Suppose $|f^\prime| \leq |g^\prime|$ on R and $f(0) = g(0) = 0$. Must $|f| \leq |g|$ on R?
[If we add the conditions that $g^\prime$ is either non-negative or non-positive on the half-line R+ and likewise on R-, then the result is an easy consequence of the Mean Value Thm. It seems as if it should be true as stated, but I don't see how to prove it.]
Without the condition of non-changing sign of $g'$, $\lvert f\rvert$ can be larger than $\lvert g\rvert$. Consider for example
$$f(x) = \int_0^x \sin^2 t\,dt = \frac{1}{2} x - \frac{1}{4}\sin (2x)$$
and
$$g(x) = \int_0^x \sin t\,dt = 1 - \cos x.$$
Clearly $\lvert f'(x)\rvert = \sin^2 x \leqslant \lvert \sin x\rvert = \lvert g'(x)\rvert$ for all $x$, yet $g$ is bounded while $f$ is unbounded.
The assumption $$|f'(x)|\leq |g'(x)|\qquad\forall x\in{\mathbb R}\tag{1}$$ involves one free variable. One hopes to obtain a general statement involving two free variables. This is similar to the fact that $$|f'(x)|\leq M\qquad\forall x\in{\mathbb R}$$ implies $$|f(y)-f(x)|\leq M\>|y-x|\qquad\forall x, \ \forall y\ .$$ The best you can obtain from $(1)$ is the following: $${\rm Var}_J(f)\leq {\rm Var}_J(g)\qquad\forall\, J\ ,$$ whereby ${\rm Var}_J(f)$ denotes the total variation of $f$ on the interval $J=[x,y]$, $\>x<y$. A fortiori $$|f(y)-f(x)|\leq{\rm Var}_{[x,y]}(g)\ .$$
This question now seems to me cute but dumb.
Let $g(x) = 2x - x^2$. Let $f = g$ for $x \leq 1$ and $f = 2 - g$ for $x > 1$. Then $|f^\prime| = |g^\prime|, f(0) = g(0) = 0$, but $|f| > |g|$ for $x > 1$.