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(Update below)

I'm studying for my Functional Analysis final, and I found an interesting exercise in one of the old exams. There is one part where I am unsure, maybe someone could point me in the right direction.

Consider the Banach space $V:= L^\infty(\Bbb{R})$ with the supremum norm. For each $n \in \Bbb{N}$, define

$$ l_n(g) = \frac{1}{2n}\int_{-n}^ng(x)\, dx, \,\,\, \forall g\in V. $$

Prove that there exists $l \in V^*$ and a subsequence $(n_k)_{k \in \Bbb{N}}$ such that $(l_{n_k})$ converges weak-$*$ to $l$ as $k \to \infty$.

If I understood the convergence in the weak-$*$ sense correctly, I need to find a linear functional $l \in V^*$ such that $l_n(g) \to l(g)$ for all $g \in V$ (or rather, we have convergence along a subsequence). My initial thought was just defining $l(g) = \lim_{n\to \infty} l_n(g)$, but I see no reason why this limit should exist for all $g$.

I then proceeded to define $l$ as above on the subspace of $V$ given by $\{g \in V \mid lim_{x\to \infty}g(x)\text{ and }\lim_{x \to -\infty}g(x) \text{ exist.}\}$, where $l_n(g)$ should converge. Using Hahn-Banach, I can then extend this functional to all of $V$.

Now, since $l_n(g)$ is a bounded sequence, by Bolzano-Weierstrass we can extract a converging subsequence, which should give exactly the subsequence $(l_{n_k})_{k\in \Bbb{N}}$ we are looking for.

Am I on the right track here? I'm particularly unsure about the characterization of weak-$*$ convergence, which is a subject I am still struggling a bit. Any help is greatly appreciated.

Update

As mentioned by @Ofek Arian in his answer, my argument does not work, because just finding for all $g \in V$ a subsequence s.t. $l_{n_k}(g)$ converges is not enough to conclude there is a subsequence s.t. $(l_{n_k})$ converges for all $g \in V$.

Based on @David Mitra's comment, the statement that the exercise wants me to prove might actually not even be true. I will try to flesh out his argument here, I hope I understood correctly.

If we look at $l_n$ as an element of $L^1$, say $l_n(x) = \frac{1}{2n}\chi_{[-n,n]}(x)$, then the weak-$*$ convergence in the exercise is the same as weak convergence of $l_n$ in $L^1$. We have that $l_n(x) \to 0$ pointwise and if a sequence of elements in $L^1$ converges pointwise (a.e.), as well as weakly in $L^1$, then the two limits must agree. (See e.g., Pointwise a.e. convergence and weak convergence in Lp). So the weak limit along any subsequence would need to be the zero function. But for $g \equiv 1$, the constant $1$ function, we have $l_n(g) = 1 \, \forall n \in \Bbb{N}$, showing that $l_n$ can't possibly converge weakly to the zero function along any subsequence.

Since the first part of the exercise was 'state the Banach-Alaoglu Theorem', I am led to believe that the intention might have been what @Ofek Arian mentioned in his comment, namely to invoke Banach-Alaoglu to deduce that the closed unit ball of $V^*$ is weak-$*$ compact, and use this fact to extract a converging subsequence. But compactness and sequential compactness need not agree in the weak-$*$ topology, and it is in fact different in the case of $(L^\infty)^*$. It's possible that this is just a mistake in the exercise...

Update 2

As @SeverinSchraven noticed, one needs to be careful when arguing that such subsequence cannot exist. It is a priori not clear, that the weak-$*$ limit of $l_n$ can be represented by a function in $L^1$, so the linked result cannot be directly applied. Nonetheless, this can be fixed, as @DavidMitra argued: If we had weak-$*$ convergence in $(L^\infty)^*$, we would get a weak Cauchy sequence (in $L^1$), and in $L^1$, weak Cauchy implies weakly convergent. See e.g., Weak limit of an $L^1$ sequence. Thus, we may actually apply the linked result as done above to argue that no such subsequence can exist.

(I hope I have written down your argument faithfully).

noam.szyfer
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  • If you want to know what is the maximum that $\ell_n(h)$ could be among all functions $h$ with $|h|_{\infty} \leq 1$ it is attained by $h(x) = \operatorname{sgn}(g(x))$ where $\operatorname{sgn}(g)(x)$ is $+1$ if $g(x)>0$, it is $-1$ if $g(x)<0$ and its value does not really matter if $g(x)=0$ so you can take it to be $0$. – Shannon Starr Jan 24 '22 at 17:40
  • Maybe I am off, but: the $l_n$ can be thought of as elements of $L_1$: $l_n={1\over2n}\chi_{[-n,n]}$. Then you would need to show the existence of a weakly converging subsequence. Since any subsequence converges pointwise to 0, any weak limit must be the zero function. But for $g=1$ and any $n$, $l_n(g)=1$. – David Mitra Jan 24 '22 at 18:50
  • @DavidMitra, do you mean that it is impossible to find such subsequence? – noam.szyfer Jan 24 '22 at 19:29
  • Yes, if my argument is valid. Maybe I am missing something... – David Mitra Jan 24 '22 at 20:52
  • I think the problem is that the weak-* convergence on $L^\infty$ and the weak convergence on $L^1$ are not the same. The weak-* limit is an element in $V^$ and a general element in there need not be (represented by) an element in $L^1$. So while the convergence would agree if $\ell\in V^$ could be represented by an $L^1$ function, they are not quite the same in general. Thus, you cannot directly apply the linked result about weak convergence. Unfortunately, I also don't see how to do it. – Severin Schraven Jan 25 '22 at 20:58
  • @SeverinSchraven Thanks, I should have seen that. But in $L_1$, weakly Cauchy sequences are weakly convergent. If we had weak* convergence in $L_\infty^*$, we'd have a weak Cauchy, thus weakly convergent, sequence in $L_1$. So, I still don't see how this result holds. – David Mitra Jan 25 '22 at 23:14
  • @DavidMitra Is it easy to see that weak Cauchy implies weakly converging in $L^1$? I remember that it holds for reflexive Banach spaces, but cannot remember the proof. – Severin Schraven Jan 26 '22 at 01:05
  • @SeverinSchraven Thank you for your comment! I found a question about $L^1$ weakly Cauchy sequences on this site and have edited my post to include the link. It contains an outline of the argument. I'd say it's not "easy to see", but difficulty often is in the eye of the beholder ;) – noam.szyfer Jan 26 '22 at 07:14
  • @SeverinSchraven I'd say its not so easy. Wojtaszczyk's Banach spaces for analysts obtains it from the characterization of relatively weakly compact sets in $L_1$ as those not containg a copy of $\ell_1$'s unit vector basis. – David Mitra Jan 26 '22 at 08:22
  • @DavidMitra Thanks for the reference. – Severin Schraven Jan 26 '22 at 16:18

1 Answers1

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Your understanding of the weak* topology is good. It is useful to think of the natural embedding $\iota: X \to {X^*}^*$, then the weak* topology is exactly the weak topology $\sigma(X^*,\iota(X))$, i.e. the weakest topology making the functionals $\chi_x: X^* \to \mathbb C$ given by $\chi_x (f) = f(x)$ continious. As for your solution, I think the definition of the subspace in which the limit exist is a bit delicate, as functions in $\mathcal L^\infty$ are defined up to a null set (i.e. of measure zero), so the regular notion of the limit in infinity is not well-defined.

Ofek Arian
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  • Thanks for the feedback! What would be the way to define it more precisely? Something like ${g \mid \exists A \subseteq \Bbb{R} \text{ with measure zero, s.t. } \lim_{|x|\to \infty} g(x) \text{ exists on } V\setminus A}$ maybe? And your comment about the natural embedding is really helpful. – noam.szyfer Jan 24 '22 at 17:50
  • Perhaps it can be difined via "there is a representative f s.t. f=g a.e. and $\lim_{|x|\to\infty} f$ exists". Anyway I'm not comfortable with the rest of your proof because the fact for every g $l_n (g)$ has a convergent subsequence does not imply the existance of a subsequence of $l_n$ s.t the limit exists for each g. Instead, I advise you to use the Alaoglu theorem: note $\Vert{l_n}\Vert_=1$ for every n, so it is a sequence in the closed unit ball of the dual space, which is weak compact. – Ofek Arian Jan 24 '22 at 18:42
  • Ah, you are completely right, my argument does not work. I'm not sure how you would proceed with weak* compactness though, since this only gives me (cover-)compactness and not sequential compactness. But I'll think about it more tomorrow. – noam.szyfer Jan 24 '22 at 19:30
  • @noam.szyfer You definitely should apply the Banach–Alaoglu theorem. According to the wikipedia page https://en.wikipedia.org/wiki/Banach%E2%80%93Alaoglu_theorem#Sequential_Banach%E2%80%93Alaoglu_theorem "the weak* topology on the closed unit ball of the dual of a separable space is metrizable, and thus compactness and sequential compactness are equivalent." I'm unsure if the examiner expected the students to know that, but at least you know the result is true – JackT Jan 26 '22 at 08:43
  • @JackT Thanks for the link, we actually saw this result in class. But also, $L^\infty$ is not separable, so what you are quoting does not apply. We actually showed in an exercise that $L^\infty$ is not separable by explicitly showing that the closed unit ball of its dual in the weak-$*$ topology is not sequentially compact. This is why Banach-Alaoglu cannot save me here. – noam.szyfer Jan 26 '22 at 09:06
  • @noam.szyfer Ah yes I'm sorry. I read the article too quickly and missed that it required separability. – JackT Jan 26 '22 at 11:25