1. Convergence F18.2

Fall 2018 Problem 2: Consider the function f : [0, 1] → ℝ defined by

f(x) = \begin{cases}x log x & if x ∈ (0, 1] \\ 0 & if x = 0. \end{cases}

Be sure to justify your answer for each of the following questions. (a) Is f Lipschitz continuous on [0, 1]?

(b) Is f uniformly continuous on [0, 1]?

(c) Suppose (p_{n}) is a sequence of polynomial functions on [0, 1], converging uniformly to f. Is the set A = {p_{n}|n ≥ 1} ∪ {f} equicontinuous?

1.1. Ideas F18.2.

• Sketch the function and note that only near 0 is interesting.

• Lipschitz is similar to bounded slope.

• Unifom continuity is the same as continuity for compact domains.

• Equicontinuous arises in the Arzelá-Ascoli theorem.

• Uniform convergence of polynomials arises in the Weierstrass theorem.

• Equicontinuity requires choosing δ for all the functions at once.

• By direct computation the slope is unbounded near 0.

• By direct computation (L’Hôpital) f is continuous at 0.

• Maybe if f and g are uniformly close this allows the choice of δ.

1.2. Write up F18.2. :

Claim: If f ∈ C[0, 1] ∩ C^{1}(0, 1) and f is Lipschitz continuous then |f′| is bounded.

Compute f′(x) = log(x) + 1 which is unbounded near x = 0 so by the claim the answer to (a) is no.

Claim: If K is compact and f ∈ C(K) then f is uniformly continuous.

The function f is clearly continuous at any x ∈ (0, 1] and computing with L’Hôpital’s rule

lim_{x→0} f(x) = lim_{x→0} \frac{log x}{x^{−1}} = lim_{x→0} \frac{x^{−1}}{−x^{−2}} = 0 = f(0)

so f is continuous and by the claim the answer to (b) is yes.

Definition If (X, d) is a metric space a family F ⊂ C(X) is equicontinuous if

∀ϵ > 0 ∃δ > 0 ∀f ∈ F, d(x, y) < δ ∃|f(x) − f(y)| < ϵ.

(i) Since {p_{n}} converges uniformly to f one has:

∀ϵ > 0 ∃N_{ϵ} ∀n > N_{ϵ}, x ∈ [0, 1] ∃|f(x) − p_{n}(x)| < \frac{1}{3}ϵ.

(ii) Since f is uniformly continuous one has:

∀ϵ > 0 ∃δ_{ϵ} ∀x, y ∈ [0, 1], |x − y| < δ_{ϵ} ∃|f(x) − f(y)| < \frac{1}{3}ϵ.

(iii) Since each {p_{n}} is uniformly continuous one has:

∀ϵ > 0 ∃δ′_{ϵ} ∀n ≤ N_{ϵ}, x, y ∈ [0, 1], |x − y| < δ′_{ϵ} ∃|p_{n}(x) − p_{n}(y)| < \frac{1}{3}ϵ.

(iv) Combining these facts one has:

∀ϵ > 0 ∃δ′′_{ϵ} = min{δ_{ϵ}, δ′_{ϵ}} ∀g ∈ A, x ∈ [0, 1] ∃|g(x) − g(x)| < ϵ

so the answer to (c) is yes.

The last inequality follows directly from (i) if g = f and from (iii) if n ≤ N_{ϵ} while otherwise

|p_{n}(x) − p_{n}(y)| ≤ |P_{n}(x) − f(x)| + |f(x) − f(y)| + |f(y) − p_{n}(y)| ≤ ϵ.

2. Convergence S15.1

(Spring 2015 Problem 1.) Let H be a separable Hilbert space. A sequence (x_{n}) in H converges in the Cesáro sense to x ∈ H if the averages of its partial sums converge strongly to x, i.e., if

\overline{x}_{N} = \frac{1}{N} ∑_{n=1}^{N} x_{n} → x as N → ∞.

(a) Prove that if (x_{n}) converges strongly to x in H, then (x_{n}) also converges in the Cesáro sense to x.

(b) Give an example of a sequence that converges in the Cesáro sense but does not converge weakly.

(c) Give an example of a sequence that converges weakly but does not converge in the Cesáro sense.

2.1. Ideas S15.1.

• For convenience subtract x from everything so that wlog x = 0.

• Recall the definitions of strong and weak convergence.

• For (a) consider and compare the sequences a_{n} = ∥x_{n}∥_{2} and \overline{a}_{n} = ∥\overline{x}_{n}∥_{2}.

• If the sequnce converges the average must eventually also.

• For (b) and (c) take H either ℝ or ℓ^{2}ℕ.

• For (b) the average can be small even if the value occasionally jumps so ℝ works.

• For (c) consider the sequence {e_{n}} ⊆ ℓ^{2}ℕ which converges weakly but not in norm to 0. Oops this also Cesáro converges.

• For (c) try telescoping {e_{n} − e_{n+1}} ⊆ ℓ^{2}ℕ.

2.2. Write up S15.1. :

Definition: If {x_{n}}_{n∈ℕ} is a sequence in a Hilbert space H with inner product ⟨·, ·⟩ then

(st) (x_{n}) converges strongly to x ∈ H if (a_{n} − ∥x_{x} − x∥) converges

to 0 in ℝ, (wk) (x_{n}) converges weakly to x ∈ H if for every y ∈ H the sequence

(⟨y, x_{n} − x⟩) converges to 0 in ℝ. Claim: A sequence (x_{n}) is a Hilbert space converges in the (weak, strong or Cesáro) sense to x iff (x_{n} − x) converges in the same sense to 0.

(a) By the lemma we may assume that (x_{n}) converges strongly to 0 and hence from the definition that (a_{n}) converges to 0 in ℝ and it will suffice to show that (\overline{a}_{n} = ∥\overline{x}_{n}∥) also does. Write â_{n} = \frac{1}{n} ∑_{m=1}^{n} a_{m}. For any ϵ > 0 choose N_{ϵ} with ∀n > N_{ϵ} ∃a_{n} < \frac{ϵ}{2} which is possible from the convergence. Write M_{ϵ} for the maximum of N_{ϵ} and 2ϵ^{−1}N_{ϵ}â_{N_{ϵ}}.

Claim:

(i) If m > M_{ϵ} then â_{m} < ϵ. (ii) 0 ≤ \overline{a}_{n} ≤ â_{n}. Part (a) of the problem follows from the claim.

Part (ii) of the claim follows from the triangle inequality. Part (i) of the claim follows from the computation that if

m > M_{ϵ} then â_{n} ≤ \frac{1}{m}[∑_{r=1}^{N_{ϵ}} a_{r} + ∑_{r=N_{ϵ}}^{m} a_{r}] ≤ \frac{1}{m}[N_{ϵ}â_{N_{ϵ}} + (m − N_{ϵ})\frac{ϵ}{2}] < \frac{N_{ϵ}â_{Nϵ}}{M_{ϵ}} + \frac{ϵ}{2} ≤ ϵ

(b) Consider H = ℝ and x_{2^{r}} = 1 but x_{n} = 0 otherwise.

(c) Consider H = ℓ^{2}ℕ with orthonormal basis (e_{n})_{n∈ℕ} and set x_{n} = e_{n} − e_{n+1} so that a_{n} = â_{n} = \sqrt{2} and \overline{a}_{n} = \frac{1}{m}.

3. Convergence S15.4

(Spring 2015 Problem 4) Let f_{n} : [0, 1] → ℝ be a sequence of measurable functions. Suppose

(i) ∫_{0}^{1} |f_{n}(x)|^{2}dx ≤ 1 for n = 1, 2, · · ·. (ii) f_{n} → 0 almost everywhere.

Show that

lim_{n→∞} ∫_{0}^{1} f_{n}(x)dx = 0.

3.1. Ideas S15.4.

• compact, eg Arzela-Ascoli

• bounded convergence

• Egorov: a.e. convergence in a set of finite measure gives almost uniform convergence.

• ∥ · ∥_{2} bound is needed because of examples like f_{n} = nχ_{[0, \frac{1}{n}]}.

• Try to use the bound by splitting into big and little.

3.2. Write up S15.4. Claim (Egorov):

∀ϵ > 0
∃N_{ϵ} ∈ ℕ, E_{ϵ} ∈ BI, λE_{ϵ} < ϵ
∀n > N_{ϵ}, x ∈ E_{ϵ}^{c}
∃|f_{n}x| < ϵ.

Using the claim compute ∀ϵ > 0, n > N_{ϵ} one has (1)

| ∫_{x∈[0,1]} f_{n}(x)dx|

(2)

≤ ∫_{x∈E_{ϵ}} |f_{n}(x)|dx + ∫_{x∈E_{ϵ}^{c}} |f_{n}(x)|dx

(3)

= ∥f_{n}χ_{E_{ϵ}}∥_{1} + ∥f_{n}χ_{E_{ϵ}^{c}}χ_{I}∥_{1}

(4)

≤ ∥f_{n}∥_{2}∥χ_{E_{ϵ}}∥_{2} + ∥f_{n}χ_{E_{ϵ}^{c}}∥_{∞}∥χ_{I}∥_{1}

(5)

≤ 1 · \sqrt{ϵ} + ϵ · 1

which approaches 0 as ϵ approaches 0.

Proof. (of Claim) Write U_{ϵ,N} = {x ∈ I|∀n > N∃|f_{n}x| < ϵ} so that U_{ϵ,N} ⊆ U_{ϵ,N+1} and if U = {x| lim_{n} f_{n}x = 0} then ∀ϵ∃U ⊆ ∪_{N}U_{ϵ,N}. By hypothesis λU^{c} = 0 so ∀ϵ > 0∃N_{ϵ}, λU_{ϵ,N_{ϵ}}^{c} < ϵ and E_{ϵ} = U_{ϵ,N_{ϵ}}^{c} satisfies the claim. □

Justifications for the (in)equalities:

(1)

(2) | ∫ f| ≤ ∫ |f|

(3) ∥fg∥_{1} ≤ ∥f∥_{p}∥g∥_{q} if 1 ≤ p, q ≤ ∞ and \frac{1}{p} + \frac{1}{q} = 1 such as (p, q) = (2, 2) or (∞, 1).

(4) ∥f_{n}∥_{2} ≤ 1 by hypothesis, ∥χ_{A}∥_{p}^{p} = μA by the definition of the integral and p-norm and ∥f_{n}χ_{E_{ϵ}^{c}}∥_{∞} ≤ ϵ by the choice of E_{ϵ}.

4. Convergence F15.1

(Fall 2015 Problem 1.) Consider the following sequences of functions parameterized by n:

• a_{n}(x) = e^{i2\sqrt{n}πx}, x ∈ [0, 1],

• b_{n}(x) = \sqrt{n}e^{−n|x|}, x ∈ ℝ,

• c_{n}(x) = ne^{−nx^{2}}, x ∈ ℝ,

• d_{n}(x) = ∑_{k=−n}^{n} e^{i2kπx}, x ∈ [0, 1]. As n tends to infinity, which sequences converge (a) almost everywhere,

(b) L^{2}-strongly, (c) L^{2}-weakly but not strongly. Explain your answer.

4.1. Ideas F15.1:

• Recall the definitions.

• Sketch the families noting interesting x values.

• (a) and (d) look nices after Fourier transformation. This helps with L^{2} but not as good for a.e.

• (b) and (c) have different decay rates.

5. Convergence F14.3

(Fall 2014 Problem 3) Let H be a separable Hilbert space with inner product ⟨·, ·⟩ : H × H → ℂ, and let {e_{n} : n ∈ ℕ} be an orthonormal basis of H. Define a metric d : B × B → ℝ on the closed unit ball B of H by

d(x, y) = ∑_{n=1}^{∞} \frac{|⟨x − y, e_{n}⟩|}{2^{n}}.

(a) Show that the sequence (x_{k}) in B converges to x ∈ B with respect to the metric d if and only if it converges weakly to x in H.

(b) Prove that (B, d) is a compact metric space.

5.1. Ideas F14.3.

• Consider pairing with basis elements to relate to weak convergence.

• Complete and totally bounded is enough for compact in a metric space.

• The Banach-Alaoglu theorem is essentially this statement.

• Sequential compactness will suffice for a metric space to be compact.

• Separability gives a countable dense set of elements of H to use in the weak topology.