Tensor product of measure spaces

Question

For a compact topological space $X$, denote by $\mathcal{M}(X)$ the Banach space of finite signed Borel (Radon) measures on $X$ with the total variation norm. This is canonically isometric to the dual space of the space of continuous functions $C(X)$.

If $X$ and $Y$ are compact spaces, then we have the isometric isomorphism $$\mathcal{M}(X) \hat{\otimes}_\pi \mathcal{M}(Y) \cong \mathcal{M}(X \times Y),$$ where on the left side, we have the completed projective tensor product (Side question: Does anybody know a reference for this statement? I only found the corresponding statement for the subspaces $L^1$, although the proof is very similar). [EDIT: this claim is incorrect in general -- see below]

Now of course, one can also complete with respect to the injective tensor product, where the norm is given by $$\varepsilon(z) = \sup_{\xi, \eta} \bigl\{ (\xi \otimes \eta)(z) \mid \xi \in \mathcal{M}(X)^\prime, \eta \in \mathcal{M}(X)^\prime, \|\xi\| = \|\eta\| = 1\bigr\}$$ for $z \in \mathcal{M}(X) \otimes \mathcal{M}(Y)$ (the algebraic tensor product). However, since the measure space are dual spaces, there is also a third alternative, namely $$\tilde{\varepsilon}(z) = \sup_{f, g} \bigl\{ z(f \otimes g) \mid f \in C(X), g \in C(Y), \|f\| = \|g\| = 1\bigr\}.$$ Notice that (at least formally), the latter a much smaller norm, since die dual space of $\mathcal{M}(X)$ are huge. It seems to me that here we have the isomorphism $$\mathcal{M}(X)\hat{\otimes}_{\tilde{\varepsilon}} \mathcal{M}(Y) \cong B\bigl(C(X) \times C(Y)\bigr),$$ with the space of bounded bilinear maps on $C(X) \times C(Y)$. Is this correct?

In any case, we have the inclusions $$\mathcal{M}(X) \hat{\otimes}_\pi \mathcal{M}(Y) \subseteq \mathcal{M}(X)\hat{\otimes}_{{\varepsilon}} \mathcal{M}(Y) \subseteq \mathcal{M}(X)\hat{\otimes}_{\tilde{\varepsilon}} \mathcal{M}(Y)$$

Now the questions:

• Is there a characterization of the middle space?
• Are there "good" (i.e. somewhat natural examples) that show that this inclusion is in fact strict?

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Edit: As Yemon Choi notices, I was possibly wrong by claiming that $\mathcal{M}(X) \hat{\otimes}_\pi \mathcal{M}(Y) \cong \mathcal{M}(X \times Y)$; in fact, it is only clear that it is an isometrically embedded subspace. Furthermore, by the comment of Matthew Daws, we have $\varepsilon = \tilde{\varepsilon}$.

Edit2: Thanks to the answers below, we obtain the following picture: First, we have $$ \mathcal{M}(X) \hat{\otimes}_\varepsilon \mathcal{M}(Y) \cong \mathcal{M}(X) \hat{\otimes}_{\tilde{\varepsilon}} \mathcal{M}(Y),$$ using the fact that the inclusion of the closed unit ball of $C(X)$ into the closed unit ball of $\mathcal{M}^\prime(X)$ is dense with respect to the weak-$*$-topology on $\mathcal{M}^\prime(X)$. From this follows that the norms $\varepsilon(z)$ and $\tilde{\varepsilon}(z)$ coincide.

Now due to the great answer by Matthew Daws below, the diagonal measure $\mu \in \mathcal{M}([0, 1]^2$ is an example of an element which is not contained in $\mathcal{M}([0, 1]) \hat{\otimes}_\varepsilon \mathcal{M}([0, 1])$, hence neither in $\mathcal{M}([0, 1]) \hat{\otimes}_\pi \mathcal{M}([0, 1])$.

In fact studying some literature tells me that $\mathcal{M}(X \times Y)$ can be identified with the subset of integral forms of $\mathrm{Bil}(C(X) \times C(Y))$. However, it is not yet clear to me yet whether every element of $\mathcal{M}(X) \hat{\otimes}_\varepsilon \mathcal{M}(Y)$ is actually integral, i.e. contained in $\mathcal{M}(X, Y)$, and whether the inclusion $$\mathrm{Bil}(C(X) \times C(Y)) \supseteq \mathcal{M}(X \times Y)$$ is really a proper inclusion.

Answer 1

Let me work with general Banach spaces (good references are Vector Measures by Diestel and Uhl, or Introduction to Tensor Products of Banach Spaces by Ryan).

For Banach spaces $E,F$ the bounded bilinear maps on $E\times F$ can be identified with the dual space of $E \hat\otimes_\pi F$ (this is a special case of the universal property of the norm $\pi$) which in turn can be identified with $B(E,F^*)$ in the obvious way. The injective tensor norm agrees with the norm induced by embedding $E^*\otimes F^*$ into $B(E,F^*)$; the closure is the approximable operators. As $M(X)$ has the (metric) approximation property we find that $M(X) \hat\otimes_\epsilon M(Y)$ is the compact operators from $C(X) \rightarrow M(Y)$.

An example where $M(X) \hat\otimes_\epsilon M(Y)$ is not the bounded bilinear maps on $C(X) \times C(Y)$.

Let $X=Y=[0,1]$ with Lebesgue measure and consider $T:C(X) \rightarrow L^1(X) \subseteq M(X)$. By considering e.g. the functions $f_n(t) = \exp(2\pi itn)$ we see that $T$ is not compact.

Furthermore, we do not have that $M(X\times Y) \subseteq M(X) \hat\otimes_\epsilon M(Y) = K(C(X),M(Y))$.

Again let $X=Y=[0,1]$ and let $\mu\in M(X\times Y)$ be $$ \langle \mu, f \rangle = \int_{[0,1]} f(t,t) \ dt \qquad (f\in C([0,1]^2) $$ so $\mu(E\times F) = |E\cap F|$ the Lesbegue measure of $E\cap F$, for $E,F\subseteq [0,1]$ measurable. This does induce a bounded linear map $T:C([0,1]) \rightarrow M([0,1])$ by $$ \langle T(f), g \rangle = \langle \mu, f\otimes g \rangle = \int_{[0,1]} f(t) g(t) \ dt \qquad (f,g \in C([0,1])). $$ But this is just the same $T$ as I considered above: the inclusion of $C([0,1])$ into $L^1([0,1]) \subseteq M([0,1])$; this is not compact.

This $\mu \in M(X\times Y)$ is also an example of an element not in $M(X) \hat\otimes_\pi M(Y)$.

I don't know an elementary way to see this; but the theory is interesting. Again see the references at the start. A vector measure $\lambda$ on a measure space $X$ is a countably additive map $\lambda:X\rightarrow E$, for a Banach space $E$. $\lambda$ is regular if $\varphi\circ\lambda\in M(X)$ is regular for each $\varphi\in E^*$. Every weakly compact operator $T:C(X)\rightarrow E$ has the form $$ T(f) = \int_X f \ d\lambda \qquad (f\in C(X)), $$ for a regular vector measure $\lambda$.

Let $\lambda_0$ be a finite positive measure on $X$. Then $\lambda$ is absolutely continuous with respect to $\lambda_0$ exactly when $\lambda_0(E)=0 \implies \lambda(E)=0$. Say that $\lambda$ has the Radon-Nikodym property if, whenever $\lambda$ is absolutely continuous with respect to $\lambda_0$, then $\lambda = \int F \ d\lambda_0$ for some Bochner integrable $F:X\rightarrow E$. (The point is that not all vector measures have this property, in contrast to the scale case). An operator $T:C(X)\rightarrow E$ comes from some member of $M(X) \hat\otimes_\pi E$ (that is, is a nuclear operator) if and only if the representing vector measure has the Radon-Nikodym property.

Again with $X=Y=[0,1]$ with $\mu$ and $T$ as before, the associated vector measure $\lambda:[0,1]\rightarrow M([0,1])$ is $$ \lambda(E) : F \mapsto |E\cap F| $$ for measurable $E,F\subseteq [0,1]$. This is absolutely continuous with respect to Lebesgue measure. If $F:[0,1]\rightarrow M([0,1])$ is a Bochner integrable function inducing $\lambda$ then $$ \langle \lambda(E), f \rangle = \int_E f(t) \ dt = \int_E \langle F(s), f \rangle \ ds \qquad (f\in C([0,1])). $$ Thus $F(s) = \delta_s$ the point-mass at $s$, for all $s\in [0,1]$. However, by the Pettis Measurability Theorem, $F$ should be essentially separably valued but clearly this is not the case, giving the required contradiction.

My guess would be that you can boost this example up to show that $M(X)\hat\otimes_\pi M(Y)$ is strictly contained in $M(X\times Y)$ unless one of $X$ or $Y$ is discrete (in which case we do have equality).