# Show that set of functions with Lipschitz constant $\leq n$ has no interior in $(C[0,1],\Vert \cdot \Vert_\infty)$

Let $$f\in C[0,1]$$ be a Lipschitz function with Lipschitz constant $$n,$$ i.e. such that $$\forall x,y \in [0,1]:|f(x)-f(y)|\leq n|x-y|$$ for a fixed $$n \in \mathbb N$$.

Here is a problem. I would like to construct a function $$g\in C[0,1]$$ with the slope higher than $$n$$ at some point and $$\Vert f-g \Vert_\infty < \epsilon$$ for any $$\epsilon>0.$$

My work: Pick $$t_0 \in (0,1)$$ and $$\delta>0$$ sufficiently small and $$h>0$$ such that $$\frac{h}{t_0}>n$$. $$g(x)=\begin{cases} \frac{h}{t_0}x+f(0) & x \in [0,t_0)\\ \frac{f(t_0+\delta)-f(0)-h}{\delta}(x-t_0)+f(0)+h& x\in [t_0,t_0+\delta)\\ f(x) & x\in [t_0+\delta,1] \end{cases}$$

The shape of $$g$$ is like $$\wedge$$ which lies above $$f$$ on the intervall $$[0,t_0+\delta)$$ and connects $$f(0)$$ with $$f(t_0+\delta)$$. Desired slope property if satisfied on first intervall. Now $$\Vert f-g \Vert_\infty=\Vert (f-g)|_{[0,t_0+\delta)} \Vert_\infty$$ Question: Are small $$t_0$$ and $$\delta$$ enough to guarantee $$\Vert f-g \Vert_\infty < \epsilon$$ ?
Also I would like to see any other your constructions!

• Maybe more clear to just take $f + \epsilon$ for a while and then switch to $f - \epsilon$. Join these up by a line which is nearly vertical. – T_M Nov 2 '18 at 19:08
• @T_M seems plausible, I will try that. Thanks! – user3342072 Nov 2 '18 at 19:15

Let $$g_\epsilon(x) = f(x)+\epsilon\sqrt x.$$ Then $$\|f-g_\epsilon\| \le \epsilon,$$ while
$$\left |\frac{g_\epsilon(x)-g_\epsilon(x)}{x-0}\right | \to \infty$$
as $$x\to 0^+.$$
The spirit of your idea is kind of ok, but you are not taking into account how $$f$$ behaves in that small interval close to $$0$$.
For a more brutal example, you may take an oscillating function, like $$\sin x$$. We can make it oscillate faster by multiplying $$x$$ by a constant, and you shrink it to a small band; say, $$h(x)=\tfrac1n\sin n^3 x$$. Then you take $$g(x)=f(x)+h(x).$$By construction, $$\|g-f\|_\infty=\|h\|_\infty=\varepsilon$$. And whenever $$x=\tfrac{(2k+1)\pi}{2n^3}$$, $$k\in\mathbb N$$, we'll have $$|h'(x)|=n^2$$, so the slope of $$g$$ will be at least $$n^2-n$$.