Weierstrass's Theorem

Theorem
There exists a real function $f: \closedint 0 1 \to \closedint 0 1$ such that:
 * $(1): \quad f$ is continuous
 * $(2): \quad f$ is nowhere differentiable.

Proof
Let $C \closedint 0 1$ denote the set of all real functions $f: \closedint 0 1 \to \R$ which are continuous on $\closedint 0 1$.

From Closed Real Interval is Compact, it follows that $\closedint 0 1$ is compact.

From Metric Space is Hausdorff, it follows that $\R$ is a Hausdorff space.

From Subspace of Hausdorff Space is Hausdorff, it follows that $\closedint 0 1$ is a Hausdorff space.

By Continuous Functions on Compact Space form Banach Space, $C \closedint 0 1$ is a Banach space under the supremum norm $\norm {\,\cdot \,}_\infty$.

By the definition of complete metric space, it follows that every Banach space is a complete metric space.

Let $X$ consist of the $f \in C \closedint 0 1$ such that:
 * $\map f 0 = 0$
 * $\map f 1 = 1$
 * $\forall x \in \closedint 0 1: 0 \le \map f x \le 1$

Lemma 1
For every $f \in X$, define $\hat f: \closedint 0 1 \to \R$ as follows:
 * $\map {\hat f} x = \begin {cases}

\dfrac 3 4 \map f {3 x} & : 0 \le x \le \dfrac 1 3 \\ \dfrac 1 4 + \dfrac 1 2 \map f {2 - 3 x} & : \dfrac 1 3 \le x \le \dfrac 2 3 \\ \dfrac 1 4 + \dfrac 3 4 \map f {3 x - 2} & : \dfrac 2 3 \le x \le 1 \end {cases}$

Lemma 2
The Contraction Mapping Theorem assures existence of a unique $h \in X$ with $\hat h = h$.

We have that $h \in X \subset C \closedint 0 1$.

Thus, by definition, $h$ is a continuous real function.

It remains to be shown that $h$ is nowhere differentiable.

To do this, we establish the following lemma:

Lemma 3
Let $a \in \closedint 0 1$ be arbitrarily selected.

It is to be shown that $h$ is not differentiable at $a$.

This is to be achieved by constructing a sequence $\sequence {t_n}$ with elements in $\closedint 0 1$, which has the following limit:
 * $\ds \lim_{n \mathop \to \infty} t_n = a$

To this end, let $n \in \N$ be arbitrary.

Let $k$ be the unique largest element of $\set {1, 2, 3, 4, \ldots, 3^n}$ such that:
 * $\paren {k - 1} 3^{-n} \le a \le k 3^{-n}$

By the Triangle Inequality:

Next, let $t_n$ be either $\dfrac {k - 1} {3^n}$ or $\dfrac k {3^n}$, such that the following equation is satisfied:
 * $\size {\map h {t_n} - \map h a} = \max \set {\size {\map h {\dfrac {k - 1} {3^n} } - \map h a}, \size {\map h a - \map h {\dfrac k {3^n} } } }$

This implies:
 * $\forall n \in \N: t_n \ne a$

Furthermore:
 * $2 \size {\map h {t_n} - \map h a} \ge 2^{-n}$

and:
 * $\size {t_n - a} \le 3^{-n}$

Hence, for any $n$:
 * $t_n \in \closedint 0 1$

and also:
 * $\ds \lim_{n \mathop \to \infty} t_n = a$

The above inequalities imply that:
 * $\dfrac {\size {\map h {t_n} - \map h a} } {\size {t_n - a} } \ge \dfrac 1 2 \paren {\dfrac 3 2}^n$

But the absolute value of this expression diverges when $n$ tends to $\infty$.

Therefore $\ds \lim_{n \mathop \to \infty} \dfrac {\map h {t_n} - \map h a} {t_n - a}$ cannot exist.

From the definition of differentiability at a point, we conclude that $h$ is not differentiable at $a$.