I need to understand a proof in the book "Lecture notes on topology and geometry" by Singer and Thorpe, from 2.5 Applications
Theorem 1: There exists a continuous real-valued function $f\in C([0,1])$ such that $f$ has a derivative in no point of $[0,1]$.
Proof: Define for all $n\in \mathbb{N}$ the set $$ C_n =\left\{f\in \mathcal{C}([0,1]): \Big|\dfrac{f(t+h)-f(t)}{h}\Big|\leqslant n,\text{for some $t$ and all $h$ with $t+h$ $\in [0,1]$} \right\} $$
Show that $C_n$ is nowhere dense for each $n$, then since $C([0,1])$ is a complete metric space we have $$ \mathcal{C}([0,1])\neq \bigcup_{n\in \mathbb{N}} C_n $$ it must exists a function $f\in \mathcal{C}([0,1])$ such that $f\notin C_n$ for all $n$ and this means it has no derivative.
- I understand why this last remark is true because if $f\notin C_n$ for all $n$ then $$ \Big|\dfrac{f(t+h)-f(t)}{h}\Big|> n $$ and then
$$\limsup_{h\rightarrow0}\Big|\dfrac{f(t+h)-f(t)}{h}\Big|=\infty$$ for all $t\in [0,1]$ thus the derivative does not exists.
- I understand how to show that $\overline C_n=C_n$ because is well explained in the book, I must considere a sequence $\{f_k\}$ in $C_n$ which converges to $f$ and show that $f\in C_n$.
I have a problem showing that $C_n$ is nowhere dense, the book says that given any $\epsilon >0$ and any $g\in C_n$ there exists $f \in C([0,1])$ such that $||f-g||<\epsilon$ and $f\notin C_n$, and suggest i should do that by approximating $g$ (using the fact that $g$ is uniformly continuous) by a piecewise linear function $g_1$ and then use a "sawtooth" function for each linear piece of $g_1$ and then "to patch". Also note that the norm is $||f||=sup|f(x)|$ for all $x\in [0,1]$
I ask for an explanation on this last remark, a good reference to understand it or any other help that you can provide.