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 $a$.
== Historical Note ==
== Sources ==
* : Chapter $\text {A}.33$: Weierstrass ()
* : $8.10$ (for the core proof)
Category:Weierstrass's Theorem
Category:Real Analysis