- Is ℕ a G
_{δ}set? What about ℚ? What about a countable dense subset of a complete metric space? - ↑ Let f : ℝ → ℂ be a function. Define the oscillation of a function in Bby ω
_{r}f(x) = sup{|f(z) − f(y)| : y,z ∈ B(x,r)}. Define the oscillation of the function at the point, x by ωf(x) = lim_{r→0}ω_{r}f(x). Show f is continuous at x if and only if ωf(x) = 0. Then show the set of points where f is continuous is a G_{δ}set (try U_{n}= {x : ωf(x) <}). Does there exist a function continuous at only the rational numbers? Does there exist a function continuous at every irrational and discontinuous elsewhere? Hint: Suppose D is any countable set, D =_{i=1}^{∞}, and define the function, f_{n}to equal zero for every xand 2^{−n}for x in this finite set. Then consider g≡∑_{n=1}^{∞}f_{n}. Show that this series converges uniformly. - Let f ∈ C([0,1]) and suppose f
^{′}(x) exists. Show there exists a constant, K, such that |f(x) −f(y)|≤ K|x−y| for all y ∈ [0,1]. Let U_{n}= {f ∈ C([0,1]) such that for each x ∈ [0,1] there exists y ∈ [0,1] such that |f(x) −f(y)| > n|x−y|}. Show that U_{n}is open and dense in C([0,1]) where for f ∈ C,Show that ∩

_{n}U_{n}is a dense G_{δ}set of nowhere differentiable continuous functions. Thus every continuous function is uniformly close to one which is nowhere differentiable. - ↑Suppose f= ∑
_{k=1}^{∞}u_{k}where the convergence is uniform and each u_{k}is a polynomial. Is it reasonable to conclude that f^{′}= ∑_{k=1}^{∞}u_{k}^{′}? The answer is no. Use Problem 3 and the Weierstrass approximation theorem do show this. - Let X be a normed linear space. We say A ⊆ X is “weakly bounded” if for each
x
^{∗}∈ X^{′},sup{|x^{∗}(x)| : x ∈ A} < ∞, while A is bounded if sup{||x|| : x ∈ A} < ∞. Show A is weakly bounded if and only if it is bounded. - Let X and Y be two Banach spaces. Define the norm
Show this is a norm on X × Y which is equivalent to the norm given in the chapter for X × Y . Can you do the same for the norm defined for p > 1 by

- Let f be a 2π periodic locally integrable function on ℝ. The Fourier series for f is
given by
where

Show

where

Verify that ∫

_{−π}^{π}D_{n}dt = 1. Also show that if g ∈ L^{1}, thenThis last is called the Riemann Lebesgue lemma. Hint: For the last part, assume first that g ∈ C

_{c}^{∞}and integrate by parts. Then exploit density of the set of functions in L^{1}. - ↑It turns out that the Fourier series sometimes converges to the function pointwise.
Suppose f is 2π periodic and Holder continuous. That is ≤ K
^{θ}where θ ∈ (0,1]. Show that if f is like this, then the Fourier series converges to f at every point. Next modify your argument to show that if at every point, x,≤ K^{θ}for y close enough to x and larger than x and≤ K^{θ}for every y close enough to x and smaller than x, then S_{n}f→, the midpoint of the jump of the function. Hint: Use Problem 7. - ↑ Let Y = {f such that f is continuous, defined on ℝ, and 2π periodic}. Define
||f||
_{Y }= sup{|f(x)| : x ∈ [−π,π]}. Show that (Y,||||_{Y }) is a Banach space. Let x ∈ ℝ and define L_{n}(f) = S_{n}f(x). Show L_{n}∈ Y^{′}but lim_{n→∞}||L_{n}|| = ∞. Show that for each x ∈ ℝ, there exists a dense G_{δ}subset of Y such that for f in this set, |S_{n}f(x)| is unbounded. Finally, show there is a dense G_{δ}subset of Y having the property that |S_{n}f(x)| is unbounded on the rational numbers. Hint: To do the first part, let f(y) approximate sgn(D_{n}(x − y)). Here sgnr = 1 if r > 0,−1 if r < 0 and 0 if r = 0. This rules out one possibility of the uniform boundedness principle. After this, show the countable intersection of dense G_{δ}sets must also be a dense G_{δ}set. - Let α ∈ (0,1]. Define, for X a compact subset of ℝ
^{p},where

and

Show that

is a complete normed linear space. This is called a Holder space. What would this space consist of if α > 1? - ↑Now recall Problem 10 about the Holder spaces. Let X be the Holder functions
which are periodic of period 2π. Define L
_{n}f= S_{n}fwhere L_{n}: X → Y for Y given in Problem 9. Showis bounded independent of n. Conclude that L_{n}f → f in Y for all f ∈ X. In other words, for the Holder continuous and 2π periodic functions, the Fourier series converges to the function uniformly. Hint: L_{n}fis given bywhere f

= f+ gwhere≤ C^{α}. Use the fact the Dirichlet kernel integrates to one to writeShow the functions, y → g

∕sinare bounded in L^{1}independent of x and get a uniform bound on. Now use a similar argument to showis equicontinuous in addition to being uniformly bounded. If L_{n}f fails to converge to f uniformly, then there exists ε > 0 and a subsequence, n_{k}such that_{∞}≥ ε where this is the norm in Y or equivalently the sup norm on. By the Arzela Ascoli theorem, there is a further subsequence, L_{nk l}f which converges uniformly on. But by Problem 8 L_{n}f→ f. - Let X be a normed linear space and let M be a convex open set containing 0.
Define
Show ρ is a gauge function defined on X. This particular example is called a Minkowski functional. It is of fundamental importance in the study of locally convex topological vector spaces. A set, M, is convex if λx + (1 − λ)y ∈ M whenever λ ∈ [0,1] and x,y ∈ M.

- ↑ The Hahn Banach theorem can be used to establish separation theorems. Let M
be an open convex set containing 0. Let xM. Show there exists x
^{∗}∈ X^{′}such that Rex^{∗}(x) ≥ 1 > Rex^{∗}(y) for all y ∈ M. Hint: If y ∈ M,ρ(y) < 1. Show this. If xM,ρ(x) ≥ 1. Try f(αx) = αρ(x) for α ∈ ℝ. Then extend f to the whole space using the Hahn Banach theorem and call the result F, show F is continuous, then fix it so F is the real part of x^{∗}∈ X^{′}. - A Banach space is said to be strictly convex if whenever ||x|| = ||y|| and x≠y,
then
F : X → X

^{′}is said to be a duality map if it satisfies the following: a.)= ||x||.b.) F(x)(x) = ||x||^{2}. Show that if X^{′}is strictly convex, then such a duality map exists. The duality map is an attempt to duplicate some of the features of the Riesz map in Hilbert space which is discussed in the chapter on Hilbert space. Hint: For an arbitrary Banach space, letShow F

≠∅ by using the Hahn Banach theorem on f(αx) = α||x||^{2}. Next show Fis closed and convex. Finally show that you can replace the inequality in the definition of Fwith an equal sign. Now use strict convexity to show there is only one element in F. - Prove the following theorem which is an improved version of the open mapping
theorem, [?]. Let X and Y be Banach spaces and let A ∈ℒ. Then the following are equivalent.
There exists a constant M such that for every y ∈ Y , there exists x ∈ X with y = Ax and

Note this gives the equivalence between A being onto and A being an open map. The open mapping theorem says that if A is onto then it is open.

- Suppose D ⊆ X and D is dense in X. Suppose L : D → Y is linear and
||Lx||≤ K||x|| for all x ∈ D. Show there is a unique extension of L,, defined on all of X with ||x||≤ K||x|| andis linear. You do not get uniqueness when you use the Hahn Banach theorem. Therefore, in the situation of this problem, it is better to use this result.
- ↑ A Banach space is uniformly convex if whenever ||x
_{n}||,||y_{n}||≤ 1 and ||x_{n}+ y_{n}||→ 2, it follows that ||x_{n}− y_{n}||→ 0. Show uniform convexity implies strict convexity (See Problem 14). Hint: Suppose it is not strictly convex. Then there existandboth equal to 1 and= 1 consider x_{n}≡ x and y_{n}≡ y, and use the conditions for uniform convexity to get a contradiction. It can be shown that L^{p}is uniformly convex whenever ∞ > p > 1. See Hewitt and Stromberg [?] or Ray [?]. - Show that a closed subspace of a reflexive Banach space is reflexive. Hint: The
proof of this is an exercise in the use of the Hahn Banach theorem. Let Y be the
closed subspace of the reflexive space X and let y
^{∗∗}∈ Y^{′′}. Then i^{∗∗}y^{∗∗}∈ X^{′′}and so i^{∗∗}y^{∗∗}= Jx for some x ∈ X because X is reflexive. Now argue that x ∈ Y as follows. If xY , then there exists x^{∗}such that x^{∗}= 0 but x^{∗}≠0. Thus, i^{∗}x^{∗}= 0. Use this to get a contradiction. When you know that x = y ∈ Y , the Hahn Banach theorem implies i^{∗ }is onto Y^{′}and for all x^{∗}∈ X^{′}, - We say that x
_{n}converges weakly to x if for every x^{∗}∈ X^{′},x^{∗}(x_{n}) → x^{∗}(x). x_{n}⇀ x denotes weak convergence. Show that if ||x_{n}− x||→ 0, then x_{n}⇀ x. - ↑ Show that if X is uniformly convex, then if x
_{n}⇀ x and ||x_{n}||→||x||, it follows ||x_{n}− x||→ 0. Hint: Use Lemma 15.2.9 to obtain f ∈ X^{′}with ||f|| = 1 and f(x) = ||x||. See Problem 17 for the definition of uniform convexity. Now by the weak convergence, you can argue that if x≠0, f→ f. You also might try to show this in the special case where== 1. - Suppose L ∈ℒand M ∈ℒ. Show ML ∈ℒand that
^{∗}= L^{∗}M^{∗}. - Let X and Y be Banach spaces and suppose f ∈ℒis compact. Recall this means that if B is a bounded set in X, then fhas compact closure in Y. Show that f
^{∗}is also a compact map. Hint: Take a bounded subset of Y^{′},S. You need to show f^{∗}is totally bounded. You might consider using the Ascoli Arzela theorem on the functions of S applied to fwhere B is the closed unit ball in X.

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