TITLE: Creating an increasing function $g$ given a continuous function $f$, such that $|f(x)-f(y)| \leq |g(x)-g(y)|$
QUESTION [7 upvotes]: 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.
Thank you.

REPLY [12 votes]: Let $f:[a, b]\to\mathbb R$ be a continuous function which is not of bounded variation, see here for an example. Then such a $g$ cannot be found. To see this, assume that 
$$|f(x) - f(y)| \le | g(x) - g(y)|$$
then by assumption, for any $M$, there is a partition $a =y_0< y_1 < y_2 < \cdots y_{n-1} < y_n  = b$ so that 
$$\sum_{k=1}^n |f(y_{k}) - f(y_{k-1})| \ge M.$$
In particular, 
$$\sum_{k=1}^n |g(y_{k}) - g(y_{k-1})| \ge M.$$
But $g$ is inceasing, so $g(b) \ge M + g(a)$. But $M$ is arbitrary, so $g$ cannot be continuous.