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Let $A_n$ be a subset of continuous functions on $[0,1]$ given by:

$A_n$ = {$f∈C[0,1]$:there exists $x∈[0,1]$ such that $|f(x)−f(y)|≤n|x−y|$ for all $y∈[0,1]$}.

Show $A_n$ is nowhere dense, and use this fact to show that there are nowhere differentiable continuous functions on $[0,1]$.

We already showed $A_n$ is closed. How to show interior of $A_n$ is empty?

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  • $\begingroup$ Does anyone know how to show that nowhere differentiable continuous functions must exist on [0,1] given that $A_n$ is nowhere dense? I believe the Baire theorem must be invoked. $\endgroup$ Commented Apr 10, 2014 at 2:27

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As it's currently written, this answer provides only the idea of the proof.

A function $f(x)=0$ belongs to $A_n$, so $A_n\ne\emptyset$. Now let's take a function $$g(x)=1-2|x-1/2|,$$ a "hat" function. Let's define it by periodicity to the whole $\Bbb R$: $$g(x)=g(x-1)=g(x+1)\quad \forall x\in \Bbb R.$$

Now take $$g_k = \frac {1}{2^k} g(2^kx)$$and$$G_N(x)=\sum_{k\ge N}g_k(x).$$

For large $N$ this function is close to zero (and, therefore, to $A_n$), it is continuous as an absolutely convergent series of continuous functions, but it can't belong to $A_n$.

We want to show that $$\forall f\in A_n\quad f+G_N\notin A_n.$$ (to be continued)

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  • $\begingroup$ I'm not sure what you did here. You produced some kind of "general" positive function that doesn't belong to $A_n$...what about every other function? I'm not familiar with this proof technique, excuse my ignorance. $\endgroup$ Commented Apr 8, 2014 at 22:36
  • $\begingroup$ This only shows $A_n$ does not contain a neighborhood of $0$. It's not hard to adapt this to arbitrary neighborhoods, but it's worth pointing out. $\endgroup$ Commented Apr 8, 2014 at 22:41
  • $\begingroup$ @AlexBecker indeed, thank you for this comment. I'll edit my post to take this into account. $\endgroup$ Commented Apr 9, 2014 at 7:07
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I'll write something to show that there are nowhere differentiable functions on $C[0,1]$.

Proof:

\begin{align*} A_n\text{ nowhere dense }&\iff\left(\overline{A_n}\right)^\circ=\varnothing\\ &\iff X\setminus(\overline{A_n})^\circ=X\\ &\iff\overline{X\setminus \overline{A_n}}=X\\ &\iff\overline{(X\setminus A_n)^\circ}=X\\ &\iff (X\setminus A_n)^\circ\text{ is dense in }X\\ &\iff(X\setminus A_n)\text{ contains a dense open subset}. \end{align*}

Let $G_n=(X\setminus A_n)^\circ$, then ${G_n}\subset(A_n)^c$

By Baire's category theorem, $\cap_nG_n$ is dense in $X$ and hence $\cap_nG_n\neq\emptyset$.

For any $f\in{\cap_nG_n}$, since $f\in(A_n)^c$ for any n, $|f(x)-f(y)|\ge n|x-y|$ for all $x,y\in[0,1]$ and all n.

Thus $f$ is nowhere differentiable.

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  • $\begingroup$ f does not belong to (An)c means for every t in [0,1] there exists sn in [0,1] such that |f(sn)-f(t)|> n|sn-t|. Please clarify how you can made that statement |f(x)−f(y)|≥n|x−y| for all x,y∈[0,1]x,y∈[0,1] and all n $\endgroup$ Commented Sep 17, 2017 at 5:23

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