
n2 x3
1
x= 1
− nx2 + x|x=0n =
→ 0.
3
3n
Problem 2.
Let 0 < a < b < 1 be fixed. Show that polynomials of the form
P (x) = c2015 x2015 + c2016 x2016 + . . . + cn xn , n ≥ 2015 (not fixed),
are uniformly dense in the space of real valued continuous functions over [a, b], but not dense in
the set of continuous functions over [0, 1].
Solution:Let A denote the set polynomials of the form
P (x) = c2015 x2015 + c2016 x2016 + . . . + cn xn , n ≥ 2015.
Clearly, sums, products and scalar multiples of polynomials of this form are also of this form, so A
is an algebra.
Furthermore, A separates points and vanishes nowehere on [a, b]. Indeed, the polynomial Q(x) =
∈ A can be used to separate points over [a, b]:
x2015
x1 6= x2 =⇒ x2015
6= x2015
.
1
2
By the same reasoning, the polynomial Q(x) = x2015 can be used to show that A vanishes nowehere:
for x1 ∈ [a, b] =⇒ x2015
6= 0 (since x1 6= 0 as a > 0).
1
By Stone-Weierstraβ, A is uniformly dense in C[a, b].
On the other hand, all polynomials in A vanish at 0: P (0) = 0. If the algebra A were dense in
C[0, 1] then for each function f ∈ C[0, 1], we could find a sequence of polynomials Pn ⇒ f , Pn ∈ A.
However Pn (0) = 0, which would imply f (0) = 0. Nonetheless, there are continuous functions f
which do not vanish at 0, for instance f (x) = x + 1. Contradiction!
Problem 3.
Consider the function f : [−π, π) → R given by
f (x) = ex .
n
π
−π
e −e
(i) Show that cn = (−1)
1−in ·
2π
(ii) Find the value of the sum
.
∞
X
n=1
n2
1
.
+1
Solution:
(i) We compute the Fourier coefficients
Z π
Z π
1
1 e(1−in)x x=π
1 e(1−in)π − e−(1−in)π
1
−inx
f (x)e
dx =
e(1−in)x dx =
|x=−π =
.
cn =
2π −π
2π −π
2π 1 − in
2π
1 − in
Using that eiπ = −1, so that einπ = e−inπ = (−1)n , the above formula simplifies to
(−1)n eπ − e−π
.
2π
1 − in
There is no need to do the calculation separately for n = 0 since the denominator of the
above fraction never vanishes.
(ii) We compute
1 eπ − e−π
√
|cn | =
.
2π n2 + 1
In particular |cn | = |c−n |, so the above gives
∞
∞
∞
X
X
(eπ − e−π )2 (eπ − e−π )2 X 1
+
.
|cn |2 = |c0 |2 + 2
|cn |2 =
2
2
2+1
4π
2π
n
n=−∞
cn =
n=1
n=1
Next, we compute
Z π
Z π
1
1
1 e2π − e−2π
f (x)2 dx =
e2x dx =
.
2π −π
2π −π
2π
2
Using Parseval’s theorem, we know
Z π
∞
X
1
2
f (x) dx =
|cn |2 .
2π −π
n=−∞
Substituting, we find
∞
X
1
1
e2π − e−π π
1 eπ + e−π π
=
−
+
·
=
−
+
· .
n2 + 1
2 (eπ − e−π )2 2
2 eπ − e−π 2
n=1
Problem 4.
Let f : R → R be an integrable 2π-periodic function with zero integral.
(i) Using Parseval’s theorem, show that the Fourier coefficients cn → 0.
(ii) If f is differentiable with continuous derivative, show that ncn → 0.
(iii) Show that
Z
Z
π
π
f (x)2 dx ≤
f 0 (x)2 dx.
−π
−π
Solution:
(i) We have
Z
1
2π
∞
X
π
2
f (x) =
−π
|cn |2 .
n=−∞
Since the series on the right converges, we conclude cn → 0 as n → ∞.
(ii) We have seen in the homework that the Fourier coefficients of f 0 equal incn . Indeed, the
coefficients of f 0 can be computed integrating by parts and using periodicity
Z π
Z π
1
1
0
0
−inx
−inx π
−inx
cn =
f (x)e
dx =
f (x)e
|x=−π −
f (x)(−in)e
dx
2π −π
2π
−π
Z
in π
=
f (x)e−inx dx = incn .
2π −π
Since f 0 is continuous, hence integrable, it follows from part (i) applied to the function f 0
that
ncn → 0.
(iii) By Parseval, we have
1
2π
Z
∞
X
π
f (x)2 dx =
−π
|cn |2 .
n=−∞
On the other hand, also by Parseval, we have
Z π
∞
∞
X
X
1
|incn |2 =
n2 |cn |2 .
f 0 (x)2 dx =
2π −π
n=−∞
n=−∞
Furthermore,
Z π
1
c0 =
f (x) dx = 0
2π −π
so c0 does not appear in either Fourier series. For n 6= 0, we have
∞
∞
X
X
2
2
2
2
|cn | ≤ n |cn | =⇒
|cn | ≤
n2 |cn |2 .
n=−∞
Hence the above identities show that
Z π
Z
2
f (x) dx ≤
−π
n=−∞
π
−π
f 0 (x)2 dx.
Problem 5.
Consider φ : R → R a nonnegative continuous function with φ(0) = 0.
Let F be the family of functions f : [0, 1] → R such that f (0) = 0 and for all x, y ∈ [0, 1] we have
|f (x) − f (y)| ≤ φ(x − y).
(i) Show that F is uniformly bounded.
(ii) Show that F is equicontinuous.
(iii) Show that any sequence of functions in F has a subsequence converging to a continuous
function.
Solution:
(i) We set y = 0. This yields that for all f ∈ F we have
|f (x)| ≤ φ(x).
Since φ is continuous, it is bounded by some constant M over the interval [0, 1]. Thus
|f (x)| ≤ φ(x) ≤ M
shows that the family F is uniformly bounded.
(ii) Fix > 0. Since φ is continuous and φ(0) = 0, it follows that there exists δ > 0 such that
if |t| < δ we have
φ(t) < .
For t = x − y, we conclude that if |x − y| < δ we have
|f (x) − f (y)| ≤ φ(x − y) < ,
for all f ∈ F. This is exactly the definition of equicontinuity.
(iii) By Arzel`
a-Ascoli, any sequence in F has a uniformly convergent subsequence. The limit
function is continuous since it is uniform limit of continuous functions (members of F are
continuous; in fact, even stronger, F is equicontinuous).
Problem 6.
Assume that f : (0, 1) → R is a function with bounded derivative |f 0 (x)| ≤ A for all x ∈ (0, 1).
Show that f is bounded.
(i) Prove that
1 1 + f
.
2
2 (ii) Conclude from (i) that f is bounded over (0, 1).
(iii) Take f (x) = sin x1 . Show f is bounded, but f 0 is unbounded over the interval (0, 1).
Therefore, the converse of (ii) does not hold.
|f (x)| ≤ A x −
Solution:
(i) Using the mean value theorem, we conclude that for all x, there exists c between 12 and x
such that
1
1
1 1 0
1 0
f (x) − f
= x−
f (c) =⇒ f (x) − f
= x − |f (c)| ≤ A x − .
2
2
2 2
2
Using the triangle inequality, we find
1
1
1
+ f
≤ A x − + f 1 .
|f (x)| ≤ f (x) − f
2
2
2
2 (ii) We have x − 12 ≤ 12 for all x ∈ (0, 1) so that above becomes
A 1 |f (x)| ≤ + f
,
2
2 showing that f is bounded by M = A2 + f 12 .
(iii) We have |f (x)| ≤ 1, so f is bounded. On the other hand, by the chain rule
1
1
f 0 (x) = − 2 cos
.
x
x
If xn =
1
2nπ
∈ (0, 1) for n ∈ Z>0 , then
f 0 (xn ) = −(2nπ)2 cos(2nπ) = −(2nπ)2 → −∞
showing that f 0 is unbounded.
Problem 7.
Assume that f : [0, 1] → R is a bounded function such that f is integrable over any interval [c, 1]
for c > 0. Using Riemann’s criterion, show that f is integrable over [0, 1].
Solution:Since f is bounded, there exists a constant M > 0 such that |f (x)| ≤ M . We fix > 0.
We exhibit a partition P of [0, 1] such that
U (P, f ) − L(P, f ) < .
Let c =
4M .
Since f is integrable over [c, 1], by Riemann’s criterion, there exists a partition
Q = {c = x0 < x1 < x2 < . . . < xn = 1}
of [c, 1] such that
U (Q, f ) − L(Q, f ) < .
2
Let P be the partition
P = {0} ∪ Q = {0 < c = x0 < x1 < . . . < xn = 1}.
Then
U (P, f ) − L(P, f ) = (sup f − inf f ) · c + (U (Q, f ) − L(Q, f )) ≤ 2M c + (U (Q, f ) − L(Q, f ))
[0,c]
[0,c]
+ = .
4M
2
By Riemann’s criterion, it follows that f is integrable.
< 2M ·
Problem 8.
Consider the series
f (x) =
∞
X
n=1
x
.
1 + n2 x2
(i) Show that the series converges uniformly over (−∞, −a) ∪ (a, ∞) for any a > 0.
(ii) Conclude that the series converges pointwise on R, and the function f is continuous over
R \ {0}.
(iii) Show that f is not continuous at x = 0, and conclude that the series does not converge
uniformly on R.
Solution:
(i) We use the Weierstraβ M -test. For x ∈ (−∞, −a) ∪ (a, ∞), we have
1
1
|x|
x
1 + n2 x2 ≤ n2 x2 = n2 |x| < n2 a .
P∞ 1
P∞
1
x
1 P∞
Since
1 n2 a = a
n=1 n2 converges, it follows that
n=1 1+n2 x2 converges uniformly
over (−∞, −a) ∪ (a, ∞).
(ii) Clearly, f (0) = 0, so we do get pointwise convergence at x = 0. For x 6= 0, there exists
a > 0, for instance a = |x|
2 , such that |x| > a. Thus x ∈ (−∞, −a) ∪ (a, ∞), and over this
interval we proved uniform, hence pointwise convergence.
In fact, the limit function f is continuous over (−∞, −a) ∪ (a, ∞) for any a > 0, being
uniform limit of continuous functions (the partial sums are clearly continuous). Since any
x ∈ R \ 0 belongs to an interval (−∞, −a) ∪ (a, ∞) for a > 0 (take a = |x|
2 ), this shows that
f is in fact continuous at all points x ∈ R \ {0}.
(iii) Assume that f is continuous at 0. Now, f (0) = 0 by direct substitution. Then
1
f
→ f (0) = 0, k → ∞.
k
We obtain a contradiction once we show that
1
1
f
≥ for all k.
k
2
To see the above inequality, we estimate
X
k
k
k
k
1
X
X
X
1
k
k
1
k
1
k
f
≥
≥
=
=
= .
2 =
2
2
2
2
n
k
k +n
k +k
2k
2k
2
1+ 2
n=1
k
n=1
n=1
n=1
Thus f is discontinuous at 0. Since uniform limit of continuous functions is continuous
and as f is discontinuous at 0, it follows that the series does not converge uniformly on R.