# 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

$X$ is a complete metric space under $\norm {\,\cdot \,}_\infty$.

$\Box$

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

$\hat \cdot: X \to X$ is a contraction mapping.

Furthermore, we have the following inequality:

$\forall f, g \in X: \norm {\hat f - \hat g}_\infty \le \dfrac 3 4 \norm {f - g}_\infty$

$\Box$

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

For every $n \in \N$ and $k \in \set {1, 2, 3, 4, \ldots, 3^n}$, the following inequality holds:

$\size {\map h {\dfrac {k - 1} {3^n} } - \map h {\dfrac k {3^n} } } \ge 2^{-n}$

$\Box$

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:

 $\ds$  $\ds \size {\map h {\frac {k - 1} {3^n} } - \map h a} + \size {\map h a - \map h {\frac k {3^n} } }$ $\ds$ $\ge$ $\ds \size {\map h {\frac {k - 1} {3^n} } - \map h {\frac k {3^n} } }$ $\ds$ $\ge$ $\ds 2^{-n}$

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 $a$.

$\blacksquare$

## Source of Name

This entry was named for Karl Weierstrass.

## Historical Note

The construction of a real function which is continuous, but nowhere differentiable, was first demonstrated by Karl Weierstrass.

The demonstration that such functions exist came as a profound shock to the mathematical community.