Space of Somewhere Differentiable Continuous Functions on Closed Interval is Meager in Space of Continuous Functions on Closed Interval/Lemma 1
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Theorem
Let $I = \closedint a b$.
Let $\map \CC I$ be the set of continuous functions on $I$.
Let $\map \DD I$ 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:
- $\ds A = \bigcup_{\tuple {n, m} \in \N^2} A_{n, m}$
Then:
- $\map \DD I \subseteq A$
Proof
Let $f \in \map \DD I$.
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:
\(\ds \size {\frac {\map f t - \map f x} {t - x} }\) | \(<\) | \(\ds \size {\map {f'} x} - \fractpart {\size {\map {f'} x} } + 1\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \floor {\size {\map {f'} x} } + 1\) | Definition of Fractional Part | |||||||||||
\(\ds \) | \(=\) | \(\ds n\) |
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:
- $\map \DD I \subseteq A$
as required.
$\blacksquare$