Space of Somewhere Differentiable Continuous Functions on Closed Interval is Meager in Space of Continuous Functions on Closed Interval/Lemma 2/Lemma 2.1

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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:

the sequence $\sequence {\map {f_{i_k} } {x_{i_k} } }_{k \mathop \in \N}$ converges to $\map f x$.

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, $f \in A_{n, m}$.

$\blacksquare$

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