# Copy of $\ell_\infty$ inside $\ell_\infty(\Gamma)$ containing given subspace

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.

• 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? Commented Jun 26 at 20:59
• @e.lipnowski Exactly. Commented Jun 27 at 4:10
• 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 $\ell_\infty([0, 1])$. Commented Jun 27 at 7:31
• 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/… Commented Jun 27 at 12:38

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.