Space of Somewhere Differentiable Continuous Functions on Closed Interval is Meager in Space of Continuous Functions on Closed Interval/Lemma 2/Lemma 2.1
Theorem
Let $I = \closedint a b$.
Let $\map \CC I$ be the set of continuous functions on $I$.
Let $d$ be the metric induced by the supremum norm.
Let:
- $\ds A_{n, m} = \set {f \in \map \CC I: \exists x \in I: \forall t \in \R: 0 < \size {t - x} < \frac 1 m \implies \size {\frac {\map f t - \map f x} {t - x} \le n} }$
Then:
- for each $\tuple {n, m} \in \N^2$, $A_{n, m}$ is closed in $\tuple {\map \CC I, d}$.
Proof
Fix $\tuple {n, m} \in \N^2$.
From Space of Continuous on Closed Interval Real-Valued Functions with Supremum Norm forms Banach Space:
- $\tuple {\map \CC I, d}$ is complete.
Hence, from Subspace of Complete Metric Space is Closed iff Complete:
- $A_{n, m}$ is closed if and only if $\tuple {A_{n, m}, d}$ is complete.
Let $\sequence {f_i}_{i \mathop \in \N}$ be a Cauchy sequence in $\tuple {A_{n, m}, d}$.
Since $\tuple {\map \CC I, d}$ is complete, $\sequence {f_i}_{i \mathop \in \N}$ converges some $f \in \map \CC I$.
We aim to show that $f \in A_{n, m}$.
Since $f_i \in A_{n, m}$ for each $i \in \N$, there exists $x_i \in I$ such that:
- $\ds \size {\frac {\map {f_i} t - \map {f_i} {x_i} } {t - x_i} } \le n$ for each $t$ with $0 < \size {t - x_i} < \dfrac 1 m$.
Note that since $I$ is bounded, $\sequence {x_i}_{i \mathop \in \N}$ is bounded.
Therefore, by the Bolzano-Weierstrass Theorem:
- there exists a convergent subsequence of $\sequence {x_i}_{i \mathop \in \N}$, $\sequence {x_{i_k} }_{k \mathop \in \N}$.
Let:
- $\ds x = \lim_{k \mathop \to \infty} x_{i_k}$
Note that we also have:
- $\ds f = \lim_{k \mathop \to \infty} f_{i_k}$
From Subset of Metric Space contains Limits of Sequences iff Closed, since $I$ is closed, $x \in I$.
From Necessary Condition for Uniform Convergence:
We therefore have:
\(\ds \size {\frac {\map f t - \map f x} {t - x} }\) | \(=\) | \(\ds \lim_{k \mathop \to \infty} \size {\frac {\map {f_{i_k} } t - \map {f_{i_k} } {x_{i_k} } } {t - x_{i_k} } }\) | ||||||||||||
\(\ds \) | \(\le\) | \(\ds n\) |
for all $t$ with $0 < \size {t - x} < \dfrac 1 m$.
That is, $f \in A_{n, m}$.
$\blacksquare$