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Problem: Let $\mathcal{M}$ be a metric space and let $\mathcal{F}$ be a bounded family of real valued functions on $\mathcal{M}$. Assume that $\mathcal{F}$ is equicontinuous. Define, for each $x \in \mathcal{M}$, $\displaystyle s(x) = \sup_{f \in \mathcal{F}} f(x)$. Prove that $s$ is continuous.

Attempt: I have no knowledge of topology so I want to prove this directly. To begin, let $x \in \mathcal{M}$. Let $\epsilon > 0$. Since $\displaystyle s(x) = \sup_{f \in \mathcal{F}} f(x)$, we can find a $g \in \mathcal{F}$ such that $\displaystyle |s(x) - g(x)| < \frac{\epsilon}{2}$. By the equicontinuity of $\mathcal{F}$, we can find a $\delta > 0$ such that if $d(x,y) < \delta$, for some $y \in \mathcal{M}$, then $\displaystyle |f(x) - f(y)| < \epsilon_1 < \frac{\epsilon}{2}$, for all $f \in \mathcal{F}$. In particular, $\displaystyle |g(x) - g(y)| < \frac{\epsilon_1}{2}$. Let $h \in \mathcal{F}$ such that $\displaystyle |s(y) - h(y)| < \frac{\epsilon}{2}$. We also have by the equicontinuity of $\mathcal{F}$ that $\displaystyle |h(x) - h(y)| < \frac{\epsilon_1}{2}$. Now we want to show that $g(y)$ and $h(y)$ are close.

And this is where I am stuck. I don't think it's necessarily true that $g$ and $h$ at $y$ would be close. Where should I go from here?

Thank you.

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You don't need to compare $g(y)$ and $h(y)$ (though they would be close after all).

You've shown that $|s(x) - g(y)| < \epsilon$. In particular, $s(x) < g(y) + \epsilon$. But $g(y) \le s(y)$ because $s$ is the supremum. So $s(x) < s(y) + \epsilon$.

Likewise, you've shown that $|s(y)-h(x)| < \epsilon$. So arguing similarly, $s(y) < s(x) + \epsilon$.

Combining these yields $|s(x) - s(y)| < \epsilon$.

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  • $\begingroup$ nice ! And it seemed confusing at first sight ! :P $\endgroup$ – dem0nakos Apr 16 '18 at 21:11
  • $\begingroup$ @Nate Eldredge: Could you help me to complete my proof? $\endgroup$ – Saeed Nov 8 '18 at 21:47
  • $\begingroup$ @Saeed: You should post it as a new question rather than asking people individually via comments. $\endgroup$ – Nate Eldredge Nov 8 '18 at 22:53
  • $\begingroup$ I think I forgot to paste the link in my comment: here is my question which I have asked math.stackexchange.com/questions/2990482/… $\endgroup$ – Saeed Nov 8 '18 at 23:02
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Hint: Let's stick with your $g$ and not hurry over to $h$ so fast. If $d(x,y)<\delta$ we have

$$s(x) < g(x) + \epsilon/2 < g(y) + \epsilon \le s(y) + \epsilon.$$

By symmetry, the inequality the other way around must also be true.

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    $\begingroup$ The "by symmetry" is a little subtle. Applying it naively at $y$, you would get the statement "there exists $\delta_y$ such that for all $z$ with $d(y,z) < \delta_y$ we have $s(y) < s(z) + \epsilon$." But this $\delta_y$ may not be the same as the original $\delta = \delta_x$, and we can't necessarily take $z=x$ in the second statement. Indeed, your statement is true without equicontinuity, but it only implies lower semicontinuity of $s$. $\endgroup$ – Nate Eldredge Apr 16 '18 at 20:58

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