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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!

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    $\begingroup$ 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. $\endgroup$ – T_M Nov 2 '18 at 19:08
  • $\begingroup$ @T_M seems plausible, I will try that. Thanks! $\endgroup$ – user3342072 Nov 2 '18 at 19:15
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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 $x\to 0^+.$

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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 $n^2-n$.

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