This is a follow-up to this answer, which in turn arose in a follow-up to this question.

Let $M([0, 1])$ denote the Banach space of signed Borel measures on $[0,1]$ equipped with the total variation norm. This space is the dual of $C([0, 1])$, via the obvious identification.

Question. Is $M([0, 1])$ isomorphic to $\ell^\infty$?

I expect the answer to be negative.

  • $\begingroup$ There are many properties distinguishing $\ell_\infty$ from $M[0,1]$. $\ell_\infty$ is injective as a Banach space. $\endgroup$ – Adrián González-Pérez Jun 6 at 11:28
  • $\begingroup$ @Adrian: post that as an answer, if you want, with some more details, possibly. $\endgroup$ – Giuseppe Negro Jun 6 at 12:40
  • $\begingroup$ @AdriánGonzález-Pérez: my previous comment was addressed to you, but probably you did not receive it because I misspelled your name. $\endgroup$ – Giuseppe Negro Jun 7 at 13:04

There are several ways to proceed depending on how much machinery you are willing to use.

$\ell_\infty$ is a von Neumann algebra and thus, by Sakai theorem, it has a unique predual. Nevertheless $M([0,1])$ has plenty of nonisomorphic preduals, for instance $M([0,1])$ and $M([0,1]^2)$ are isomorphic while $C([0,1]^2)$ and $C([0,1])$ are not.

In the particular case of $\ell_\infty$ you can see that it has a unique predual directly. Use that $E^\ast = X_1 \oplus_\infty X_2$ implies that $E = E_1 \oplus_1 E_2$ with $E_1$ and $E_2$ the preduals of $X_1$ and $X_2$ respectively. Iterating will give you that $E$ satisfying that $E^\ast = \ell_\infty$ has a nested family $\ell^1(1) \subset \ell_1(2) \subset ... \ell_1(N)$ dense in $E$. Therefore $E \cong \ell_1$.

There other strategies, what I sugested in the comments was using injectivity.

A Banach space $E$ is called injective iff it satisfies a Hanh-Banach theorem if you use as an endpoint space. I.e.: for every $Y \subset X$ and bounded map $\phi: Y \to E$ there is a bounded extension $\tilde{\phi}: X \to E$ with the same norm.

Observation $\ell_\infty$ is trivially injective because we can apply Hanh-Banach in each of the coordinates.

I think, although I do not have a reference at hand, that all injective Banach spaces are $1$-complemented subsets of spaces of the form $C(K)$, where $K$ is totally disconnected.

It is easy to check that a Banach space is injective iff whenever it sits inside a larger space, it is complemented. Therefore it is enough to see that $M([0,1])$ sits as a subset of a larger space in a way that is not complemented. Probably there are examples of this in the literature, but I cannnot came up with one. The closest thing I can think of is:

  • Use that $M([0,1]) \cong \mathbb{C} \oplus_1 \mathbb{C} \oplus_1 M(0,1)$, to reduce the problem to the open interval.
  • Since $(0,1)$ is homeomorphic to $\mathbb{R}$, $M(0,1)$ is isomorphic to $M(\mathbb{R})$.
  • Take $M(\mathbb{R}) \subset L^1(\mathbb{R})^{\ast \ast}$, the double dual of $L^1$. Assume that there is a projection $P: L^1(\mathbb{R})^{\ast \ast} \to M(\mathbb{R})$ to reach contradiction.
  • The unit ball $B$ of $B(L^1(\mathbb{R})^{\ast \ast}, M(\mathbb{R}))$ is comapct in the point weak-$\ast$ topology given by the predual $M(\mathbb{R}) = C_b(\mathbb{R})/n$, where $n$ is the preannihilator of $M(\mathbb{R})$. The projections are a closed convex subset $B_0 \subset B$. There is an action of $\mathbb{R}$ over $B_0$ given by $t \mapsto \tau_t \circ P \circ \tau_{-t}$ where $\tau_t$ is the translation. By amenability, we have a fixed point $P_0 \in B_0$ which would be an equivariant projection (a projection commuting with the translation action). If the original $P$ would preserve $1_{\mathbb{R}}$, i.e. $$\langle P(\psi),1_{\mathbb{R}} \rangle = \langle \psi, 1_{\mathbb{R}} \rangle$$ we would get a contradiction right away since an invariant mean $m \in L^1(\mathbb{R})^{\ast \ast}$ would give a $\mathbb{R}$-invariant and nonzero finite measure $P(m)$. I do not know whether you can take $P$ preserving $1$ without loss of generality.

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.