What is the bound on a function $f(x)$ given that $|f'(x)| \leq |g'(x)|$? Suppose you know that $|f'(x)| \leq |g'(x)|$ for some interval on the real line. Is it possible to infer a bound on $|f(x)|$ in terms of $g(x)$?
Maybe you can use the Mean Value Theorem to do so? If so, what additional assumptions do you need about $f$ and $g$?

If $f'$ is integrable, you can say
$$ \lvert f(x)-f(y) \rvert = \left\lvert \int_y^x f' \right\rvert \leq \int_y^x \lvert f' \rvert \leq \int_y^x \lvert g' \rvert = V_y^x(g) , $$
the total variation of $g$. You can see this in play in a pair with $g'$ taking on positive and negative values, $f' = \lvert g' \rvert$. Then $f$ can get quite large while $g$ wanders near $0$: $g'(x) = \cos{x}$, for example.
In the case that $f'$ is not necessarily integrable, the mean-value inequality implies the weaker bound
$$ \lvert f(x)-f(y) \rvert \leq \lvert x-y \rvert \sup_{[y,x] \cup [x,y]} \lvert f' \rvert \leq \lvert x-y \rvert \sup_{[y,x] \cup [x,y]} \lvert g' \rvert  $$
(the intervals being written slightly facetiously just because $x$ may be to either side of $y$).
I doubt much more can be said, since the values of $g$ do not control the values of $g'$ on whole intervals.
