LEC ANA2 9

Proposition 169.1(Proposition).

Let $(X,\Vert \cdot \Vert )$ be a NLS. Then $X$ is complete iff for every $\{x_n{\}}_{n=1}^{\infty }\subseteq X$,

$$\sum _{n=1}^{\infty }\Vert x_n\Vert <\infty ⟹\sum _{n=1}^{\infty }{x}_n\in X.$$

Proof.

$(⟹)$ Suppose $X$ is complete. Let $\{x_n\}\subseteq X$ such that ${\sum }_{n=1}^{\infty }\Vert x_n\Vert <\infty$. It suffices to prove ${\sum }_{n=1}^N{x}_n$ is Cauchy, that is

$$\Vert \sum _{k=n}^m{x}_n\Vert \to 0.$$

But, $\Vert {\sum }_{k=n}^m{x}_k\Vert \le {\sum }_{k=n}^m\Vert x_n\Vert \to 0$ by hypothesis.

$(⟸)$ Suppose $\{x_n\}\subseteq X$ is Cauchy. There exists $n_1$ such that $\Vert x_n-x_{n_1}\Vert <1$ for $n\ge n_1$. Similarly, there exists $n_k$ such that $\Vert x_n-x_{n_k}\Vert <1/2^k$ for $n\ge n_k$. For this subsequence

$$\begin{array}{r}\Vert x_{n_{k+1}}-x_{n_k}\Vert <1/2^k\\ ⟹\sum _{k=1}^{\infty }\Vert x_{n_{k+1}}-x_{n_k}\Vert <\infty .\end{array}$$

By hypotheses, ${\sum }_{k=1}^N(x_{n_{k+1}}-x_{n_k})$ converges, say to $x$. Then, $x_{n_{N+1}}\to x+x_{n_1}$. Thus, $\{x_n\}$ has a convergent subsequence. Since $\{x_n\}$ is Cauchy, $\{x_n\}$ converges.□

Proposition 169.2(Proposition).

A complete NLS is not a countable union of proper closed subspaces.

Proof.

By Corollary 168.9, It suffices to prove that if $K$ is a proper closed subspace of $X$, then $K$ is nowhere dense. Since $K\ne X$, there exists $y\notin K$. Clearly, $y/\Vert y\Vert$ is not in $K$ either, so WLOG we can assume $\Vert y\Vert =1$. If $x\in K$, then $x+ϵy\notin K$ for all $ϵ>0$. Since $\Vert x-(x+ϵy)\Vert =\Vert ϵy\Vert =ϵ$, this implies $x\notin K^{\circ }$, so $K$ is nowhere dense.□

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Nowhere differentiable functions are not meagre in C[0, 1]

Lemma 169.3(Lemma).

Let $\mathcal{P}$ be the set of all piecewise linear functions in $C[0,1]$. $\mathcal{P}$ is dense in $C[0,1]$.

Proof.

Suppose$f\in C[0,1]$. Let $ϵ>0$. Let $\delta$ be chosen by the uniform continuity of $f$. Choose $0=x_0<x_1<\cdots <x_n=1$ such that the distance between two consecutive $x_i$‘s is less than $\delta$. Let $p\in \mathcal{P}$ be the piecewise linear interpolation on $(x_0,f(x_0)),\dots ,(x_n,f(x_n))$.

If $x\in [0,1]$, $x\in [x_i,x_{i+1}]$ for some $i$.

$$\begin{array}{rl}|f(x)-p(x)|& \le |f(x)-f(x_{i+1})|+|f(x_{i+1})-p(x)|.\end{array}$$

$|f(x)-f(x_{i+1})|<ϵ$ by construction. $|f(x_{i+1})-p(x)|=|p(x_{i+1})-p(x)|\le |p(x_{i+1})-p(x_i)|<ϵ$.□

9920c5

Theorem 169.4(Theorem).

The subset of nowhere differentiable continuous functions in $C[0,1]$ is of second category.

Proof.

By Corollary 168.9, we know that $C[0,1]$ is of second category. Thus, it suffices to show that the subset $D$ of $C[0,1]$ consisting of functions which are differentiable at at least one point in $[0,1]$ is of first category.

Define

$$E_N=\{f\in C[0,1]:\exists x_0\in C[0,1]\text{ such that }|f(x)-f(x_0)|\le N|x-x_0|\forall x\in [0,1]\}.$$

Note that if $f$ is differentiable at $x_0$, then there exists $N$ such that $|f(x)-f(x_0)|\le N|x-x_0|$ for all $x\in [0,1]$: we can bound $|f(x)-f(x_0)|/|x-x_0|$ on $(x_0-ϵ,x_0+ϵ)\setminus \{x_0\}$ for some $ϵ>0$ using the fact that its limit exists as $x\to 0$, and it is easy to bound it on $[0,x_0-ϵ]\cup [x_0+ϵ,1]$ using the fact that $f$ is continuous. It follows that $D\subseteq {\bigcup }_{N=1}^{\infty }{E}_N$. We only need to prove each $E_N$ is closed and $E_N^{\circ }=\varnothing$.

Each $E_N$ is closed in $C[0,1]$

Suppose $\{f_n\}\subseteq E_N$ and $f_n\to f$ in $C[0,1]$. For each $f_n\in E_n$, there exists $x_n\in [0,1]$ such that $|f_n(x)-f_n(x_n)|\le N|x-x_n|$ for all $x\in [0,1]$. $[0,1]$ is compact; $\{x_n\}$ must have a convergent subsequence; WLOG assume it is $\{x_n\}$. Let $\{x_n\}\to x_0$. It suffices to show

$$|f(x)-f(x_0)|\le N|x-x_0|$$

for all $x\in [0,1]$.

$$\begin{array}{rl}|f(x)-f(x_0)|& \le |f(x)-f_n(x)|+|f_n(x)-f_n(x_0)|+|f_n(x_0)-f(x_0)|\end{array}$$

Since convergence in $B[0,1]$ implies uniform convergence, the first and third terms can be individually bounded by $ϵ$ for $n\ge n_0$.

$$\begin{array}{rl}|f_n(x)-f_n(x_0)|& \le |f_n(x)-f_n(x_n)|+|f_n(x_n)-f_n(x_0)|\\ & \le \underset{\to N|x-x_0|}{\underbrace{N|x-x_n|}}+\underset{\to 0}{\underbrace{N|x_n-x_0|}}\\ ⟹& |f(x)-f(x_0)|\le N|x-x_0|+4ϵ.\end{array}$$

$E_N^{\circ }$ is empty for each $N$

We have to show that for all $f\in E_N$, for all $ϵ>0$, there exists $g\in E_N^c$ such that $\Vert g-f\Vert <ϵ$. Let $f\in E_N$. Let $ϵ>0$. Choose $p\in \mathcal{P}$ such that $\Vert f-p{\Vert }_{\infty }<ϵ$, which we can do by Lemma 3. Let $M>\max \{N+|m_i|\}$, where the $m_i$‘s are the slopes of the linear parts of $p$.

Choose $\delta$ such that $0<\delta <ϵ/M$. Choose $0=x_0<x_1<\cdots <x_n$ such that $|x_{i+1}-x_i|<\delta$.

Define $h$ by $h(x_i)=0$, $h((x_i+x_{i+1})/2)=ϵ$, and the linear interpolation through these points elsewhere. Then, the slopes of $h$ are $>M$.

Take $g=p+h$. Then,

$$|g(x)-g(y)|\ge N|x-y|$$

for all $x,y$. Thus, $g\notin E_N$, and $\Vert f-g{\Vert }_{\infty }<2ϵ$: $\Vert f-(p+h)\Vert \le \Vert f-p\Vert +\Vert h\Vert <ϵ$.□