Homework# 8 solutions
3. a. Pointwise, but not uniformly on R. P gk(x) is not continuous. It has jumps at positive integers.
b. Uniformly (Weierstrass M-test with 1/k2). Continuous.
c. Converges uniformly on [a, b] ⊂ (0, π) (use the Dirichlet test by com- puting 2 sin(x)Sn(x) = Pn
k=1(−1)k2 sin(x) cos(kx) = Pn
k=1(−1)k[sin((k + 1)x) − sin((k − 1)x)]). Converges pointwise on R \ {(2m + 1)π : m ∈ Z}.
P∞
k=1gk(x) diverges at {(2m + 1)π : m ∈ Z}.
d. Pointwise, but not uniformly. Continuous.
4. a.& b. Observe that for x ∈ [1, 2]
∞
X
n=1
x (1 + x)n ≤
∞
X
n=1
xn (1 + x)n ≤
∞
X
n=1
(2 3)n. Weierstrass test ⇒ converges uniformly on [1, 2].
c. Unform convergence ⇒ interchange of limit and summation.
8. No. The example given in the class provides a counterexample, i.e.,
fn(x) =
n2x, 0 ≤ x < 1/n
2n − n2x, 1/n ≤ x < 2/n 0, 2/n ≤ x ≤ 1.
fn→ 0 pointwise for all x ∈ [0, 1], but not uniformly. Note that the example fn(x) = xn for x ∈ [0, 1] doesn’t work since fn does not converge to a continuous function.
19. Splitting
(sin nx
n2 )x3 = sin nx n3/2
x3
√n. Let φn(x) = x3/√
n and fn(x) = sin nx/n3/2. For any fixed bounded set {φn(x)} are decreasing and uniformly bounded. By the Weierstrass test
1
P fn(x) uniformly converges on any set. So by Abel’s test, P fn(x)φn(x) converges uniformly on any bounded set. So the convergent function is con- tinuous on any bounded set and therefore it is continuous in R.
20. a. Straightforward. b. Obvious!
c. This is proved in Rudin’s book (see Theorem 7.18 ??).
26. Easy application of the contraction mapping theorem. Define Φ(f ) = A(x) +
Z 1 0
k(x, y)f (y)dy on M = C([0, 1], R) and check all the conditions.
37. Using the intermediate value theorem. If f (x) is not a constant, then f (x1) and f (x2) different are rational numbers for some x1 < x2. Any ir- rational numbers between f (x1) and f (x2) must be in the range of (x1, x2).
This is a contradiction.
45. a. K is compact ⇒ K can be covered by a finite number of balls D(xj, δ).
It suffices to check Cauchy’s criterion: for any ε, any x ∈ K, x ∈ D(xj, δ) for some xj and
ρ(fn(x), fm(x))
≤ ρ(fn(x), fn(xj)) + ρ(fm(x), fm(xj)) + ρ(fn(xj), fm(xj))
< ε
for all n, m large, where ρ(fn(x), fn(xj)) < ε/3 and ρ(fm(x), fm(xj)) < ε/3 by the equicontinuity and ρ(fn(xj), fm(xj)) < ε/3 by the pointwise convergence.
b. It is easy to see that fn(x) converges pointwise to 0. However max
x∈[0,1]|fn(x)| = fn(1 n) = 1.
So the convergence is not uniformly. From a., we can conclude that fn(x) is not equicontinuous. You can check this by yourselves.
2
47. Let K be a dense subset of A on which fn(x) converges (pointwise sense).
Now for any x ∈ A and any δ > 0, we can find y ∈ K such that d(x, y) < δ and equicontinuity implies ρ(fn(x), fn(y)) < ε/3. Then
ρ(fn(x), fm(x))
≤ ρ(fn(x), fn(y)) + ρ(fm(x), fm(y)) + ρ(fn(y), fm(y))
< ε
for all n, m large, where ρ(fn(y), fm(y)) < ε/3 is ensured by the convergence of fn on K. In the proof of Theorem 5.6.2, one can replace the condition of pointwise compactness by that fn converges on a countable dense subset of A.
57. Please refer to the solution on the back of the textbook. It is clear enough.
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