In Brezis' book, look at the maximum principle for the heat equation in $\Omega \times (0,T)$ where $\Omega$ is an open bounded subset of $\mathbb{R}^{n}$: $$u_{t}-\Delta u=0$$ $$ (x,t) \in \Omega \times (0,T)$$ $$u=0 \in \partial \Omega.$$ He proves that: $$ \max _{\overline{\Omega} \times[0, T]} u=\max _P u $$ where $P=(\overline{\Omega} \times\{0\}) \cup(\Gamma \times[0, T])$ is called the parabolic boundary of the cylinder $\Omega \times(0, T)$.
He sets $$v(t,x)=u(x,t)+\epsilon \lvert x\rvert^{2}$$ and he supposes that there is $(a,b)\notin P$ such that: $$ \max _{\overline{\Omega} \times[0, T]}v=v(a,b) $$ and if $b=T$ $$\partial_{t}v(a,b)\geq 0 \tag{1}$$ and if $b<T$ then $$\partial_{t}v(a,b)=0. \tag{2}$$
Question: I don't see why we have $(1)$ and $(2)$. I tried to use the definition of the derivative using limits but I just get that the derivate is non-negative? Thanks.
Reference: Brezis, H., & Brézis, H. (2011). Functional analysis, Sobolev spaces and partial differential equations (Vol. 2, No. 3, p. 5). New York: Springer.
By assumption, the function $f\colon[0,T]\to\mathbb R,\,t\mapsto v(a,t)$ has a global maximum in $t=b$. Thus $f(b)\geq f(b-h)$ for all $h\geq 0$, which implies $$ \lim_{h\searrow 0}\frac{f(b)-f(b-h)}h\geq 0. $$ Hence $\partial_t f(b)\geq 0$ (note that $b=0$ is not possible if $(a,b\notin P$, hence $b-h\in [0,T]$ for sufficiently small $h$).
If $b<T$, you can also take the limit from the right side to conclude $\partial_t f(b)=0$.