The dual to the space of continuous functions is not isomorphic to $\ell^\infty$

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.

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