Showing a set is nowhere dense in $C([0,1])$

Question

Let $E_n$ be the set of all $f \in C\big([0,1]\big)$ for which there exists $x_0 \in [0,1]$ (depending on $f$) such that \begin{align*} \lvert\, f(x)-f(x_0)\rvert \leq n\lvert x-x_0\rvert, \end{align*} for all $x \in [0,1]$. Why is $E_n$ nowhere dense in $C\big([0,1]\big)$.

I have been able to show that $E_n$ is closed in $C\big([0,1]\big)$, but haven't been able to show why the interior is empty. Any advice?...

Answer 1

Let $g(x)=\lvert x\rvert$ in $[-1,1]$, and extend $g$ to be periodic in $\mathbb R$, with period $2$. Set $$ g_{k,\ell}(x)=\frac{1}{k} g(\ell x). $$ It is not hard to see that $$ g_{k,\ell}\in E_n \quad\text{iff}\quad n\ge \frac{\ell}{k}. $$ Fix now $n\in\mathbb N$. We shall show that $E_n$, which is a closed subset of $C[0,1]$, has empty interior.

Let $f\in E_n$ and $\varepsilon>0$. We shall show that $B(f,\varepsilon)\not\subset E_n$. Let $k,\ell\in\mathbb N$, so that $$ \frac{1}{\ell}<\varepsilon\quad\text{and}\quad \frac{k}{\ell}>2n. $$ Then $$ f+g_{k,\ell}\in B(f,\varepsilon) \quad\text{and}\quad f+g_{k,\ell}\not\in E_n. $$

Answer 2

I assume you mean $C[0,1]$ with sup-norm.

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Hint: Try to show that for each $n$ and arbitrary $\varepsilon>0$ there exists $g\notin E_n$ such that $\|g\|_\infty<\varepsilon$.

Try to show that $f\in E_n$ and $g\notin E_{2n}$ implies $f+g\notin E_n$.

Using these two facts you should be able that if $f\in E_n$ then in any ball $B(f,r)$ there is a function which does not belong to $E_n$.