Weierstrass's Theorem

From ProofWiki
Jump to navigation Jump to search

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.


Sources


This page may be the result of a refactoring operation.
As such, the following source works, along with any process flow, will need to be reviewed.
When this has been completed, the citation of that source work (if it is appropriate that it stay on this page) is to be placed above this message, into the usual chronological ordering.
In particular: are the lemmata given?
If you have access to any of these works, then you are invited to review this list, and make any necessary corrections.
To discuss this page in more detail, feel free to use the talk page.
When this work has been completed, you may remove this instance of {{SourceReview}} from the code.


Retrieved from "https://proofwiki.org/w/index.php?title=Weierstrass%27s_Theorem&oldid=612283"