Space of Somewhere Differentiable Continuous Functions on Closed Interval is Meager in Space of Continuous Functions on Closed Interval

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $I = \closedint a b$.

Let $\map \CC I$ be the set of continuous functions on $I$.

Let $\map {\mathcal D} I$ be the set of continuous functions on $I$ that are differentiable at a point.

Let $d$ be the metric induced by the supremum norm.


Then $\map {\mathcal D} I$ is meager in $\struct {\map \CC I, d}$.


Corollary

there exists a function $f \in \map \CC I$ that is not differentiable anywhere.


Proof

Let:

$\ds A_{n, \, m} = \set {f \in \map \CC I: \text {there exists } x \in I \text { such that } \size {\frac {\map f t - \map f x} {t - x} } \le n \text { for all } t \text { with } 0 < \size {t - x} < \frac 1 m}$

and:

$\displaystyle A = \bigcup_{\tuple {n, \, m} \in \N^2} A_{n, \, m}$

Lemma 1

$\map \DD I \subseteq A$

$\Box$


Lemma 2

for each $\tuple {n, \, m} \in \N^2$, $A_{n, \, m}$ is nowhere dense in $\struct {\map \CC I, d}$.

$\Box$


From Lemma 2:

$A$ is the countable union of nowhere dense sets.

So, by the definition of a meager space:

$A$ is meager in $\struct {\map \CC I, d}$.

By Subset of Meager Set is Meager Set and Lemma 1, we have:

$\map {\mathcal D} I$ is meager in $\struct {\map \CC I, d}$.

$\blacksquare$