Prove that $f(x) > g(x)$ where both functions are convex and have the same value and slope at $0$ Let $f: [-a,a] \to \mathbb{R}$ and $g: [-a,a] \to \mathbb{R}$ be two non-negative, convex and smooth functions. We further know $f(0) = g(0)=0$ and $f'(0) = g'(0)=0$. I'd like to show $$f(x) \ge g(x).$$ To this end, I think it is sufficient to show $f''(x) \ge g''(x)$ for $\forall x \in [-a,a]$. Am I right? If yes, is this a well-known result in analysis?  


Indeed, in your setting, $f''\geqslant g''$ implies $f\geqslant g$. 

To show this, note that, since $f(0)=g(0)=0$, for every nonnegative $x$,  $$f(x)=\int_0^xf'(t)\,\mathrm dt,\qquad g(x)=\int_0^xg'(t)\,\mathrm dt.$$ Using the hypothesis that $f'(0)=g'(0)=0$, an integration by parts of these identities yields $$f(x)=\int_0^x(x-t)f''(t)\,\mathrm dt,\qquad g(x)=\int_0^x(x-t)g''(t)\,\mathrm dt.$$ Since $x-t\geqslant0$ for every $t$ in $(0,x)$, this shows that $f(x)\geqslant g(x)$. Surely you can adapt the argument to the case when $x\leqslant0$... :-)
