A topological space X is a Baire space if a countable intersection of open dense sets is still dense in X. Then Baire category theorem consists in:

Theorem: (Baire) A complete metric space is a Baire space.

It is a surprising consequence of Baire category theorem that almost all continuous functions are nowhere differentiable in the following sense: Let C([0,1]) be the set of continuous functions [0,1] \rightarrow \mathbb{R} endowed with the sup norm || \cdot || : f \mapsto \max\limits_{x \in [0,1]} |f(x)|; then

Property: The set of continuous nowhere differentiable functions is dense in C([0,1]).

Proof. For all n>0, let

U_n= \{ f \in C([0,1]) \mid \forall x \in [0,1], \exists y \in B(x,1/n), |f(x)-f(y)| >n |x-y| \}.

For convenience, let V_n be the complement of U_n in C([0,1]). We first show that U_n is open, or equivalently that V_n is closed.

Let (f_k) be a sequence in V_n converging to some f \in C([0,1]). So for all k>0, there exists x_k \in [0,1] such that for all y \in B(x_k,1/n), |f_k(x_k)-f_k(y)| \leq n|x_k-y|. By compactness, there is a subsequence (x_{\sigma(k)}) converging to some x \in [0,1]. Let y \in B(x,1/n); for N>0 large enough, |x_{\sigma(k)}-y|<1/n and |x_{\sigma(k)}-x|<1/n for all k \geq N. Then, for k \geq N:

\begin{array}{lcl} |f(x)-f(y)| & \leq & |f(x)-f_{\sigma(k)}(x)| +|f_{\sigma(k)}(x)-f_{\sigma(k)}(x_{\sigma(k)})| \\ & & +|f_{\sigma(k)}(x_{\sigma(k)})-f_{\sigma(k)}(y)|+|f_{\sigma(k)}(y)-f(y)| \\ & \leq & 2 ||f-f_{\sigma(k)}||_{\infty}+n|x-x_{\sigma(k)}|+n|x_{\sigma(k)}-y| \end{array}

When k \rightarrow + \infty, we get |f(x)-f(y)| \leq n|x-y|. Therefore, f \in V_n hence V_n is closed.

Then, we show that U_n is dense in C([0,1]). Let f \in C([0,1]) and \epsilon>0. Because f is uniformly continuous (according to Heine theorem), there exists \delta>0 such that for all x,y \in [0,1], |x-y|<\delta implies |f(x)-f(y)|<\epsilon. Now let 0=x_0<x_1< \dots < x_{n-1} <x_n=1 such that |x_{i+1}-x_i|<\delta for 0 \leq i \leq n.

Introduce g \in C([0,1]) piecewise linear such that g(x_i)=f(x_i), g'(x)>n and |g(x)-f(x_i)|<\epsilon/2 for all x \in [x_i,x_{i+1}] and 0 \leq i \leq n+1. In particular, g \in U_n. Let x \in [0,1] and 0 \leq i \leq n such that x \in [x_i,x_{i+1}]. Then

|f(x)-g(x)| \leq |f(x)-f(x_i)|+|g(x_i)-g(x)| \leq \epsilon,

hence ||f-g||_{\infty} \leq \epsilon. Therefore, U_n is dence in C([0,1]).

Because C([0,1]) is a Banach space, U:= \bigcap\limits_{n \geq 1} U_n is dense in C([0,1]) according to Baire theorem. However, any fonction of U is nowhere differentiable, consequently the set of nowhere differentiable functions is itself dense in C([0,1]). \square

Corollary: The set of continuous nowhere locally monotonic functions is dense in C([0,1]).

A monotonic function being almost everywhere differentiable, a nowhere differentiable function is nowhere locally monotonic and the assertion follows from the property. However, we can give a direct proof using Baire category theorem:

Sketch of proof. For p,q \in \mathbb{Q} with p<q, let I_{p,q}= \{f \in C([0,1]) \mid f \ \text{increasing on} \ [p,q] \}. If f \in I_{p,q} and \epsilon>0, then there exists a piecewise linear function f_{\epsilon} oscillating around f such that ||f-f_{\epsilon}||_{\infty} < \epsilon and f_{\epsilon} \notin I_{p,q}. Therefore, I_{p,q} has empty interior. Moreover, I_{p,q} is closed: let (f_n) be a sequence in I_{p,q} converging to some f \in C([0,1]); then, for all x<y \in [p,q], f(x)-f(y) = \lim\limits_{n \to + \infty} f_n(x)-f_n(y) \leq 0, so f \in I_{p,q}.

The same argument can be used for D_{p,q} = \{f \in C([0,1]) \mid f \ \text{decreasing on} \ [p,q] \}.

According to Baire category theorem, the set \bigcup\limits_{p<q \in \mathbb{Q}} I_{p,q} \cup D_{p,q} of continuous functions locally monotonic at some point has empty interior. \square