# Modifying critical point of function on manifold ODE theory

Assume $$M$$ is compact Riemannian manifold and $$g(t)$$ is a time dependent metric. Assume $$\Psi:M\times [0,C)\rightarrow \mathbb{R}$$ is smooth and $$C<\infty$$. Suppose we have $$(\partial_t\Psi -\Delta_{g(t)} \Psi)\geq 0$$. If $$\Psi\geq 0$$ on $$M\times 0$$, then, $$\Psi \geq 0$$ everywhere.

Proof is as follows:

It suffices to assume that case $$\Psi>0$$. Now, assume for contradiction $$\Psi<0$$ somewhere. Since $$M$$ is compact, we can find $$(x'',t'')$$ such that $$\Psi(x'',t'')=0$$ and $$\Psi(x,t)\geq 0$$ for all $$x\in M$$ and all $$t\in [0,t'']$$. Then on this point, $$\partial_t \Psi$$ $$\leq 0$$ and $$\Delta \Psi \geq 0$$.

I am confsed as to why compactness implies the existence of $$(x'',t'')$$ as above and why that implies $$\partial_t \Psi\leq 0$$.

Note.

1. For fixed $$x$$, there exists $$t_x$$ such that $$\Psi(x,t_x)=0$$.

2. For fixed $$t$$ there exists unique $$x_t$$ such that $$\Psi(x_t,t)\geq \Psi(x,t)$$ for all $$x\in M$$.

3. There are proofs but using maximal time, I don't understand why such time exists by compactness.

• I think the main idea is to use $1.$ in your Note (which I think follows by the IVT), then compactness, to see that there must be a point $x''$, with $t_{x''}$ smallest possible, for which $\Psi(x'',t_{x''}) = 0$. At this point $\Psi$ achieves a local minimum with respect to the first coordinate, i.e. keeping $t_{x''}$ fixed and varying only the first coordinate (so that the hessian is positive definite) and is decreasing with respect to time. Commented Jul 12, 2023 at 7:47
• @porridgemathematics Hi. Yes, note 1 follows from IVT. The issue I have is using compactness to prove existence of smallest possible $t_{x''}$. Could you elaborate on that in an anser? Thanks
– user1040289
Commented Jul 12, 2023 at 14:37
• I might type up an answer later, but I think the idea is to show that $M \rightarrow \mathbb{R}_{\geq 0} : x \rightarrow t_{x}$ is continuous using compactness, and then use compactness again to show that this function achieves a minimum on $M$. Commented Jul 12, 2023 at 15:17
• I do not see why your points 1. and 2. are true, but I also do not think they are relevant to the statement you want to show. Commented Jul 12, 2023 at 15:58
• @s.harp For fixed $x$, $\Psi(x,.)$ is continuous with connected domain. For fixed t, $\Psi(.,t)$ is continuous with compact domain.
– user1040289
Commented Jul 12, 2023 at 21:38

Let $$X:=\{t\in [0,C)\mid \exists x: \Psi(t,x)<0\}$$, which is an open set and so $$t'':=\inf (X)$$ is not contained in $$X$$, in particular you have $$\Psi(t, x)\geq 0$$ for all $$x\in M$$ and $$t\in[0,t'']$$.
Now let $$(t_n, x_n)$$ be so that $$t_n\to t''$$ and $$\Psi(t_n, x_n)<0$$ for all $$n$$. By compactness of $$M$$ the sequence $$x_n$$ can be assumed to be convergent to some $$x''$$ and then $$\Psi(t'',x'')\leq 0$$ by continuity of $$\Psi$$. But we already have $$\Psi(t'',x'')\geq0$$, so $$\Psi(t'',x'')=0$$.