Let $u\in\mathcal{C}^1(\mathbb{R}\_+\times[0,1],\mathbb{R})$ such that, for any $t\geq 0$, for all $x\_0\in[0,1]$ satisfying $u(t,x\_0)=\sup\_{x\in[0,1]}u(t,x)$, we have


$$\partial\_t u(t,x\_0)\leq 0.$$


Is the function $\sup\_{x\in[0,1]}u(\cdot,x)$ non-increasing?



Condition for the maximum to be non-increasing
$\newcommand{\de}{\delta}\newcommand{\ep}{\varepsilon}$The answer is yes.


Indeed, let $m(t):=\sup\_{x\in[0,1]}u(t,x)=\max\_{x\in[0,1]}u(t,x)$. For each real $t\ge0$, choose any $x\_t\in[0,1]$ such that $u(t,x\_t)=m(t)$.


Assume first that
\begin{equation\*}
 (D\_1u)(t,x\_t)<0 \tag{1}
\end{equation\*}
for all $t\in[0,1]$, where $(D\_1u)(t,x):=\partial\_t u(t,x)$.
Then for each $t\in(0,1]$ there is some $\de\in(0,t)$ such that for all $s\in(t-\de,t)$ we have $u(s,x\_t)>u(t,x\_t)$ and hence
\begin{equation\*}
 m(s)\ge u(s,x\_t)>u(t,x\_t)=m(t). 
\end{equation\*}
So, the lower left derivative of $m$ is $\le0$ on $(0,1]$. Also, the function $m$ is continuous, since $u\in\mathcal{C}^1(\mathbb{R}\_+\times[0,1],\mathbb{R})\subset\mathcal{C}(\mathbb{R}\_+\times[0,1],\mathbb{R})$. So (cf. e.g. Titchmarsh, The Theory of Functions, 2nd ed., Example (IV) in Section 11.3), $m$ is nonincreasing (assuming (1)).


If we only have $(D\_1u)(t,x\_t)\le0$ for all $t\in[0,1]$, then (1) holds, for any real $\ep>0$, with $u\_\ep$ in place of $u$, where $u\_\ep(t,x):=u(t,x)-\ep t$. So, by what has been proved, the function $m\_\ep$ defined by
\begin{equation}
 m\_\ep(t):=\sup\_{x\in[0,1]}u\_\ep(t,x)=m(t)-\ep t
\end{equation}
is nonincreasing, for any real $\ep>0$. So, $m$ is nonincreasing, as desired.


