Continuity Proof "If $f(x,w)$ is continuous and its domain is a cartesian product, $\max_{w}f(x,w)$ is continuous."

Question

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.)"

Answer 1

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 $W$).

Answer 2

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 $W=[0,w_1]$.