Meager sets of periodic functions

Jordan Bell

February 6, 2015

The following is often useful.¹¹ 1 Walter Rudin, Real and Complex Analysis, third ed., p. 68, Theorem 3.12.

Theorem 1.

If $(X,\mu)$ is a measure space, $1\leq p\leq\infty$ , and $f_{n}\in L^{p}(\mu)$ is a sequence that converges in $L^{p}(\mu)$ to some $f\in L^{p}(\mu)$ , then there is a subsequence of $f_{n}$ that converges pointwise almost everywhere to $f$ .

Proof.

Assume that $1\leq p<\infty$ . For each $n$ there is some $a_{n}$ such that

\left\|f_{a_{n}}-f\right\|_{p}<2^{-n}.

Then

\sum_{n=1}^{\infty}\left\|f_{a_{n}}-f\right\|_{p}^{p}<\sum_{n=1}^{\infty}2^{-% np}=\frac{2^{-p}}{1-2^{-p}}<\infty.

Let $\epsilon>0$ . We have

\left\{x\in X:\limsup_{n\to\infty}|f_{a_{n}}(x)-f(x)|>\epsilon\right\}\subset% \bigcap_{N=1}^{\infty}\bigcup_{n=N}^{\infty}\left\{x\in X:|f_{a_{n}}(x)-f(x)|>% \epsilon\right\}.

For any $N$ , this gives, using Chebyshev’s inequality,

\begin{split}&\displaystyle\mu\left(\left\{x\in X:\limsup_{n\to\infty}|f_{a_{n% }}(x)-f(x)|>\epsilon\right\}\right)\\ \displaystyle\leq&\displaystyle\sum_{n=N}^{\infty}\mu\left(\left\{x\in X:|f_{a% _{n}}(x)-f(x)|>\epsilon\right\}\right)\\ \displaystyle\leq&\displaystyle\epsilon^{-p}\sum_{n=N}^{\infty}\left\|f_{a_{n}% }-f\right\|_{p}^{p}.\end{split}

Because $\sum_{n=1}^{\infty}\left\|f_{a_{n}}-f\right\|_{p}^{p}<\infty$ , we have $\sum_{n=N}^{\infty}\left\|f_{a_{n}}-f\right\|_{p}^{p}\to 0$ as $N\to\infty$ , which implies that

\mu\left(\left\{x\in X:\limsup_{n\to\infty}|f_{a_{n}}(x)-f(x)|>\epsilon\right% \}\right)=0.

This is true for each $\epsilon>0$ , hence

\mu\left(\left\{x\in X:\limsup_{n\to\infty}|f_{a_{n}}(x)-f(x)|>0\right\}\right% )=0,

which means that for almost all $x\in X$ ,

\lim_{n\to\infty}|f_{a_{n}}(x)-f(x)|=0.

Assume that $p=\infty$ . Let

E_{k}=\{x\in X:|f_{k}(x)|>\left\|f_{k}\right\|_{\infty}\}.

The measure of each of these sets is $0$ , so for

E=\bigcup_{k}E_{k}

we have $\mu(E)=0$ . For $x\not\in E$ ,

|f(x)-f_{k}(x)|\leq\left\|f-f_{k}\right\|_{\infty}\to 0,\qquad k\to\infty,

showing that for almost all $x\in X$ , $f_{k}(x)\to f(x)$ . ∎

The following results are in the pattern of $A$ being a strict subset of $X$ implying that $A$ is meager in $X$ .

We first work out two proofs of the following theorem.

Theorem 2.

For $1<p\leq\infty$ , $L^{p}(\mathbb{T})$ is a meager subset of $L^{1}(\mathbb{T})$ .

Proof.

For $n\geq 1$ , let

C_{n}=\left\{f\in L^{1}(\mathbb{T}):\left\|f\right\|_{p}\leq n\right\}.

Let $n\geq 1$ . If a sequence $f_{k}\in C_{n}$ converges in $L^{1}(\mathbb{T})$ to some $f\in L^{1}(\mathbb{T})$ , then there is a subsequence $f_{a_{k}}$ of $f_{k}$ such that for almost all $x\in\mathbb{T}$ , $f_{a_{k}}(x)\to f(x)$ , and so $f_{a_{k}}(x)^{p}\to f(x)^{p}$ . Applying the dominated convergence theorem gives

\frac{1}{2\pi}\int_{\mathbb{T}}|f(x)|^{p}dx=\lim_{k\to\infty}\frac{1}{2\pi}% \int_{\mathbb{T}}|f_{a_{k}}(x)|^{p}dx=\lim_{k\to\infty}\left\|f_{a_{k}}\right% \|_{p}^{p}\leq n^{p},

hence $\left\|f\right\|_{p}\leq n$ , showing that $f\in C_{n}$ . Therefore, $C_{n}$ is a closed subset of $L^{1}(\mathbb{T})$ On the other hand, let $f\in C_{n}$ and let $g\in L^{1}(\mathbb{T})\setminus L^{p}(\mathbb{T})$ . Then $f+\frac{1}{k}g\to f$ in $L^{1}(\mathbb{T})$ , and for each $k$ we have $f+\frac{1}{k}g\not\in C_{n}$ , as that would imply $g\in L^{p}(\mathbb{T})$ . This shows that $f$ does not belong to the interior of $C_{n}$ . Because $C_{n}$ is closed and has empty interior, it is nowhere dense. Therefore

L^{p}(\mathbb{T})=\bigcup_{n=1}^{\infty}\left\{f\in L^{1}(\mathbb{T}):\left\|f% \right\|_{p}\leq n\right\}

is meager in $L^{1}(\mathbb{T})$ . ∎

Proof.

The open mapping theorem tells us that if $X$ is an $F$ -space, $Y$ is a topological vector space, $\Lambda:X\to Y$ is continuous and linear, and $\Lambda(X)$ is not meager in $Y$ , then $\Lambda(X)=Y$ , $\Lambda$ is an open mapping, and $Y$ is an $F$ -space.²² 2 Walter Rudin, Functional Analysis, second ed., p. 48, Theorem 2.11.

Let $j:L^{p}(\mathbb{T})\to L^{1}(\mathbb{T})$ be the inclusion map. For $f\in L^{p}(\mathbb{T})$ ,

\left\|j(f)\right\|_{1}=\left\|f\right\|_{1}\leq\left\|f\right\|_{p},

showing that the inclusion map is continuous. On the other hand, $j$ is not onto, so the open mapping theorem tells us that $j(L^{p}(\mathbb{T}))=L^{p}(\mathbb{T})$ is meager in $L^{1}(\mathbb{T})$ . ∎

Suppose that $X$ is a topological vector space, that $Y$ is an $F$ -space, and that $\Lambda_{n}$ is a sequence of continuous linear maps $X\to Y$ . Let $L$ be the set of those $x\in X$ such that

\Lambda x=\lim_{n\to\infty}\Lambda_{n}x

exists. It is a consequence of the uniform boundedness principle that if $L$ is not meager in $X$ , then $L=X$ and $\Lambda:X\to Y$ is continuous.³³ 3 Walter Rudin, Functional Analysis, second ed., p. 45, Theorem 2.7.

For $n\geq 1$ , define $\Lambda_{n}:L^{2}(\mathbb{T})\to\mathbb{C}$ by

\Lambda_{n}f=\sum_{|k|\leq n}\hat{f}(k),\qquad f\in L^{1}(\mathbb{T}).

Define

L=\left\{f\in L^{2}(\mathbb{T}):\textrm{$\lim_{n\to\infty}\Lambda_{n}f$ exists% }\right\}.

The sequence $t\mapsto\sum_{k=1}^{n}\frac{e^{ikt}}{k}$ is a Cauchy sequence in $L^{2}(\mathbb{T})$ , hence converges to some $f\in L^{2}(\mathbb{T})$ , which satisfies

\hat{f}(k)=\begin{cases}\frac{1}{k}&k\geq 1\\ 0&k\leq 0.\end{cases}

Then

\Lambda_{n}f=\sum_{k=1}^{n}\frac{1}{k}\to\infty,\qquad n\to\infty,

meaning that $f\in L^{2}(\mathbb{T})\setminus L$ . This shows that $L\neq L^{2}(\mathbb{T})$ . Therefore, the above consequence of the uniform boundedness principle tells us that $L$ is meager.