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

## 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:

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

Then:

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

## Proof

### Lemma 2.1

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

$\Box$

Fix $\tuple {n, \, m} \in \N^2$.

By Lemma 2.1, $A_{n, \, m}$ is closed in $\struct {\map \CC I, d}$.

Therefore by the second definition of nowhere dense, we aim to show that:

$A_{n, \, m}$ contains no non-empty open sets of $\struct {\map \CC I, d}$.
the open sets of $\struct {\map \CC I, d}$ are precisely the unions of the open balls of $\struct {\map \CC I, d}$.

Therefore, it suffices to show that:

$A_{n, \, m}$ contains no non-empty open balls of $\struct {\map \CC I, d}$.

Let $f \in \map \CC I$ and $\epsilon > 0$.

Consider the open ball $\map {B_\epsilon} f$.

We aim to show that:

there exists $g \in \map {B_\epsilon} f$ such that $g \not\in A_{n, \, m}$.

This ensures that:

$\map {B_\epsilon} f$ is not contained in $A_{n, \, m}$.
there exists a continuous piecewise linear function $p: I \to \R$ such that $\map d {f, p} < \dfrac \epsilon 2$.
$p$ is differentiable on $I$ except at finitely many points $\set {c_1, c_2, \ldots, c_n}$.

Further:

the derivative of $p$ takes finitely many values.

Therefore:

there exists a real number $M > 0$ such that $\size {\map {p'} x} < M$

for each $x$ such that $p$ is differentiable.

Let $k \in \R$ be such that:

$k > \dfrac {2 \paren {M + n} } \epsilon$

Let $\phi: I \to \R$ be a piecewise linear function such that:

$\map \phi x = \begin{cases}k x - n & \frac n k \le x \le \frac {n + 1} k, \, n \text { is an even integer,} \\ -k x + \paren {1 + n} & \frac n k \le x \le \frac {n + 1} k, \, n \text { is an odd integer,}\end{cases}$

then:

$\size {\map \phi x} \le 1$ and $\size {\map {\phi'} x} = k$ for each $x$ for which $p$ and $\phi$ are differentiable.

Now, define $g: I \to \R$ to have:

$\map g x = \map p x + \dfrac \epsilon 2 \map \phi x$

for each $x \in I$.

We then have:

 $\ds \norm {g - p}_\infty$ $=$ $\ds \frac \epsilon 2 \norm \phi_\infty$ $\ds$ $=$ $\ds \frac \epsilon 2 \sup_{x \mathop \in I} \size {\map \phi x}$ Definition of Supremum Norm $\ds$ $\le$ $\ds \frac \epsilon 2$ since $\size {\map \phi x} \le 1$

We therefore have:

 $\ds \norm {g - f}_\infty$ $\le$ $\ds \norm {g - p}_\infty + \norm {p - f}_\infty$ Supremum Norm is Norm $\ds$ $<$ $\ds \epsilon$

So $g \in \map {B_\epsilon} f$.

We now show that $g \not \in A_{n, \, m}$.

Let $x \in I$ be a point at which $\phi$ and $p$ are both differentiable.

Then $g$ is differentiable at $x$, with:

$\size {\map {g'} x} = \size {\map {p'} x + \dfrac \epsilon 2 \map {\phi'} x}$

Since $\size {\map {p'} x} < M$ and $k > \dfrac {2 \paren {M + n} } \epsilon$, we have:

 $\ds \size {\map {p'} x + \frac \epsilon 2 \map {\phi'} x}$ $\ge$ $\ds \size {\size {\map {p'} x} - \frac \epsilon 2 \size {\map {\phi'} x} }$ $\ds$ $=$ $\ds \size {\size {\map {p'} x} - \frac {k \epsilon} 2}$ $\ds$ $>$ $\ds \size {\size M - \size {M + n} }$ $\ds$ $=$ $\ds n$

From the definition of the derivative, there exists some real $\delta > 0$ such that:

$\size {\dfrac {\map g t - \map g x} {t - x} } > n$

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

In particular, we cannot have:

$\size {\dfrac {\map g t - \map g x} {t - x} } \le n$

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

Now, let $x \in I$ be a point for which one of $\phi$ or $p$ are not differentiable.

Recall that there are only finitely such $x$.

Since the set:

$\set {t \in I: 0 < \size {t - x} < \dfrac 1 m}$

is not finite, it contains a point $t^*$ such that $\phi$ or $p$ are differentiable, for which we have:

$\size {\dfrac {\map g {t^*} - \map g x} {t^* - x} } > n$

So, for any $x \in I$ there exists $t$ with $0 < \size {t - x} < \dfrac 1 m$ such that:

$\size {\dfrac {\map g t - \map g x} {t - x} } > n$

We conclude that $g \not \in A_{n, m}$.

$\blacksquare$