Condition for the maximum to be non-increasing 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?
 A: $\newcommand{\de}{\delta}\newcommand{\ep}{\varepsilon}$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.
