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Suppose $\mathcal{E} \subset C_0([a, b], \mathbb R)$ is a family of functions.

  1. Show that $g(x) = \{\sup f(x) : f \in \mathcal{E}\}$ is continuous does not imply that $\mathcal{E}$ is equicontinuous.

  2. If $g(x) = \{\sup f(x) : f \in \mathcal{F}\}$ is continuous for every $\mathcal{F} \subset \mathcal{E}$, is $\mathcal{E}$ equicontinuous?

I need a hint for these two questions. I don't even know where to start.

Martin Sleziak
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Aden Dong
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  • @julien: the artificial continuity of the sup in my counterexample for question 1 kind of begs the hypothesis of question 2. I see nothing wrong with the questions the way they are. – robjohn Feb 28 '13 at 08:24
  • I gave the question a more descriptive title. – Nate Eldredge Feb 28 '13 at 16:11
  • @NateEldredge: When altering a title, make sure that part of the question was not in the title. Questions should be self-contained, but often they are not. I have placed the old title into the question. – robjohn Feb 28 '13 at 16:25
  • @robjohn: Oops! Thanks for the fix. – Nate Eldredge Feb 28 '13 at 22:05

1 Answers1

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Hint: Consider $$ f_n(x)=\left\{\begin{array}{} 0&\text{for }x\lt\frac12-\frac1{2n}\\ 1-n\left(1-2x\right)&\text{for }\frac12-\frac1{2n}\le x\le\frac12\\ 1&\text{for }x\gt\frac12\\ \end{array}\right. $$

robjohn
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    For hint 2, I guess you aim at a contradiction with the continuity of the sup of the subsequence. I confess I can't see it... – Julien Feb 28 '13 at 14:28
  • @julien: and it is well you can't :-) I tried to flesh out my approach, and hit a snag which motivated the counterexample in Hint 2. – robjohn Feb 28 '13 at 15:50
  • I like your new hint 2, +1. So in this case, your sequence is nonincreasing and for any $\mathcal{F}$, we have $g_{\mathcal{F}}=\sup_\mathcal{F}f=f_{n}$ with $n=\min{k;;;f_k\in \mathcal{F}}$, so it's indeed continuous. – Julien Feb 28 '13 at 16:13
  • So are you saying that if $\sup_{\mathcal F}|f|$ is continuous for all $\mathcal F$, then $\mathcal E$ is equicontinuous? – Julien Feb 28 '13 at 16:21
  • @julien: Consider $$ f_n(x)=\left{\begin{array}{} 0&\text{for }x\lt\frac12-\frac1n\ 1-n\left(\frac12-x\right)&\text{for }\frac12-\frac1n\le x\le\frac12\ 1&\text{for }x\gt\frac12\ \end{array}\right. $$ I guess this would be a counterexample for both 1 and 2. – robjohn Feb 28 '13 at 16:51
  • Yes, great. Thanks! – Julien Feb 28 '13 at 21:46