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?

  • $\begingroup$ No, this reasoning is not entirely valid since $A$ is not a fixed compact set (it varies with $x$ ; you should denote it by $A(x)$ or $A_x$). So your argument does show that $g$ is well-defined but not that $g$ is continuous. $\endgroup$ Nov 14 '18 at 11:01
  • $\begingroup$ @EwanDelanoy Fixed the notation. What part of the proof does $A$'s non-constancy cause to fail? $\endgroup$
    – actinidia
    Nov 14 '18 at 11:03
  • $\begingroup$ I already answered that in my former comment : it's the part that says that $g$ is continuous. $g$ is not defined on $A_x$, it is defined on $[0,1]$. $f$ is defined on all of $[0,1]^2$, and also by restriction on $A_x$. Note that $f$ is a two-variables function while $g$ is a one-variable function. $\endgroup$ Nov 14 '18 at 12:10
  • 2
    $\begingroup$ Uniform continuity of $f$ can be shown to imply continuity of $g$. $\endgroup$
    – random
    Nov 14 '18 at 12:34
  • $\begingroup$ This question is relevant (math.stackexchange.com/questions/2740283/…). You can show that the family $\{f_x\}$ is equicontinuous (using uniform continuity of $f$), and note $g$ is the supremum of $\{f_x\}$ $\endgroup$
    – Matthew C
    Nov 14 '18 at 14:52

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.


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