Prove that $u_{max}(t)=\sup_{x\in I}u(x,t)$ is locally Lipschitz Let $I=[a,b]$ be a closed and bounded interval. Let $u:I\times(0,T)\to\mathbb{R}$ be a $C^1$ (continuously differentiable) function. Let $u_{max}:(0,T)\to\mathbb{R}$ be the function defined by
\begin{align}
u_{max}(t):=\sup_{x\in I}u(x,t)
\end{align}
I would like to show that:

The function $u_{max}$ is locally Lipschitz on $(0,T)$.

My attempt
Fix a $t\in(0,T)$. It suffices to show that there exists an open neighbourhood of $t$ in $(0,T)$ such that $u_{max}$ is Lipschitz in this neighbourhood.
To begin with, we utilise the fact that $u$ is locally Lipschitz (since it is $C^1$). Thus for any $p\in I$, there exists an open neighbourhood $U_{(p,t)}$ of $(p,t)$ (without loss of generality, we suppose
\begin{align}
U_{(p,t)}=(p-r,p+r)\times(t-\delta,t+\delta)
\end{align}
is an open square) such that $u$ is Lipschitz on $U_{(p,t)}$, meaning $\exists C_{(p,t)}>0$ such that on $U_{(p,t)}$,
\begin{align}
|u(x,\sigma)-u(y,\tau)|\leq C_{(p,t)}(|x-y|+|\sigma-\tau|) & & (1)
\end{align}
The open squares then form an open cover of $I\times\{t\}$. By compactness, we extract a finite subcover consisting of open squares
\begin{align}
U_{(p_i,t)}=(p_i-r_i,p_i+r_i)\times(t-\delta_i,t+\delta_i),\qquad i=1,\ldots,N_t
\end{align}
From this we can choose
\begin{align}
\delta_t:&=\min_{1\leq i\leq N_t}\delta_i \\
C_t:&=\max_{1\leq i\leq N_t}C_{(p_i,t)}
\end{align}
Now I wish to show that the function $u_{max}$ is Lipschitz on the open neighbourhood $\mathcal{O}_t:=(t-\delta_t,t+\delta_t)$:
\begin{align}
|u_{max}(\sigma)-u_{max}(\tau)|\leq C'_t|\sigma-\tau| & & (2)
\end{align}
where $C_t'>0$ is a possibly different constant. The key point here is that $C'_t>0$ should be universally work for any $\sigma,\tau\in\mathcal{O}_t$.

My question

How to prove the inequality (2)?

Note that one can prove, by using (1), that for $0<|h|<\delta_t$,
\begin{align}
|u_{max}(t+h)-u_{max}(t)|\leq C_t|h|
\end{align}
(As a side note, this implies that $u_{max}$ is continuous at $t$.) Thus
\begin{align}
|u_{max}(\sigma)-u_{max}(\tau)|&\leq|u_{max}(\sigma)-u_{max}(t)|+|u_{max}(\tau)-u_{max}(t)| \\
&\leq C_t|\sigma-t|+C_t|\tau-t|
\end{align}
Now if $\sigma,\tau$ are on different sides of $t$ (e.g. $\sigma<t<\tau$), then $|\sigma-t|+|\tau-t|=|\sigma-\tau|$ and we can choose $C_t':=C_t$. But now I'm stuck for the other case where $\sigma,\tau$ lie on the same side of $t$.

Any comment, hint or answer is welcomed and appreciated.
 A: It turns out that $u_{max}$ is indeed Lipschitz continuous on the open neighbourhood $\mathcal{O}_t$ of $t$, in fact with Lipschitz constant given directly by $C_t$.
To begin with, let $x\in I$ be arbitrary. Let $1\leq i=i(x)\leq N_t$ be the index such that $x\in(p_i-r_i,p_i+r_i)$. For any $\sigma,\tau\in\mathcal{O}_t$, we note that
\begin{align}
u(x,\sigma)&=\big(u(x,\sigma)-u(x,\tau)\big)+u(x,\tau) \\
&\leq|u(x,\sigma)-u(x,\tau)|+\sup_{y\in I}u(y,\tau) \\
&\leq C_{(p_i,t)}\big(|x-x|+|\sigma-\tau|\big)+u_{max}(\tau) \\
&\leq C_t|\sigma-\tau|+u_{max}(\tau)
\end{align}
The R.H.S. is independent of $x\in I$. Since this inequality is true for any $x\in I$, we may take the supremum of the L.H.S. to conclude that
\begin{align}
u_{max}(\sigma):=\sup_{x\in I}u(x,\sigma)
\leq C_t|\sigma-\tau|+u_{max}(\tau)
\end{align}
or equivalently
\begin{align}
u_{max}(\sigma)-u_{max}(\tau)\leq C_t|\sigma-\tau|
\end{align}
Exchanging the roles of $\sigma$ and $\tau$ does not affect the argument. Therefore we have
\begin{align}
\left|u_{max}(\sigma)-u_{max}(\tau)\right|\leq C_t|\sigma-\tau|
\end{align}
This proves that $u_{max}$ is Lipschitz continuous on $\mathcal{O}_t$ with Lipschitz constant $C_t$.

A naive estimate which does not work is to begin the estimate of $\left|u_{max}(\sigma)-u_{max}(\tau)\right|$ by taking points $x_{\sigma},x_{\tau}$ at which $u(\cdot,\sigma)$ and $u(\cdot,\tau)$ respectively achieve the maximum on $I$, and then consider the inequalities
\begin{align}
\left|u_{max}(\sigma)-u_{max}(\tau)\right|
&=|u(x_{\sigma},\sigma)-u(x_{\tau},\tau)| \\
&\leq|u(x_{\sigma},\sigma)-u(x_{\sigma},\tau)| 
+|u(x_{\sigma},\tau)-u(x_{\tau},\tau)|
\end{align}
or something similar. This is motivated by the naive intuition that when $\sigma,\tau$ are close to each other, the points $x_{\sigma},x_{\tau}$ at which the maxima are attained should also be close to each other. Unfortunately this is not always true in general and therefore such estimate fails to provide an ingredient to the complete proof.
