Continuity Proof "If $f(x,w)$ is continuous and its domain is a cartesian product, $\max_{w}f(x,w)$ is continuous." I posted several questions about the continuity of $max$ functions.
Let me assume that the function $f(x,y,z,w)$ is continuous. Is the $\max_{w} f(x,y,z,w)$ continuous?
Continuity proof for compact domain
Continuity proof "If $f(x,y,z,w)$ is continuous and domains of $x,y,z,w$ are all convex set, then $\max_{w} f(x,y,z,w)$ is continuous."
I combined the answers about the continuity and the condition that the domain is convex or compact is not sufficient for showing that the $\max_{w} f(x,w)$ is continuous.
However, in Continuity proof for compact domain, Hans Engler gave me a comment that $\max_{w} f(x,w)$ is continuous if the domain is a Cartesian product.
So I want to know how to prove: "If $f(x,w)$ is continuous and $x\in [0,x_1]\times[0,x_2]\times[0,x_2]\times[0,x_2]\times[0,x_4]$, $w\in[0,w_1]$, $\max_{w}f(x,w)$ is continuous. ($x_1,x_2,x_3,x_4$ and $w_1$ is real number.)"
 A: Let $K$, $W$ be compact metric spaces (with distances $d_K$ and $d_W$) and $f\colon K\times W\to\Bbb R$ continuous. Then
$$
M(x)=\max_{w\in W}f(x,w)
$$
is continuous on $K$.
Proof. Since $K\times W$ is compact, $f$ is uniformly continuous. Given $\epsilon>0$ there exists $\delta>0$ such that
$$
x,y\in K,\quad d_K(x,y)\le\delta\implies |f(x,w)-f(y,w)|\le\epsilon\quad\forall w\in W.
$$
Then
$$
f(x,w)\le f(y,w)+\epsilon\le M(y)+\epsilon\quad\forall w\in W.
$$
Taking the maximum with respect to $x$ we get
$$
M(x)\le M(y)+\epsilon.\tag{1}
$$
Similarly, we can prove that
$$
M(y)\le M(x)+\epsilon.\tag{2}
$$
From (1) and (2) we finally get
$$
|M(x)-M(y)|\le\epsilon.
$$
In your case, $K=[0,x_1]\times[0,x_2]\times[0,x_2]\times[0,x_2]\times[0,x_4]$ and $W=[0,w_1]$.
A: Good news, this should be true!
We denote the domain for $x$ by $X$ and the domain for $w$ by $W$.
We also define
$$
g(x):= \max_{w \in W} f(x,w).
$$
For proving continuouity of $g$ at a point $\bar x$,
let $x_n$ be a sequence such that $x_n\to \bar x$.
We have to show that $g(x_n)\to g(x_0)$.
Let $w_n$ be such that $g(x_n)=f(x_n,w_n)$.
Since $w_n$ is from a compact domain, it has a convergent subsequence $w_{n_k}$
with limit $w$.
Then it follows that
$$
g(x_0)\geq f(x_0,w)
= \lim_k f(x_{n_k},w_{n_k})
= \lim_k g(x_{n_k}).
$$
By using the same argument for a suitable subsequence of $x_n$, it follows that
$$
g(x_0)\geq \limsup_n g(x_n).
$$
Let $w_0$ be such that $g(x_0)=f(x_0,w_0)$.
It remains to show that $g(x_0)\leq \liminf_n g(x_n)$.
But this is true due to the following calculation.
$$
g(x_0)
= f(x_0,w_0)
= \liminf_n f(x_n,w_0)
\leq
\liminf_n g(x_n).
$$
This completes the proof.
some comments:
Note that we only need the compactness of $W$ (for the existence of maximizers and converging subsequence).
For $X$ we do not need compactness.
We also dont need any convexity (neither for $X$ nor for $W$).
