Proof of weak maximum principle for heat equation

Question

I am having trouble with the last step of the proof of the weak maximum principle for the heat equation, as found in

Salsa, Sandro, Partial differential equations in action. From modelling to theory

th. 2.4 pp. 36-38.

Here is a sketch of the proof.

Let $w_t - D\Delta w = q \leq 0$ on the space-time cylinder $Q_T$ with parabolic boundary $\partial_p Q_T$ and let $0 < \varepsilon < T$.

Defining the auxiliary function $u = w - \varepsilon t \leq w$, we have $$u_t - D\Delta u = q-\varepsilon < 0$$ and we prove that $$ \max_{\overline{Q}_{T-\varepsilon}} u = \max_{\partial_p Q_{T-\varepsilon}} u$$ by a calculus argument: the Hessian is negative semidefinite and the time derivative is non-negative at the maximum, leading to a contradiction.

Moreover we prove that $$ \max_{\overline{Q}_{T-\varepsilon}} w \leq \max_{\partial_p Q_T} w +\varepsilon T $$ essentially by subset properties w.r.t. maxima and the definition of $u$.

So far, so good.

Here comes the tricky part, at least for me. It goes

Since $w$ is continuous in $\overline{Q}_{T-\varepsilon}$ we deduce that $$ \max_{\overline{Q}_{T-\varepsilon}} w \to \max_{\overline{Q}_T} w \quad\text{as}\; \varepsilon\to 0 $$

therefore in the limit above we get $$ \max_{\overline{Q}_T} w \leq \max_{\partial_p Q_T} w $$ which concludes the proof since the opposite inequality is true for subsets.

My comments

• $w$ is not only continuous, but also uniformly continuous on the compact $\overline{Q}_T$ and all its subsets, by Heine-Cantor theorem, but I cannot figure out how to use this property with the maxima.

• I have also explored the possibility that they meant 'uniformly convergent' instead of 'continuous', but this seems only applicable to other proof strategies, in which only $ \overline{Q}_T $ and not $ \overline{Q}_{T-\varepsilon} $ is used, like for instance in Evans (1998) ch. 7 th. 8. For that line of reasoning, I am having trouble applying the triangle inequality to prove that $$ \max_{\overline{Q}_T} u \to \max_{\overline{Q}_T} w \quad\text{as}\; \varepsilon\to 0 $$ although this is slightly off topic.

• For $\varepsilon\to 0$ the sets $\overline{Q}_{T-\varepsilon}$ are increasing, therefore we can build a corresponding sequence of maxima which is increasing and so converges to its supremum over $\varepsilon$, but here again how can I prove that $$ \sup_{\varepsilon>0} \max_{\overline{Q}_{T-\varepsilon}} w = \max_{\overline{Q}_T} w$$

Any help will be greatly appreciated.

Answer 1

$\def\Q#1{\bar Q_{#1}}\def\M#1{\max_{\Q{#1}}w}$Compactness and continuity makes this fairly easy. We know that the maximum over $\Q T$ is attained at some point $(x_0,t_0)$. If $t_0 < T$ then $\M {T - \epsilon} = \M{T}$ for all $\epsilon \in [0,T-t_0]$, so we are done; so we just need to consider the case $t_0 = T$.

Given an arbitrary $\epsilon > 0$, the continuity of $w$ provides a $\delta > 0$ such that $|w(x,t) - \M T| < \epsilon$ whenever $|(x,t) - (x_0,T)| < \delta$. Thus if we take a time $t > T - \delta$ we can conclude $$\M t \ge w(x_0,t) >\M T - \epsilon.$$ On the other hand we have $\M t \le \M T$ since $\Q t \subset \Q T$; so we have shown that $\M t \to \M T$ as $t \nearrow T$.