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.

  • $\begingroup$ I have a question about the end of the proof by Salsa, could you help? $\endgroup$
    – sound wave
    Jun 11, 2021 at 19:58
  • $\begingroup$ @soundwave I am no expert, but if I can help I will. Feel free to ask. $\endgroup$ Jun 17, 2021 at 18:37
  • $\begingroup$ Thanks for reply, before the "My comments" section you wrote "which concludes the proof since the opposite inequality is true for subsets.", but I don't understand what it means, could you elaborate more? Many thanks (in Salsa's book it is just written "which concludes the proof" without any further explanation) $\endgroup$
    – sound wave
    Jun 19, 2021 at 6:56
  • 1
    $\begingroup$ Another question: why is the time derivative $u_t$ non-negative at the maximum? $\endgroup$
    – sound wave
    Jun 19, 2021 at 23:17
  • 1
    $\begingroup$ Suppose $\max_{\overline {Q}_{T-\varepsilon}} u=u(x_0,t_0)\notin \partial_p Q_{T-\varepsilon}$. If $t_0<T-\varepsilon$ then $u_t(x_0,t_0)=0$ because it is a maximum on an internal point. If $t_0=T-\varepsilon$ then $u_t(x_0,t_0)\geq 0$ because the maximum is on the boundary and $u$ is non decreasing w.r.t. time. So $u_t(x_0,t_0)\geq 0$ $\endgroup$ Jun 23, 2021 at 19:39

1 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 $t \nearrow T$.

  • $\begingroup$ Thanks, the second part of your proof is clear; re the first part, I do not understand why for $t_0<T$ the maxima are equal. Is it because the corresponding sets are increasing w.r.t. time? $\endgroup$ Aug 18, 2017 at 15:28
  • $\begingroup$ @GiovanniMariani: If the maximum over $[0,T]$ is obtained at $t_0 < T$, then of course the maximum over a smaller $[0,t]$ (with $t \ge t_0$) will again be obtained at the same point. $\endgroup$ Aug 19, 2017 at 5:31
  • $\begingroup$ @AntonyCarapetis I see, it comes from the fact that $(x_0,t_0)\in \overline{Q}_{t_0}$ then $\max_{\overline{Q}_T} w=w(x_0,t_0)\leq \max_{\overline{Q}_{t_0}} w \leq \max_{\overline{Q}_T} w$ $\endgroup$ Aug 19, 2017 at 9:39

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