Space of Somewhere Differentiable Continuous Functions on Closed Interval is Meager in Space of Continuous Functions on Closed Interval/Lemma 1

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:


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

and:


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

Then:
 * $\map {\mathcal D} I \subseteq A$

Proof
Let $f \in \map {\mathcal D} I$.

Then, $f$ is differentiable at some $x \in I$.

Let:


 * $n = \floor {\size {\map {f'} x} } + 1$

where $\floor \cdot$ is the floor function.

Then:


 * $\size {\map {f'} x} < n$

From the definition of the derivative, there exists $\delta > 0$ such that for all $t$ with $0 < \size {t - x} < \delta$, we have:


 * $\ds \size {\frac {\map f t - \map f x} {t - x} - \map {f'} x} < 1 - \fractpart {\size {\map {f'} x} }$

From the Reverse Triangle Inequality, we then have:


 * $\ds \size {\size {\frac {\map f t - \map f x} {t - x} } - \size {\map {f'} x} } < 1 - \fractpart {\size {\map {f'} x} }$

and so:

for all $t$ with $0 < \size {t - x} < \delta$.

Pick $m \in \N$ such that $\frac 1 m < \delta$.

We then have:


 * $\ds \size {\frac {\map f t - \map f x} {t - x} } \le n$

for $t$ with $0 < \size {t - x} < \frac 1 m$.

That is:


 * $f \in A_{n, \, m} \subseteq A$

so:


 * $\map {\mathcal D} I \subseteq A$

as required.