To complete a proof I need to know if the following is true:

Given a non-empty set $\Gamma$ and a separable subspace $Y$ of $\ell_\infty(\Gamma)$, there exists a subspace $A$ of $\ell_\infty(\Gamma)$ such that it is isometric to $\ell_\infty$ and with $Y \subset A$.

I think that for $\Gamma$ "big enough" it must be true because there is "a lot of room" to form different isometric copies of $\ell_\infty$, however I cannot see how to "put" $Y$ inside of any of this copies.

Any help would be appreciated.

  • $\begingroup$ Clarifying question: When you say $\ell_\infty$ without an argument, you mean $\ell_\infty(\mathbb N)$, so that the question is only nontrivial if $\Gamma$ is uncountable? $\endgroup$ Commented Jun 26 at 20:59
  • $\begingroup$ @e.lipnowski Exactly. $\endgroup$ Commented Jun 27 at 4:10
  • 3
    $\begingroup$ One minor observation: if one wishes for $A$ to be weak$^\ast$-closed, then this is not possible in general. Indeed, $Y = C[0, 1]$ is a separable subspace of $\ell_\infty([0, 1])$, but the weak$^\ast$-closure of $Y$ is the entirety of $\ell_\infty([0, 1])$. $\endgroup$
    – David Gao
    Commented Jun 27 at 7:31
  • $\begingroup$ I've opened a new related threat with an (apparently) more accesible result that can also solve my problem. Here's the link: mathoverflow.net/questions/474046/… $\endgroup$ Commented Jun 27 at 12:38

1 Answer 1


A proof of a positive answer for the question was given to me by professor Antonio Avilés.

Let's take $D \subset Y$ a countable dense subset of $Y$ and for every $y \in D$ and $m \in \Bbb N$ let's take some $\gamma_{y, m} \in \Gamma$ such that $$\| y \|_{\Gamma} - \dfrac{1}{m} \leq |y (\gamma_{y,m})| \leq \| y \|_{\Gamma} \quad (1)$$ where I'm denoting by $\| \|_{\Gamma}$ the sup norm in $\ell_\infty(\Gamma)$.

Then, if we denote by $Q = \{\gamma_{y,m}: y \in D, m \in \Bbb N\}$ we have that $\ell_\infty (Q)$ is isometric to $\ell_\infty$ and that the map $u: Y \rightarrow \ell_\infty (Q)$ given by $$u(x)(\delta) = x(\delta), \forall \delta \in Q$$ for every $x \in Y$ is a linear isometric embedding.

The linearity is clear and it is also clear that $$\|u(x)\|_Q = \sup_{\delta \in Q} |u(x)(\delta)| = \sup_{\delta \in Q} |x(\delta)| \leq \|x\|_\Gamma \quad (2)$$ For the other inequeality, let $x \in Y$ fixed and take any $\varepsilon > 0$. Then, there exists some natural number $m$ such that $1/m < \varepsilon /2$ and some $y \in D$ such that $$\| x - y \|_\Gamma < \dfrac{\varepsilon}{4} \quad (3)$$ Then we have that $$|y(\gamma_{y,m})| \le |y(\gamma_{y,m}) - x(\gamma_{y,m})| + |x(\gamma_{y,m})| \leq \| x-y\|_\Gamma + |x(\gamma_{y,m})| \quad (4)$$ and so by (1), (3) and (4) \begin{align} \|x\|_\Gamma & \leq \|x-y\|_\Gamma + \|y\|_\Gamma < \|x-y\|_\Gamma + |y(\gamma_{y,m})| + \dfrac{1}{m} \leq |x(\gamma_{y,m})| + 2 \|x-y\|_\Gamma + \dfrac{\varepsilon}{2} < \\ & < \sup_{\delta \in Q} |x(\delta)| + \varepsilon = \|u(x)\|_Q + \varepsilon \end{align}

As the $\varepsilon > 0$ is arbitrary we obtain that $$\|x\|_\Gamma \leq \|u(x)\|_Q \quad (5)$$ By (2) and (5) it is proven that $u$ is a linear isometric embedding.

Let now $\pi_\gamma: \ell_\infty(\Gamma) \rightarrow \Bbb R$ be the projection over $\gamma \in \Gamma$, that is, $$\pi_\gamma(x) = x(\gamma), \forall x \in \ell_\infty(\Gamma)$$ Then, if we denote by $\tilde Y = u (Y) \subset \ell_\infty (Q)$, we can define for every $\gamma \in \Gamma$ the maps $\pi_\gamma \circ u^{-1}: \tilde Y \rightarrow \Bbb R$ and extend them to $\tilde \pi_\gamma: \ell_\infty(Q) \rightarrow \Bbb R$ in such a way that:

  • $\|\tilde \pi_\gamma\| \leq 1$
  • If $\gamma \in Q$, then $\tilde \pi_\gamma$ is the projection over $\gamma$ on $\ell_\infty(Q)$, that is, $\tilde \pi_\gamma(z) = z(\gamma),\, \forall z \in \ell_\infty(Q)$

We can do that because if $\gamma \not \in Q$ then the extension is guaranteed by the Hahn-Banach theorem (taking into account that $\|\pi_\gamma \circ u^{-1}\| \leq \|\pi_\gamma\| \|u^{-1}\| = \|\pi_\gamma\| \leq 1$). And, if $\gamma \in Q$, then it is easy to see that the given projection have norm less than one and it is in fact an extension of $\pi_\gamma \circ u^{-1}$ because if $z \in \tilde Y$, then there exists some $x \in Y$ such that $z=u(x)$ and so $$\tilde \pi_\gamma(z) = z(\gamma) = u(x)(\gamma) = x(\gamma) = (\pi_\gamma \circ u^{-1}) (u(x)) = (\pi_\gamma \circ u^{-1}) (z)$$ Lastly, let's define $w: \ell_\infty(Q) \rightarrow \ell_\infty(\Gamma)$ by $$w(z)(\gamma) = \tilde \pi_\gamma(z),\, \forall z \in \ell_\infty(Q), \gamma \in \Gamma$$ We have that $w$ is clearly linear and satisfies that given $z \in \ell_\infty(Q)$:

  • $\|w(z)\|_\Gamma = \sup_{\gamma \in \Gamma} |\tilde \pi_\gamma(z)| \leq \sup_{\gamma \in \Gamma} \|\tilde \pi_\gamma\| \|z\|_Q \leq \|z\|_Q$
  • $\|z\|_Q = \sup_{\delta \in Q} |z(\delta)| = \sup_{\delta \in Q} |z(\delta)| = \sup_{\delta \in Q} |\tilde \pi_\delta(z)| \leq \sup_{\gamma \in \Gamma} |\tilde \pi_\gamma(z)| = \|w(z)\|_\Gamma$

So $w$ is also a linear isometric embedding.

Now, is we define $A = w(\ell_\infty(Q))$, as $\ell_\infty(Q)$ is linearly isometric to $\ell_\infty$ and $w$ is a linear isometry, we have that $A$ is linearly isometric to $\ell_\infty$ and $$w(\tilde Y) \subset A \subset \ell_\infty(\Gamma) \quad (6)$$ But, $$w(\tilde Y) = Y \quad (7)$$ because for every $x \in Y$ we have that $$w(u(x))(\gamma) = \tilde \pi_\gamma (u(x)) = (\pi_\gamma \circ u^{-1}) (u(x)) = \pi_\gamma (x) = x(\gamma),\, \forall \gamma \in \Gamma$$ So, by (6) and (7) we conclude that $Y \subset A \subset \ell_\infty (\Gamma)$ as wanted.


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