A differential inequality needed to prove a theorem about odd-dimensional souls - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-18T15:59:29Z http://mathoverflow.net/feeds/question/75341 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/75341/a-differential-inequality-needed-to-prove-a-theorem-about-odd-dimensional-souls A differential inequality needed to prove a theorem about odd-dimensional souls Kris Tapp 2011-09-13T18:53:22Z 2011-09-14T10:05:23Z <p>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:</p> <p>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.</p> <p>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). </p> http://mathoverflow.net/questions/75341/a-differential-inequality-needed-to-prove-a-theorem-about-odd-dimensional-souls/75384#75384 Answer by Ilya Bogdanov for A differential inequality needed to prove a theorem about odd-dimensional souls Ilya Bogdanov 2011-09-14T10:05:23Z 2011-09-14T10:05:23Z <p>We may assume $g(0)=g'(0)=0$.</p> <p>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 $g(t)=h(t)=f(t)=0$.</p>