6

This question is following this one.

Take a subset $\mathcal{F} \subset C[0,1]$ and consider $$g(x)= \sup{ \{f(x)\mid f\in \mathcal{F}} \}.$$ Added: $g$ may take infinite values.

What property(ies) should have $\mathcal{F}$ in order that it exists a sequence $(f_n)$ of elements of $\mathcal{F}$ that converges pointwise to $g$?

I know that the question is open. Answers can be in the form of examples. Please don't shoot to quick to close it because it is an open question!

Community
  • 1
mathcounterexamples.net
  • 70,018
  • First of all, $\mathcal F$ better be uniformly bounded. Otherwise $g$ might be infinite at some point. – Giuseppe Negro Nov 02 '15 at 19:05
  • @GiuseppeNegro: When I read the question, I assumed that $g$ would be permitted to assume the value $\infty$ (extended real numbers). OP, can you clarify whether this is allowed or not? It's an interesting question either way. –  Nov 02 '15 at 19:10
  • 1
    Indeed $g$ may take infinite values. I edited my question accordingly. – mathcounterexamples.net Nov 02 '15 at 19:14
  • I speculate that if $\mathcal F$ is closed under the "join" operation, i.e. if $f \vee h \in \mathcal F$ whenever $f,h \in \mathcal F$, then we can find such a sequence. I haven't had the chance to try to write it formally, but I think I convinced myself mentally that if this condition holds, then we can find an increasing sequence $f_n \in \mathcal F$ which converges pointwise to $g$ on, say, $[0,1] \cap \mathbb Q$, but I don't know if this will suffice to obtain convergence on all of $[0,1]$. I'll think about it more tonight, but if anyone has a counterexample, you'll save me some time :-) –  Nov 02 '15 at 20:14
  • One observation that may be useful is that while $g$ need not be continuous, it is at least lower semicontinuous. –  Nov 02 '15 at 20:15
  • 1
    $\mathcal{F}$ doesn't have to be closed under join, it can just be upper-directed: for all $f, g \in \mathcal{F}$, there is $h \in \mathcal{F}$ such that $f \le h$ and $g \le h$ (pointwise). $h$ might be, say, $\max(f,g) + 1$. – BrianO Nov 02 '15 at 20:18
  • @BrianO: Is it indeed sufficient for $\mathcal F$ to be upper-directed? If so, I'm very interested to see a proof. Under that assumption, I was able to construct an increasing sequence $f_n \in \mathcal F$ which converges pointwise to $g$ on any countable subset of $[0,1]$ (e.g. $[0,1]\cap \mathbb Q$), but that does not seem to imply convergence everywhere in $[0,1]$, or at least I haven't been able to show that it does. –  Nov 03 '15 at 05:33
  • @Bungo It seems/I confess I read your comment hastily, and thought you got it working if $\mathcal{P}$ was closed under $\max$. Let me think about it, I haven't since my earlier possibly bogus comment :) – BrianO Nov 03 '15 at 06:49

1 Answers1

2

As the comments hint at, it suffices to assume that $\mathcal{F}$ is upper-directed, i.e. for $f,g \in \mathcal{F}$, there should be some $h \in \mathcal{F}$ with $h \geq \max \{f, g\}$.

Indeed, it is well-known that $C([0,1])$ is separable when equipped with the sup-norm. Hence, so is any subset, in particular so is $\mathcal{F}$. Thus, let $\{g_n \mid n \in \Bbb{N}\} \subset \mathcal{F}$ be dense with respect to the sup-norm.

Using the directedness of $\mathcal{F}$, we find (inductively) for each $n \in \Bbb{N}$ some $h_n \in \mathcal{F}$ with $h_n \geq \max \{h_{n-1}, g_1, \dots, g_n\}$. In particular, $(h_n)_n$ is a non-decreasing sequence, so that $h := \lim_n h_n$ exists. Since $h_n \in \mathcal{F}$, we trivially have $h_n \leq g$ for all $n$ and thus $h \leq g$.

It thus suffices to show $g \leq h$, for which it suffices to show $f \leq h$ for each $f \in \mathcal{F}$. But by construction, we have $g_n \leq h$ for each $n$. Since $\{g_n \mid n \in \Bbb{N}\} \subset \mathcal{F}$ is dense, we easily get $f \leq h$ for each $f \in \mathcal{F}$ as desired.

Thus, in the end (assuming directedness) everything boiled down to a separability statement.

PhoemueX
  • 35,087
  • 1
    Very nice! For completeness, here is a proof that $C[0,1]$ is separable: http://math.stackexchange.com/questions/643886/a-proof-that-c0-1-is-separable –  Nov 03 '15 at 22:37
  • Yes that is nice. I didn't see the separability. If $\mathcal{P}$ is equicontinuous (and upper-directed) then the result follows but... so what :) — that's too strong a condition. – BrianO Nov 04 '15 at 01:05
  • 1
    I had second thoughts about whether an arbitrary subset of a separable space is separable. (I vaguely recalled that the subset might have to be open for this to work.) But it's true for any subset of any separable metric space. Here's a proof: http://math.stackexchange.com/a/516909/169852 –  Nov 04 '15 at 22:12