If $f\leq g$ and $f,g$ are positive, increasing convex functions, then $f'\leq g'$?

Question

Suppose $f(x)\leq g(x)$ for all $x\in\mathbb R$ and $f,g$ are positive, increasing convex functions.

Question: Can I conclude that $f'(x)\leq g'(x)$?

Answer 1

No. Let $g(x)=\exp(x)$, let $f(x)=g(x)/2$ for $x<0$ and let $f(x)=1/2+xs$ for an $s$ chosen so that the graph of $f$ is tangent to that of $g$. (Think geometrically: if $s=1/2$, the graph of $f$ does not intersect that of $g$, if $s=100000$ it does, at two places, and for a unique intermediate value of $s$, it grazes that of $g$. As you change $s$ the graph of $f$ swings upward until it bumps the graph of $g$.) Let $x_0$ be the point where $f(x_0) = g(x_0)$ and $f'(x_0) = g'(x_0)$, which can be obtained (in principle) by solving obvious nonlinear equations. For values of $x\in[0,x_0)$, we have $f'(x)>g'(x)$.

Added next day: Here is a simpler example using the same principle. Let $f(x)=\exp(1+x)$ and let $g(x)=\exp(\exp(x))$. Both are strictly convex, and $f\le g$ pointwise because $1+x\le \exp(x)$ pointwise. But $f'(0) = g'(0)$, so for $x<0$, $f'(x)<g'(x)$.