$f:[0,1] \times [0,1] → \mathbb R$ is continuous. Prove that $g(x) = \max\{f(x,y) : y \in [0,1]\}$ is defined and continuous.

Question

Let $f:[0,1] \times [0,1] → \mathbb R$ be a continuous function. Prove that $$g(x) = \max\{f(x,y) : y \in [0,1]\}$$ makes sense (in that the maximum exists) and is continuous.

I said that for each $x$, we can consider $A_x = \{x\} \times [0,1] \subset \text{dom } f $. $A_x$ is compact, and since $\max$, which is acting on a continuous function $f$, is itself continuous,we know that (a) $g$ is continuous, and (b) the image $g(A_x)$ is compact, which means it attains a maximum and minimum. Therefore the maximum exists for each $x$, and by (a) $g$ is continuous.

Edit: It seems that the fact that $A_x$ isn't fixed makes the proof fail. What would a working proof look like?

Answer 1

The continuity with respect to $y$ is irrelevant as long as the functions $x\mapsto f(x,y)$ are equicontinuous with respect to $y$: Given $x_0$ and an $\epsilon>0$, there is a $\delta>0$ such that $$|x-x_0|<\delta\qquad\Rightarrow\qquad|f(x,y)-f(x_0,y)|<\epsilon\quad\forall y\ .\tag{1}$$

Let an $x_0\in[0,1]$ and an $\epsilon>0$ be given. Choose a $\delta>0$ such that $(1)$ holds. If $|x-x_0|<\delta$ then $$f(x,y)\leq f(x_0,y)+\epsilon \leq g(x_0)+\epsilon \qquad\forall y\ .$$ It follows that $g(x)\leq g(x_0)+\epsilon$. Similarly $$f(x_0,y)\leq f(x,y)+\epsilon\leq g(x)+\epsilon\quad\forall y\ ,$$ and therefore $g(x_0)\leq g(x)+\epsilon$. In all we have proven that $|x-x_0|<\delta$ implies $|g(x)-g(x_0)|\leq\epsilon$, as required.

Answer 2

Here's an alternative approach:

Let $x_0\in[0,1]$, and fix $\varepsilon>0$. Let $y_0$ be the $y$ value for which $f_{x_0}$ attains its max, that is, $g(x_0)=f(x_0,y_0)$. Since $f$ is uniformly continuous, there is some $\delta>0$ such that $\|{(x,y)-(a,b)}\|<\delta$ implies $|f(x,y)-f(a,b)|<\frac{\varepsilon}{3}$. Let $x\in[0,1]$ and assume $|x-x_0|<\delta$. Let $y_x$ be such that $f(x,y_x)=g(x)$. It follows that \begin{align*} |g(x)-g(x_0)|&=|f(x,y_x)-f(x_0,y_0)|\\&\leq|f(x_0,y_x)-f(x,y_x)|+|f(x_0,y_0)-f(x,y_0)|+|f(x,y_0)-f(x_0,y_x)|\\ &<\frac{\varepsilon}{3}+\frac{\varepsilon}{3}+|f(x,y_0)-f(x_0,y_x)| \end{align*} Notice I used the fact that $\|(x_0,y_x)-(x,y_x)\|=\|(x_0,y_0)-(x,y_0)\|=|x-x_0|$. Now if $|f(x,y_0)-f(x_0,y_x)|\geq0$, we get $$|f(x,y_0)-f(x_0,y_x)|=f(x,y_0)-f(x_0,y_x)\leq f(x,y_x)-f(x_0,y_x)<\frac{\varepsilon}{3}$$ Otherwise $$|f(x,y_0)-f(x_0,y_x)|=f(x_0,y_x)-f(x,y_0)\leq f(x_0,y_0)-f(x,y_0)<\frac{\varepsilon}{3}$$ In either case we get that $|g(x)-g(x_0)|<\varepsilon$, as needed.