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I'm trying to prove the next problem:

Let $C([0,1],\mathbb{R})$ the space of continuous function $f:[0,1]\to \mathbb{R}$ with the supremum(uniform convergence) metric and let $\mathbb{B}\subset C([0,1],\mathbb{R})$ be the subset of continuous nowhere differentiable functions. I have to show that B contains a countable intersection of dense open sets.

In order to do that, we consider the set: $$A_{n}:=\{f\in C([0,1],\mathbb{R}): \forall t\in [0,1]\space \exists h \space s.t \mid \frac{f(t+h)-f(t)}{h}\mid > n \}$$

And then, if we prove:

  1. $A_{n}$ is open in $C([0,1],\mathbb{R})$

  2. $A_{n}$ is dense in $C([0,1],\mathbb{R})$

Then we can conclude that $\mathbb{B}$ contains a countable intersection of open dense subsets. Finally, this means that the set $\mathbb{B}$ is dense because of the Baire's category theorem.

I've already proven 1) and 2) but I cant get to the conclusion.

It is probably a very elemental thing. I hope you can help.

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  • $\begingroup$ I don't know the answer, but I have a helpful comment: the set $\mathbb B$ doesn't contain a non-empty open set. Indeed, the set $\mathcal P$ of polynomials is dense on $C([0,1],\mathbb R)$. But then, since $\mathcal P\subset \mathbb B^c$, if $\mathbb B$ contained a non-empty open set $U$, then the closure $\overline P$ wouldn't reach any of the points in $U$, a contradiction. So, $\mathbb B$ contains an intersection of open sets but none of them is contained in $\mathbb B$ $\endgroup$ – André Porto Nov 10 '18 at 21:19
  • $\begingroup$ Now I know the answer. Just posted it. $\endgroup$ – André Porto Nov 10 '18 at 22:17
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    $\begingroup$ For what it's worth, it is not difficult to show that the continuous nowhere differentiable functions form a dense subset (even a $c$-dense subset) of the continuous functions, once you have the existence of one such function, by using the same method I outlined in this answer. However, the Baire category method allows you to do this without needing in advance a nowhere differentiable continuous function, and it gives a stronger result (being co-meager is stronger than being $c$-dense). $\endgroup$ – Dave L. Renfro Nov 10 '18 at 23:20
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It only lasts to see that $\displaystyle\bigcap_{n\in\mathbb N} A_n \subset \mathbb B$.

Fix $f\in\displaystyle\bigcap_{n\in\mathbb N} A_n$ and we will prove below that $f\in\mathbb B$.

Fix $t\in[0,1]$. For each $n\in \mathbb N$, there exists $h_n$ such that $$ \left|\dfrac{f(t+h_n)-f(t)}{h_n}\right| > n. $$ Therefore,

$$ \lim_{n\to +\infty}\left|\dfrac{f(t+h_n)-f(t)}{h_n}\right|=+\infty. $$

Now observe that

Claim. There exists a subsequence of $(h_n)_{n\in\mathbb N}$ converging to $0$.

Suppose the contrary. Then, there exists $\delta>0$ such that $|h_n|\geq \delta$ for any $n\in\mathbb N$. Then $$ \left|\dfrac{f(t+h_n)-f(t)}{h_n}\right|\leq \dfrac{|f(t+h_n)-f(t)|}{\delta},\ \forall n \in \mathbb N, $$ and since $f$ is continuous in $[0,1]$, it is bounded by some $M>0$, and so $$ \left|\dfrac{f(t+h_n)-f(t)}{h_n}\right|\leq \dfrac{2M}{\delta},\ \forall n \in \mathbb N, $$ but this contradicts the fact that the limit is infinite, and the claim is proved

Fix the subsequence $(h_{n_m})_{m\in\mathbb N}$ given by the Claim. We have that $h_{n_m}\to 0$ and $$ \lim_{m\to +\infty}\left|\dfrac{f(t+h_{n_m})-f(t)}{h_{n_m}}\right|=+\infty, $$ so using this classical equivalence on limits (from real analysis):

$\displaystyle\lim_{t\to a} f(t)=L \Leftrightarrow$ $\displaystyle\lim_{n\to+\infty} f(t_n)=L$, for any sequence $t_n\to a$,

it follows that the limit $\displaystyle\lim_{h\to0}\left|\dfrac{f(t+h)-f(t)}{h}\right|$ doesn't exist, and consequently, $\displaystyle\lim_{h\to0}\dfrac{f(t+h)-f(t)}{h}$ doesn't exist as well, so $f$ is not differentiable at $t$. Since $t\in[0,1]$ was fixed arbitrarily, it follows that $f$ is nowhere differentiable and so $f\in\mathbb B$.

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  • $\begingroup$ Thanks so much! As soon as I can I'll vote this answer up! $\endgroup$ – asd123 Nov 10 '18 at 22:40
  • $\begingroup$ But still dont know where is the Baire category theorem applied here, I mean, its suppose to have something to do with this problem cause my teacher said this was a pretty cool application of it. $\endgroup$ – asd123 Nov 10 '18 at 22:42
  • $\begingroup$ One of the forms of Baire's category theorem is that for any sequence $(U_n)_{n\in\mathbb N}$ of open dense sets, the intersection $\displaystyle\bigcap_{n\in\mathbb N}U_n$ is a dense set (it may be not open, but surely dense). This is the case of the sequence $(A_n)_{n\in\mathbb N}$ that you defined. Each $A_n$ is open and dense. By Baire's category theorem, $\displaystyle\bigcap_{n\in\mathbb N}A_n$ is a dense set. Since we proved that $\displaystyle\bigcap_{n\in\mathbb N}A_n\subset\mathbb B$, it follows that $\mathbb B$ is dense (any set that contains a dense set is dense). $\endgroup$ – André Porto Nov 11 '18 at 0:39
  • $\begingroup$ Thanks so much you cleared my mind :) $\endgroup$ – asd123 Nov 11 '18 at 3:13
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The Baire category theorem states that for a sequence $(U_n)_{n \in \mathbb{N}}$ of open dense sets, the set $U = \bigcap_{n=1}^\infty U_n$ is also dense.

Note that in our situation $\bigcap_{n=1}^\infty A_n \subset \mathbb{B}$ and $\bigcap_{n=1}^\infty A_n$ is dense. In order to have $\bigcap_{n=1}^\infty A_n \subset \mathbb{B}$ you should modify the definition of $A_n$ as follows: $$A_{n}:=\{f\in C([0,1],\mathbb{R}): \forall t\in [0,1]\space \exists |h| \in (0,1/n) \, s.t \mid \frac{f(t+h)-f(t)}{h}\mid > n \}.$$ Now if $f \in \bigcap_{n=1}^\infty A_n$, then for any $t \in [0,1]$ and any $n \in \mathbb{N}$ there exists $0 < |h_n| < 1/n$ with $$\Big|\frac{f(t+h_n)-f(t)}{h_n}\Big| > n.$$ Thus $f$ is not differentiable in $t$!

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