I need a solution to this problem (which is really a calculus problem) in order to prove a rigidity result for open nonnegatively curved manifolds with odd-dimensional souls:


Suppose that $f,g:\mathbf{R}\rightarrow \mathbf{R}$ are smooth functions. Assume that $f(0)=0$ and that $g(t)$ has a global maximum at $t=0$. Assume for all $t\in\mathbf{R}$ that:
$$f'(t)^2 \leq f(t)^2 + g''(t).$$
Prove that $f$ and $g$ must both be constant functions.


By the way, this is simple to prove if $g$ is analytic. In this case, the fact that $g$ has a global maximum implies that $g''(t)\leq 0$ on an interval about $t=0$. So you have $f'(t)^2\leq f(t)^2$ which implies that $f(t)=0$ on this interval (establishing the result on an interval is enough). 



A differential inequality needed to prove a theorem about odd-dimensional souls
We may assume $g(0)=g'(0)=0$.


For $t\in(0,1)$, we have 
$$
 \int\_0^t (f(s)^2+g''(s))\,ds \geq \int\_0^t f'(s)^2\,ds 
 \geq \left(\int\_0^t f'(s)\,ds\right)^2 \left(\int\_0^t 1^2\,ds\right)^{-1}
 =f(t)^2/t\geq f(t)^2,
$$
hence $g'(t)\geq f(t)^2-\int\_0^t f(s)^2\,ds$. Let $h(t)=\int\_0^t f(s)^2\,ds$; we have $g'(t)\geq h'(t)-h(t)$. Since $h$ is monotonous we get $g(t)\geq h(t)-\int\_0^t h(s)\,ds\geq h(t)-th(t)\geq 0$. So, on our interval we have $g(t)=h(t)=f(t)=0$.


