About duality of Bochner spaces

Question

I have a question about Bochner spaces, especially valued in Lebesgue $L^p$ spaces. I have been reading Analysis in Banach Spaces Volume I by Tuomas Hytönen, Jan van Neerven, Mark Veraar and Lutz Weis. I am confused about the dual of $L^1([0,1],L^1([0,1]))$. From Theorem 1.3.10, we have that the natural embedding $L^{\infty}([0,1],L^{\infty}([0,1]))$ into $L^1([0,1],L^1([0,1]))'$ given by \begin{align*} \phi : h \in L^{\infty}([0,1],L^{\infty}([0,1])) \mapsto \left( \phi \in L^1([0,1],L^1([0,1])) \mapsto \int_{[0,1]}\int_{[0,1]}\phi(x,y)\overline{h(x,y)} \mathrm{d} y \mathrm{d} x\right) \end{align*} is an isomorphism if and only if $L^{\infty}[0,1]$ has the Radon-Nikodym Property. But,the example 1.3.24 shows that it does not have this property. Hence, it means that the embedding above is not an isomorphism. However, I would expect that one can show that for every $p$ in $[1,\infty]$, $L^p([0,1],L^p([0,1]))$ is isomorphic to $L^p([0,1] \times [0,1])$ with the "natural" maps $\psi_p$ such that, for $h$ a simple function $\psi_p(h)=h$ (I am not able to check that). Then, using that the embedding $\tilde{\phi}:L^{\infty}([0,1]^2) \rightarrow (L^1([0,1]^2)'$ is an isomorphism and factorizing the maps, I would hope to be able to show that the duality result can be transferred to $L^{\infty}([0,1],L^{\infty}([0,1])) \cong L^1([0,1],L^1([0,1]))'$ by rewriting $\phi=\psi_1^* \circ \tilde{\psi} \circ \psi_{\infty}$ (where $\psi_1^*$ is the dual operator associated with $\psi_1$), but that would contradict the previous result... I am not sure where this fails. Does someone have an idea about that? In particular, is there a condition to guarantee that the there is a "natural" isomorphism between $L^p(X,L^p(Y,\mathbb{R}))$ for measured spaces $X$ and $Y$?

Answer 1

I will answer your question in a low-brow way, by showing that the "obvious map" doesn't work. In particular $L^\infty([0,1]^2)$ is not the same as $L^\infty([0,1],L^\infty([0,1]))$.

Let $f(x,y) = 1$ if $|x-y| < \frac15$ and $0$ otherwise. It is clear that this function is in $L^\infty([0,1]^2)$.

Let $g_x(y) = f(x,y)$. Then $x\mapsto g_x$ is a mapping from $[0,1]\to L^\infty([0,1])$. (Note that this mapping satisfies, whenever $x\neq x'$, that $\|g_x - g_{x'}\|_\infty = 1$. So $x\mapsto g_x$ is everywhere discontinuous.)

Now, remember that elements of the Bochner space $L^\infty([0,1],L^\infty([0,1]))$ are strongly measurable functions from $[0,1]\to L^\infty([0,1])$, which requires that they be pointwise limits of a sequence of simple functions. Simple functions have range a finite set, and hence strongly measurable functions have separable range. (This is probably in the book you cited.) But the range of $x\mapsto g_x$ is an uncountable discrete set (any pair of distinct points have distance 1). So $x\mapsto g_x$ is not strongly measurable.

(This is a general problem when you look at strongly measurable functions into non-separable spaces.)