Weakly*-Dense Subspaces of $\ell^\infty$

Question

Identify $\ell^\infty=(\ell^1)^*.$ I'm interesting in whether there exists a closed, weakly$^*$-dense subspace $F$ of $\ell^\infty$ s.t. $c_0\not\subseteq F.$

My hope is that the answer is no, but I concocted the following space. It is clearly closed, but I'm struggling to show that it is not w$^*$-dense.

Consider the closed subspace $F$ in $\ell^\infty$ by $$F=\left\{f\in\ell^\infty\middle|\lim_{k\rightarrow\infty}f_k=f_1\right\}$$ Is $F$ weakly$^*$-dense in $\ell^\infty?$

Update: As David Gao has pointed out, $F$ is in fact weakly$^*$-dense. Moreover, there are no proper closed subspaces of either $c_0$ or $F$ which are weakly$^*$-dense, so they are both minimal in the poset of closed, w$^*$-dense subspaces of $\ell^\infty.$

But its norm-closure contains $c_{00},$ and so $c_0,$ no? @Mittens — Miles Gould, Jan 26 '25 at 18:17
That is why you are adding a constant different from $0$ at the tail. The sequence that I propose are of the form $(x_1,\ldots,x_n,a,a,\ldots,a,\ldots)$ with $a\neq0$. These set does not contain $\mathcal{c}_0$. — Mittens, Jan 26 '25 at 18:18
@Mittens Ah, for a fixed $a.$ I thought you meant for all nonzero $a.$ I see. — Miles Gould, Jan 26 '25 at 18:20
Correct. This is also a subset of the space $\mathcal{c}_0\oplus \mathbb{R}$ — Mittens, Jan 26 '25 at 18:21
@Mittens I suppose a better question is: Does $c_0$ admit a proper closed w$^*$-dense subspace? — Miles Gould, Jan 26 '25 at 18:28
I think $c_{00}$ is dense in $c_0$ in the $\sigma(\ell_\infty,\ell_1)$ (and also $\sigma(c_0,\ell_1)$. — Mittens, Jan 26 '25 at 18:31
@MilesGould For your original question: I suppose you already figured out $F$ is in fact weak$^\ast$ dense? So no, norm-closed, weak$^\ast$ dense subspaces of $\ell^\infty$ need not contain $c_0$. For the question you’re asking in comments: No, no norm-closed, proper subspace of $c_0$ can be weak$^\ast$ dense in $\ell^\infty$. Indeed, if $F\subset c_0$ is a norm-closed proper subspace, then there exists nonzero $\varphi\in(c_0)^\ast$ s.t. $\varphi(F)=0$. But then $\varphi\in\ell^1$, so $F$ is contained in the kernel of $\varphi$, which is weak$^\ast$ closed, so $F$ is not weak$^\ast$ dense. — David Gao, Jan 26 '25 at 19:20
I haven't figured out whether $F$ is weakly$^*$-dense yet. I realized that their example is an affine space, not a vector space. @DavidGao — Miles Gould, Jan 26 '25 at 19:22
@MilesGould For any $(x_k) \in \ell^\infty$, $(x_1, \cdots, x_n, x_1, x_1, \cdots)$ is in the space $F$ you defined in your question. As $n \to \infty$, $(x_1, \cdots, x_n, x_1, x_1, \cdots) \to (x_k)$ weak$^\ast$. — David Gao, Jan 26 '25 at 19:26

score 4 · Accepted Answer · answered Jan 26 '25 at 19:37

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Let me just turn my comments into an answer to resolve this question:

The closed subspace $F$, as defined in the original question, is weak$^\ast$ dense in $\ell^\infty$. To see this, note that for any $(x_k) \in \ell^\infty$, the sequence $(x_1, \cdots, x_n, x_1, x_1, \cdots)$ belongs to $F$ and, as $n \to \infty$, $(x_1, \cdots, x_n, x_1, x_1, \cdots) \to (x_k)$ weak$^\ast$. As $c_0 \not\subset F$, this means norm-closed weak$^\ast$ dense subspaces of $\ell^\infty$ need not contain $c_0$.

For the question asked in comments, namely, whether any norm-closed proper subspace of $c_0$ can be weak$^\ast$ dense in $\ell^\infty$, the answer is no. Indeed, if $F \subset c_0$ is a norm-closed proper subspace, then there exists nonzero $\varphi \in (c_0)^\ast$ s.t. $\varphi(F) = 0$. But then $\varphi \in (c_0)^\ast = \ell^1 = (\ell^\infty)_\ast$, so $F$ is contained in $\text{ker}(\varphi)$, which is a weak$^\ast$ closed proper subspace of $\ell^\infty$. So $F$ is not weak$^\ast$ dense.

answered Jan 26 '25 at 19:37

David Gao

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Very succinct, thank you! I suppose I was about half-right, in the sense that $c_0$ is minimal, but not uniquely minimal. – Miles Gould Jan 26 '25 at 19:40
I have another question: Does $\ell^1=(c_0)^$ have a proper closed, w$^$-dense subspace? – Miles Gould Jan 26 '25 at 19:55
@MilesGould Of course. Just take ${(a_k) \in \ell^1: \sum_k a_k = 0}$, for example. (In general, if $V = W^\ast$ is not reflexive, then you can always find a nonzero linear functional $\varphi \in V^\ast \setminus W$. The kernel of $\varphi$ is then proper closed and weak$^\ast$ dense.) – David Gao Jan 26 '25 at 19:59
That's a great tool, thank you! – Miles Gould Jan 26 '25 at 20:09
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Nice answer. The argument used to answer the question in the comments can be applied to obtain the following result. Let $X$ be a normed space and $J\colon X \to X^{}$ the canonical embedding. Let $F \subseteq J(X)$ be a norm closed and weak$^{*}$ dense subspace of $X^{}$. Then $F = J(X)$. For this question one takes $X = c_{0}$ and identifies $X^{**}$ and $J(X)$ with $\ell^{\infty}$ and $c_{0}$ respectively. – Dean Miller Jan 26 '25 at 21:12
@DeanMiller What if $J(X)=c$ and $X^{**}=\ell^\infty?$ Then $F=c_0$ violates, no? – Miles Gould Jan 27 '25 at 01:33
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@MilesGould The double dual of $c$ is $\ell^\infty$, but the canonical embedding into the double dual $c \to \ell^\infty$ is not the usual embedding $c \to \ell^\infty$. This is because, while the dual of $c$ is $\ell^1$, the underlying set of $\ell^1$ has a point at infinity, instead of just being $\mathbb{N}$ (taking limit is a bounded linear functional on $c$). So $J(c)$, if you think it through, is actually the space $F$ you defined in your question instead of the usual copy of $c$ in $\ell^\infty$. And $F$ does not contain $c_0$ and is indeed minimal. – David Gao Jan 27 '25 at 02:41
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@MilesGould (In fact, a more elaborate argument shows that the minimal norm-closed, weak$^\ast$ dense subspaces of $\ell^\infty$ are exactly spaces of the form $J(X)$ for $X^{\ast\ast} = \ell^\infty$, or equivalently spaces of the form $J(X)$ for $X^\ast = \ell^1$. And there are lots of such spaces.) – David Gao Jan 27 '25 at 02:58
Ah right, it will be something like $c^{**}=\ell^\infty\oplus\mathbb{C}$ under the $c_0$ identification, right? – Miles Gould Jan 27 '25 at 03:43
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@MilesGould I’m not sure what you meant by “under the $c_0$ identification”, but yeah, you can take $c^{\ast\ast} = \ell^\infty \oplus \mathbb{C}$, in which case $c$ sits as the subspace ${(f, r) \in \ell^\infty \oplus \mathbb{C}: \lim f = r}$. – David Gao Jan 27 '25 at 03:50
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(Oh, I just realized. A correction to one of my previous comments: the proof I had in mind for the claim that any minimal norm-closed, weak$^\ast$ dense subspace of $\ell^\infty$ is of the form $J(X)$ for $X^{\ast\ast}=X$ made an implicit assumption that the unit ball of $F$ is also weak$^\ast$ dense in the unit ball of $\ell^\infty$. This is fine if $F$ is also a $\ast$-subalgebra, by Kaplansky density theorem, but I’m not sure if this is true for general such $F$. Instead, the correct claim should have been any minimal norm-closed, weak$^\ast$ dense subspace of $\ell^\infty$ is of the form… – David Gao Jan 27 '25 at 04:45
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… $\phi(J(X))$ where $\phi:X^{\ast\ast}\to\ell^\infty$ is a not necessarily isometric isomorphism that is weak$^\ast$-weak$^\ast$ continuous and which restricts to an isometry from $X$ to $F$. It could be the case that this already implies $\phi$ is isometric? I don’t know enough about Banach space theory to be sure.) – David Gao Jan 27 '25 at 04:46
If I update the question, could you express this proof in a second answer? My original goal was to see if the minimal closed, w$^$-dense subspaces were a suitable generalization of preduals, so I'm hoping that I can show that there are instances where such an isomorphism $\phi$ is not isometric, especially in the case where $X$ is C$^$ and not W$^*.$ – Miles Gould Jan 27 '25 at 16:35
@MilesGould Do you mean generalization of double preduals instead of preduals…? And while I now realized there are instances in which $\phi$ is not isometric - as an example, $F = {f \in \ell^\infty: \lim f = \alpha f_1}$ for any $\alpha > 1$, these examples are rather artificial, and in all such cases that I can think of, you actually do have $X^{\ast\ast}$ is isometrically isomorphic to $\ell^\infty$, just that $\phi$ is not that isometric isomorphism. While there are spaces which are isomorphic but not isometrically isomorphic to $\ell^\infty$, for example $L^\infty([0, 1])$, none of… – David Gao Jan 27 '25 at 21:30
… them that I can think of satisfy the requirement that $\phi$ has to be weak$^\ast$-weak$^\ast$ continuous and restrict to an isometry on $X$ (or even have a double predual at all). And I’m not sure what you meant by $X$ is $C^\ast$ and not $W^\ast$? As long as $\phi(J(X))$ is a $C^\ast$-subalgebra of $\ell^\infty$, $\phi$ will be isometric. Anyway, I’d suggest you ask a new question instead of editing the current one. It is generally the policy of MSE to discourage editing a question to ask a substantially different question. You can always link to the current question to provide context. – David Gao Jan 27 '25 at 21:33
I’m a bit confused: If $\phi$ restricts to an isometry on $X,$ why would its larger behavior on $X^{**}$ be relevant? It seems we only care about the image of $X.$ By the way, I posted the question and bountied it, so feel free to post this proof and I’ll reward it. – Miles Gould Feb 02 '25 at 04:27
@DeanMiller After trying to recreate the proof, it seems to only hold when the codomain $Y^{*}$ has a unique isometric predual, like in the case of $\ell^\infty,$ since it is a W$^$-algebra. I was only able to prove it for a general $Y,$ for $J(X)$, where $X^\cong Y^.$ – Miles Gould Feb 03 '25 at 17:00
@MilesGould Are you referring to my comment or one of David's comments regarding the proof? – Dean Miller Feb 04 '25 at 04:13

Weakly*-Dense Subspaces of $\ell^\infty$

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