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I want to prove that every dual space is weak*-sequentially complete.

Let $X$ be a normed linear space and let $(f_n)$ be a weak* Cauchy sequence in $X^*$. Thus for all $x\in X$, $(f_n(x))$ is a Cauchy sequence in $\mathbb K$. Thus for all $x\in X$, $\lim\limits_{n\to \infty}f_n(x)$ exists. If I define $f(x)=\lim\limits_{n\to \infty}f_n(x)$ for all $x\in X$, then $f$ is linear. But how to show that it is bounded. Had $X$ been given a Banach space I could have done it by using Banach-Steinhauss theorem. But now how to proceed? Any hint is appreciated.

Martin Sleziak
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Anupam
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2 Answers2

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Hint : You can use the following corollary of Banach-Alaoglu Theorem :

Let $ X $ be a normed space. Then every bounded subset of $ X^* $ is relatively weakly$^*$ compact.

Jakobian
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Math Fanatic
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    It's not clear how this helps, because we would still need to show that the sequence $f_n$ is bounded. – Nate Eldredge Mar 25 '19 at 03:35
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    And as David Mitra's example shows, in fact $f_n$ does not have to be bounded, and the desired statement is actually false. – Nate Eldredge Mar 25 '19 at 03:38
  • @NateEldredge This result is true for Banach space $ X $ (In David Mitra's example $ X $ is not complete). – Math Fanatic Mar 25 '19 at 05:56
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    As the other answer shows, this hint alone is not sufficient to prove the claim without completeness of $X$. – daw Jan 29 '25 at 20:05
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    @BastamTajik Stop editing this answer, the theorem holds very generally, and no separability is needed! In most generality it's known as the bipolar theorem and holds for locally convex spaces. The classical version is obtained by observing that the closed ball $B_{X^\ast}$ of $X^\ast$ is the bipolar of the closed ball $B_X$ of $X$. – Jakobian Jan 29 '25 at 20:18
  • I don't understand. I just gave a counter example for why it doesn't hold for a non-separable normed vector space. And people(maybe you). Take ${\ell}^{\infty} (\mathbb{N})$ and call its continuous dual $X^{}$. Take the sequence of projection operators $l_k(x):=x_k$. Easy to see that it's a bounded sequence by operator norm in $X^{}$ but there exists no subsequence of $l_k$ that is Weak-* convergent for all $x \in X$. So please stop asking "stop editing" and think about this counter example instead. No offense but I felt offended by the way you talk. Best regards. @Jakobian – Bastam Tajik Jan 30 '25 at 12:54
  • @BastamTajik just because you have doubts about some theorem holding is not a good reason for it to have its context modified. Since we are not talking about weak* sequential compactness, your example is not actually a counter-example. There doesn't have to exist a convergent subsequence but a convergent subnet. The answer is talking about compactness and not sequential compactness. I hope we can have a respectful exchange, unhindered by emotional burdens. – Jakobian Jan 30 '25 at 14:28
  • Sequential compactness and compactness are the same thing for a metric space. I'm all for mutual respect and patience. Thanks @Jakobian of course I assume axiom of countable choice. – Bastam Tajik Jan 30 '25 at 14:52
  • @BastamTajik $\ell^\infty$ with weak*-topology is no longer a metrizable space – Jakobian Jan 30 '25 at 15:00
  • The topology of $X^$ is always that of the operator norm induced one in all my comments. $l_k$ is a bounded sequence, by operator norm on $X^$ and weak-* convergence is defined by the norm on the complex plain. So let go of the weak-* topology all together please. @Jakobian – Bastam Tajik Jan 30 '25 at 16:12
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    Stop editing the answer. If you believe it is wrong, you can (a) leave a comment, (b) downvote the answer, (c) post a new answer, and/or (d) vote to delete the answer. – Xander Henderson Jan 30 '25 at 16:23
  • @BastamTajik Banach-Alaoglu theorem makes sense when topology is present, you can't "let it go", that's not how it works. – Jakobian Jan 30 '25 at 16:40
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    I guess there's no point in this discussion. I agree with @XanderHenderson But the counter example is right up there for anyone not content. – Bastam Tajik Jan 30 '25 at 16:45
  • @BastamTajik This may help. – Dean Miller Feb 10 '25 at 21:19
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Here is an essentially complete answer. If the normed linear space $X$ is of the second category, for example a Banach space, then the result is true. Otherwise it is false in general.

Here is a proof when $X$ is of the second category. Let $(f_{n})_{n\in\mathbb{N}}$ be a weak$^{*}$ Cauchy sequence in $X^{*}$. Then for each $x\in X$ the sequence $(f_{n}(x))_{n\in\mathbb{N}}$ is Cauchy and so the set $\{f_{n}(x) : n\in \mathbb{N} \}$ is bounded for every $x\in X$. It follows that the sequence $(f_{n})_{n\in\mathbb{N}}$ is norm bounded in $X^{*}$ by the uniform boundedness theorem, which is applicable since $X$ is of the second category. Then the sequence $(f_{n})_{n\in\mathbb{N}}$ has a weak$^{*}$ limit point $f \in X^{*}$ by the Banach-Alaoglu theorem. Hence $f(x)$ is a limit point of $(f_{n}(x))_{n\in\mathbb{N}}$ for every $x\in X$. Noting that for each $x\in X$ the Cauchy sequence $(f_{n}(x))_{n\in\mathbb{N}}$ converges to its limit point $f(x)$, it follows that $(f_{n})_{n\in\mathbb{N}}$ converges to $f$ in the weak$^{*}$ topology.

If it is no longer assumed that $X$ is of the second category then it cannot be guaranteed that the weak$^{*}$ Cauchy sequence is norm bounded. This is where the argument breaks down as the following counterexample shows. Let $c_{00}$ be the normed space of real-valued sequences with finite support equipped with the $\ell_{1}$-norm. For each $n\in\mathbb{N}$ define $f_{n}\colon c_{00} \to \mathbb{R}$ by $f_{n}(x) := \sum_{k=1}^{n} k\, x(k)$. We have $f_{n} \in (c_{00}, \|\cdot\|_{\ell_{1}})^{*}$ with $\| f_{n} \| = n$ for each $n\in\mathbb{N}$. By the definition of $c_{00}$ the sequence $(f_{n}(x))_{n\in\mathbb{N}}$ is eventually constant for every $x\in c_{00}$. Hence $(f_{n})_{n\in\mathbb{N}}$ is weak$^{*}$ Cauchy. Now suppose for a contradiction there is some $f\in (c_{00}, \|\cdot\|_{\ell_{1}})^{*}$ such that $(f_{n})_{n\in\mathbb{N}}$ converges to $f$ in the weak$^{*}$ topology. For each $k\in\mathbb{N}$ define $e_{k}\colon \mathbb{N} \to \mathbb{R}$ by $e_{k}(n) := \delta_{kn}$. Then for each $k\in\mathbb{N}$ the sequence $(f_{n}(e_{k}))_{n\in\mathbb{N}}$ converges to $k$ and so we have $f(e_{k}) = k$. But then $f$ is an unbounded linear functional which is a contradiction. Therefore the sequence $(f_{n})_{n\in\mathbb{N}}$ is weak$^{*}$ Cauchy but not convergent.

On a final note, the sequence in the counterexample above was not norm bounded. However, if the normed space is not complete then there can exist weak$^{*}$ convergent sequences which are not norm bounded. See here for an example.

Dean Miller
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