A differential inequality needed to prove a theorem about odd-dimensional souls

2011-09-13T18:53:22Z

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).

Answer by Ilya Bogdanov for A differential inequality needed to prove a theorem about odd-dimensional souls

2011-09-14T10:05:23Z

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 $g(t)=h(t)=f(t)=0$.

A differential inequality needed to prove a theorem about odd-dimensional souls - MathOverflow

A differential inequality needed to prove a theorem about odd-dimensional souls

Answer by Ilya Bogdanov for A differential inequality needed to prove a theorem about odd-dimensional souls