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.
For fixed $x$, there exists $t_x$ such that $\Psi(x,t_x)=0$.
For fixed $t$ there exists unique $x_t$ such that $\Psi(x_t,t)\geq \Psi(x,t)$ for all $x\in M$.
There are proofs but using maximal time, I don't understand why such time exists by compactness.
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 $\Psi(t'',x'')=0$.