Here is the statement of the scalar weak maximum principle on Riemanninan manifolds with boundary.
Theorem (Scalar weak maximum principle):
Let $M$ be a compact manifold with boundary and $g_t$ a smooth family of Riemannian metrics on $M$ for $t\in [0,T]$. Suppose that for $u\in C^2(M\times[0,T])$ and $U\in C^1([0,T])$ we have
\begin{align}
\partial_t u &\leq \Delta_{g_t} u + f(u,t)\,, && \text{on } M\times[0,T]\tag{1}\\
u &\leq U &&\text{on }(M\times\{0\})\cup (\partial M\times[0,T])\tag{2}\\
\partial_t U(t) &\geq f(U(t),t) && t\in [0,T]\tag{3}
\end{align}
where $f:\mathbb{R}\times[0,T]\longrightarrow \mathbb{R}$ is Lipschitz in the first variable and continuous in the second. Then, $u \leq U$ on $M\times[0,T]$.
The standard proof can be found in the book The Ricci Flow: An Introduction by Chow and Knopf. Here is an alternative proof I found in lecture notes for a course by Richard Bamler (found here). His notes don't go into details, so I have tried filling them out myself.
proof: First assume that we have strict inequalities in (2) and (3). Choose $t^*\in [0,T]$ maximal so that $u(\cdot, t) \leq U(t)$ for all $t\in [0, t^*]$. Since (2) is assumed to be a strict inequality, we have that $t^*>0$. Suppose that $t^*<T$. Then there exists $x^*\in M\setminus \partial M$ such that $u(x^*,t^*) = U(t^*)$. Since $x^*$ is a local maximum of $u(\cdot, t^*)$, we have $\nabla u(x^*,t^*) = 0$ and $\Delta u(x^*,t^*)\leq 0$. Moreover $\partial_t u(x^*,t^*)\geq \partial_tU(t^*)$. Then combining (1) – (3) and evaluating at $(x^*,t^*)$ gives
$$
\partial_t U(t^*) \leq \Delta u(x^*,t^*) + f(u(x^*,t^*),t^*) \leq f(U(t^*),t^*) < \partial_tU(t^*)
$$
which is a contradiction.
Define $U_\epsilon(t) = U(t) + \epsilon t + \epsilon^2$. Then $u(x, t) < U_\epsilon(t)$ for $(x,t)\in \partial_{\text{par}}(M\times[0,T])$. Since $f$ is Lipschitz, we have for some $C>0$
\begin{align*}
f(U_\epsilon(t), t) &\leq f(U(t),t) + C(\epsilon t + \epsilon^2)\\
&\leq \partial_tU(t) + C(\epsilon t+\epsilon^2)\\
&= \partial_t U_\epsilon(t) + C(\epsilon t+\epsilon^2) -\epsilon
\end{align*}
For sufficiently small $\epsilon$ choose $\tau\in (0,T]$ independent of $\epsilon$ such that $\tau < 1/C-\epsilon $. Then whenever $t\leq\tau$ we have
$$
C(\epsilon t+\epsilon^2)-\epsilon \leq C(\epsilon \tau+\epsilon^2)-\epsilon < C(\epsilon (1/C-\epsilon)+\epsilon^2)-\epsilon = 0\,.
$$
Thus, for $t\in [0,\tau]$ we have $f(U_\epsilon(t),t)<\partial_tU_\epsilon(t)$. By the previous case we therefore have that $u \leq U_\epsilon$ on $M\times [0,\tau]$. Letting $\epsilon \rightarrow 0$, we have $u \leq U$ on $M\times [0,\tau]$. Clearly the subset of $u-U\leq 0$ is closed in $M\times [0,T]$, thus we may assume $\tau$ is chosen maximally. If $\tau < T$ then we can repeat the above procedure with the time shift $t\mapsto t-\tau$ and $T\mapsto T-\tau$. Thus $\tau = T$.
Question: Is this proof correct?
My issue is the inequality $\tau < 1/C$. This seems to put a constraint on the maximum time $T$. For example, if $1/C<T$, then it wouldn't be possible to have $\tau = T$.
Best Answer
Since $[0,T]$ is compact, it can be covered by a finite number of intervals $[0,\tau_{1}],[\tau_{1},\tau_{2}],\ldots, [\tau_{k-1},\tau_{k}]$, each one with length less than $1/C$ (say $1/2C$). What was done for the first interval, can be repeated in each interval as is mentioned in the proof, by the time shifting $t \mapsto t-\tau$. For example, one that you have the proof in the interval $[0,\tau_{1}]$, you work in the interval $[\tau_{1},\tau_{2}]$ and your initial conditions will be the initial conditions at time $t-\tau_{1}=0$, that is, for $t=\tau_{1}$. Since these new initial conditions have the same properties has the initial conditions when you worked in the first interval, the proof would work in this case.