Max operator over a lipschitz function is still lipschitz? I have a function $f:\mathbb{R}\times\mathbb{R}\to\mathbb{R}$ which is $L$-Lipschitz (over the Manhattan distance). Is $\max_y f(x,y)$ still Lipschitz? 
In my opinion, it is true, and I can prove it, but I'm not sure whether the proof is correct or not, because to me seems too easy. 
$$ |\max_{y_1} f(x_1,y_1) - \max_{y_2} f(x_2,y_2)  | =|\max_{y_1} f(x_1,y_1) - \max_{y_2}\bigl(f(x_1,y_2) - f(x_1,y_2) + f(x_2,y_2)\bigr)  | $$ 
$$ \leq| \max_{y_1} f(x_1,y_1) - \max_{y_2} f(x_1,y_2) + L |x_1 -
 x_2| |  = L|x_1 - x_2|$$
 A: (1) Define $f_y : \mathbb{R}\rightarrow \mathbb{R}$ by $f_y(x)=f(x,y)$.
 Hence $$|f_y(x)-f_y(z)|=|f(x,y)-f(z,y)|\leq L |x-z|$$ so that each
 $f_y$ is $L$-Lipschitz.
(2) Note that $F(x):=\max_y\ f(x,y)$ is a supremum of $f_y$.
(3) Hence if $F(x)>F(z)$, then $$F(x)-F(z)\leq  F(x) - f_y(z)$$ where
$|F(x)-f_y(x)|  <\varepsilon$. That is, $$ F(x)-F(z) \leq \varepsilon
+ |f_y(x)-f_y(z)| \leq \varepsilon + L|x-z|$$
A: I think the problem is already solved by this link. However, I think your proof is correct with minor modification.
Define $F(x):=\max_y f(x,y)$, then we have:
$$
\begin{align}
|F(x)-F(z)| &= \max\left\{ \max_{y_1}f(x,y_1) - \max_{y_2}f(z,y_2), 
   \quad\max_{y_2}f(z,y_2)-\max_{y_1}f(x,y_1) \right\}\\
&= \max\left\{ \max_{y_1}\left\{f(x,y_1)+f(z,y_1)-f(z,y_1)\right\} - \max_{y_2}f(z,y_2),\right.\\
& \qquad\qquad \left. \max_{y_2}\left\{f(z,y_2)+f(x,y_2)-f(x,y_2)\right\}-\max_{y_1}f(x,y_1) \right\}\\
&\leq \max\left\{ \max_{y_1}\left\{f(x,y_1)-f(z,y_1)\right\} +\max_{y_1}f(z,y_1) - \max_{y_2}f(z,y_2),\right.\\
& \qquad\qquad \left. \max_{y_2}\left\{f(z,y_2)-f(x,y_2)\right\} +\max_{y_2}f(x,y_2) -\max_{y_1}f(x,y_1) \right\}\\
&\leq L\cdot d(x,z)
\end{align}
$$
The first equality converts absolute value to a $\max$ operator, the first inequality is due to that separate maximum of two function is larger than the maximum of their sum.
