0

I'm preparing for my Functional Analysis final by solving old exam exercises. I'm working on a two-part exercise, where I think I have a solution for the first part, but am unable to do the second.

Consider the Banach space $(L^1(0,1), \|\cdot\|_1)$. For $g:(0,1) \to \Bbb{R}$, define $$S_g = \{f \in L^1(0,1) \mid |f(x)| \leq |g(x)| \text{ for a.e. }x \in (0,1)\} $$ 1) Prove that, for every $g \in L^1(0,1)$, $S_g$ is weakly closed in $L^1(0,1)$.

Consider now the space of locally integrable functions $$L^1_{loc}(0,1) = \{g :(0,1) \to \Bbb{R} \text{ measurable}, \forall K \subset (0,1) \text{ compact, } \|g\vert_K\|_1 < \infty\}.$$ 2) Let $g \in L^1_{loc}(0,1)$ and prove that if $S_g$ is weakly compact, then $g \in L^1(0,1)$. (You can use that $S_g$ is weakly closed also if $g \in L^1_{loc}$.)

My attempt at 1) Let $(f_n)_{n\geq 1}$ be a sequence in $S_g$ such that $f_n \to f$ strongly in $L^1$. Then, we know that there exists a subsequence $f_{n_k}$ such that $f_{n_k} \to f$ pointwise almost everywhere. Thus, taking the limit as $k \to \infty$ in $$ |f_{n_k}(x)| \leq |g(x)| \text{ a.e.}$$ gives $|f(x)| \leq |g(x)|$ a.e., so that $f \in S_g$ which is therefore strongly closed. We can verify that $S_g$ is also convex, so using the fact that a convex subset of a Banach space is closed if and only if it is weakly closed we deduce that $S_g$ is weakly closed as well.

Is my reasoning correct?

As for 2), my idea was to find some sequence $(g_n)_{n \geq 1}$ in $S_g$ that converges weakly (along a subsequence) to $g$, then use the fact that $S_g$ is weakly closed to deduce that $g \in S_g \subseteq L^1(0,1).$ I tried to work with the compact sets $K_n = [\frac{1}{n}, 1-\frac{1}{n}]$ (for $n$ large enough), and consider $g_n = \chi_{K_n}g$. Then, $g_n \to g$ pointwise. The problem is, that i don't see why this sequence should be bounded, which would allow me to extract a subsequence by weak compactness. Am I onto something, or should I use a different strategy? Any hint is appreciated. Note that we did not cover anything about locally integrable functions in my course (and they apparently did last year...) so it is possible that I'm missing some classical useful fact here.

noam.szyfer
  • 1,498
  • Did you study weak compactness vs. uniform integrability in $L_1$? – David Mitra Jan 25 '22 at 16:41
  • Unfortunately, we did not. Uniform integrability was not a part of my course at all. – noam.szyfer Jan 25 '22 at 16:45
  • The Dunford-Pettis theorem looks promising here. See this (Theorem 3). – David Mitra Jan 25 '22 at 16:48
  • Perhaps this is a sledgehammer... – David Mitra Jan 25 '22 at 16:50
  • 1
    Oh, wait... Can't you just use the fact that weakly compact sets are norm bounded? – David Mitra Jan 25 '22 at 16:54
  • I don't think we showed this, but we showed that weakly convergent sequences are strongly bounded (Banach Steinhaus), so I suspect that the proof should be somewhat similar. (They probably showed it last year though). But then I get that my $g_n$ are strongly bounded, hence weakly bounded and I can extract a converging subsequence? – noam.szyfer Jan 25 '22 at 17:01
  • You are assuming $S_g$ is weakly compact, so your $(g_n)$ has a weakly convergent subsequence, no? – David Mitra Jan 25 '22 at 17:02
  • Ah yes, I don't need the sequence to be bounded if I already know $S_g$ is compact. I think i was in $\Bbb{R}^n$ mode, so I thought that i need a bounded sequence. So I basically get my weakly convergent subsequence for free...? – noam.szyfer Jan 25 '22 at 17:06
  • 1
    Yes. (You do need to know that weakly compact sets in your space are weakly sequentially compact [this is true in any Banach space, actually].) – David Mitra Jan 25 '22 at 17:08
  • Thank you David! I don't think I saw that exact Theorem, but I will look it up. The exercise is a lot clearer to me know, your help is really appreciated. Thanks for taking the time! Feel free to write a short answer, then I'll accept it if you want. (Correction: We did see Eberlein/Smulyan, but a bit hidden amongst other things... Good to be reminded!) – noam.szyfer Jan 25 '22 at 17:14
  • You're welcome! (The theorem I mention is the Eberlein-Šmulian theorem. Here, we use the easier part of it. In my answer below is a link to a proof of that.) – David Mitra Jan 25 '22 at 17:29

1 Answers1

2

For 2):

You have assumed $S_g$ is weakly compact. We may also deduce that $S_g$ is weakly sequentially compact (see this e.g.). So $(g_n)$ has a weakly convergent subsequence and its limit lies in the set $S_g$. Of course, you still need to argue (as can be done) that this limit must be $g$.

At first glance, your argument for 1) looks fine

David Mitra
  • 74,748