Given a continuous function $f$ over an interval, must there exist a continuous, increasing function $g$ such that for all $x,y$
$$|f(x)-f(y)| \leq |g(x)-g(y)|$$
I've tried assuming the opposite, that all $g$'s fail for some pair $x,y$ to reach a contradiction, since for every pair $x,y$, there is some $g$ that satisfies the inequality. Similarly for a finite number of pairs $x,y$ there is always a $g$ to satisfy. But I haven't gotten far with that nor used the continuity of $f$.
So I tried using the $\varepsilon$-$\delta$ definition on $f$ around a point since I might be able to connect $g$ to the values of $\varepsilon$, but I haven't found a way to do that either.
In the end, if a $g$ does exist (which I'm somewhat convinced is true), I'd like to find the "minimal" $g_0$ such that for all satisfying $g$'s and $x,y$
$$|g_0(x)-g_0(y)| \leq |g(x)-g(y)|$$
For example, $f=x^2$ and $g_0=x|x|$. Of course if there's a $g_0$, any $g_0+C$ is another one.