We assume that $max_{\mathbb{R}^n \times[0,T]} f(x,t) = f(x_0,t_0)\quad\text{for some }(x_0,t_0)\in\mathbb{R}^n \times[0,T]$
Then, my book says that If $t_0>0$, then, by the usual maximum principle, we have $D_xf(x_0,t_0)=0\quad f_t(x_0,t_0)\geq 0\quad \Delta_x f(x_0,t_0)\leq 0$.
Question: Can you tell me what the usual maximum principle is used here? I think the differential of $f$ with respect to time at $(x_0,t_0)$ is equal to $0$, but is it really? Perhaps, $\Delta_x f(x_0,t_0)\leq 0$ because of $f$ having a local max at $(x_0,t_0)$. Is it right? Please help me. Of course, there are several assumption for $f$, but I do not think that is important here. (Just assume that $f$ is smooth enough.)
I will assume that $f$ is sufficiently regular for everything that follows to be well defined.
It is clear that $D_x f(x_0,t_0) =0$, since $x_0$ maximizes $\tilde f_{t_0} : x\in\mathbb R^n \mapsto f(\cdot,t_0)$. Similarly, if $t$ is an interior point of $(0,T]$, it follows that $f_t(x_0,t_0)=0$, otherwise if $t_0=T$, then we have by definition $f(x_0,t) - f(x_0,t_0) \le 0 $ for all $t\in [T-\varepsilon,T]$ for some $\varepsilon >0$. Hence $f_t(x_0,t_0) := \lim_{h\to 0^+} \frac{f(x_0, T-h) - f(x_0,T)}{-h}\ge 0 $.
Now we will show that $\Delta_x f(x_0,t_0)\le 0$ : we have just seen that $D_x f(x_0,t_0) = 0$, hence by the second derivative criterion, we have $$\frac{\partial^2 f}{\partial x_1^2}(x_0,t_0) \le 0,\ \ldots,\ \frac{\partial^2 f}{\partial x_n^2}(x_0,t_0)\le 0 $$ Which directly implies $\Delta_x f(x_0,t_0) := \sum_{i=1}^n \frac{\partial^2 f}{\partial x_i^2}(x_0,t_0) \le 0$.
More generally, the maximum principle tells you that if $L$ is an elliptic operator and $f_t + Lf \le 0$ in $U\times (0,T]$ for some open, connected and bounded $U\subseteq\mathbb R^n$, then $f$ attains its maximum on $\partial U\times (0,T]$. By considering the elliptic operator $L\equiv -\Delta$ and small neighborhoods of $(x_0,t_0)$ together with the condition $f_t(x_0,t_0)\ge 0$, it would be possible to argue by contradiction to conclude that either $f$ is locally constant or $Lf(x_0,t_0)\ge 0$, but here the direct computation is more straightforward in my opinion.