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

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).

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Hi Kris! If we assume WLOG that $g(0) = 0$, I think you can also show easily enough that if the Taylor series of $g$ is not the zero series (i.e. if $d^ng/dt^n(0)\neq 0$ for some $n$) then the result holds, by writing $g$ in the form $t^n h$ for some smooth function $h$ with $h(0) \neq 0$. I'm not sure about what to do in the remaining case where $g$ has infinite order of vanishing at $0$, though. – Jack Huizenga Sep 13 '11 at 20:11
Maybe it helps to remove $g$ by writing the inequality as $\int_0^t(t-s)f^\prime(s)^2\,ds\le\int_0^t(t-s)f(s)^2\,ds$. – George Lowther Sep 13 '11 at 23:49
$f^\prime/f$ cannot be either bounded or tending to infinity in an interval $(0,\epsilon)$, also implying that its derivatives vanish to all orders at 0. I think a counterexample can be constructed by letting $f^\prime(t)/f(t)$ constist of a sequence of very large spikes as $t\to0$. – George Lowther Sep 14 '11 at 0:13
If one defines $g(t)$ to be a 1-periodic function with $g(t)=t^2-t$ on $[0,1]$, then, even $g'(t)$ has discontinuities at integers, $g''(t)=2$. Also, $g(t)$ has its global maximum at integers, in particular, at $t=0$. And, of course, the inequality is true for non-constant $f(t)=\sin t$. I wonder how difficult is to make the $g(t)$ "smoother". – Wadim Zudilin Sep 14 '11 at 1:59

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