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!
Best Answer
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 $x\to 0^+.$