Let $\psi(x,y)$ be a continuous function in two real variables, and then define $$f(x) = \sup_{t \in [-x,x]} \psi(x,t)$$ It seems to me that the function $f$ should also be continuous for $x \geq 0$, but I'm not sure how to prove it. Is this true and how would you prove it?

  • $\begingroup$ Assume $x \geq 0$ $\endgroup$ Aug 4 '19 at 19:26
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    $\begingroup$ What have you tried? $\endgroup$ Aug 4 '19 at 21:33
  • $\begingroup$ My only thought is that if you could show that $g(x,x^\prime) = \sup_{t \in [-x^\prime, x^\prime]} \psi(x,t)$ is continuous in $x$ and $x^\prime$ separately, then $f(x) = g(x,x)$ will be continuous as a corollary. To be honest, I posted this on the assumption that a theorem to this effect existed somewhere and I could cite it, and I was hoping someone could give me a reference. $\endgroup$ Aug 4 '19 at 22:23

HINT: use uniform continuity.

Fix $x_0$ and let $\epsilon > 0$. Choose $t_0 \in \left[-x, x\right]$ such that $f(x_0) = \psi(x_0, t_0)$. Since $\psi$ is continuous, there exists a $\delta > 0$ such that

$$|\psi(x, t) - \psi(x_0, t_0)| < \epsilon$$

for all $(x, t)$ in the closed ball of radius $\delta$ (let's say with respect to the norm $\|\cdot\|_\infty$) centered on $(x_0, t_0$). Hence

$$f(x) \ge \psi(x, t_0 - \operatorname{sgn}(t_0)\delta) > f(x_0) - \epsilon$$

for all $x$ such that $|x - x_0| < \delta$ (note that $t_0 - \operatorname{sgn}(t_0)\delta \in [-x, x]$).

Let $M \gg x_0$. By compactness and continuity, $\psi$ is uniformly continuous on $E = [-M, M] \times [-M, M]$. Thus we could have chosen $\delta$ to be the "universal delta" for $\psi$ over $E$. Hence, analogously,

$$f(x_0) > f(x) - \epsilon$$

and therefore $|f(x) - f(x_0)| < \epsilon$.


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