Compute the dual of the $L^1$ space of $L^1$-valued functions (Lebesgue-Bochner space)

Question

What is the dual of the space $X=L^1(\mathbb R_{-}, Y)$, where $Y=L^1(\mathbb R)$?

How could the paring between an element of the dual and an element of $X$ be represented?

Answer 1

Perhaps surprisingly, this space is strictly larger than $L^\infty(\mathbb{R}_-, Y^*)$. More generally, the dual of $L^p(\mathbb{R}_-, Y)$, $1\le p<\infty$, is strictly larger than $L^{p'}(\mathbb{R}_-, Y^*)$. I reproduce its description from the book Martingales in Banach spaces by Pisier.

Let $\mathcal B$ be the Borel $\sigma$-algebra on $\mathbb{R}_-$. Let $\{\mathcal A_n\}$ be a collection (called a filtration) such that:

• each $\mathcal A_n$ is a finite subalgebra of $\mathcal B$
• $\mathcal A_0\subset \mathcal A_1\subset \cdots$
• The $\sigma$-algebra generated by $\bigcup \mathcal A_n$ is $\mathcal B$

For example, one can take the interval $[-2^n, 0]$, divide it in $4^n$ equal subinterval and let $\mathcal A_n$ be the algebra generated by these subintervals.

The space $L^p((\mathbb{R}_-, \mathcal A_n); Y)$ of $\mathcal A_n$-measurable $Y$-valued functions is isometrically isomorphic to $L^{p'}((\mathbb{R}_-, \mathcal A_n); Y^*)$, thanks to the finiteness of $\mathcal A_n$; this is essentially the identity $(Y\oplus_p Y)^* = Y^*\oplus_{p'} Y^*$.

A sequence of functions $M_n \in L^p((\mathbb{R}_-, \mathcal A_n); Y^*)$ is a martingale if $M_n=E[M_{n+1} | \mathcal A_n]$ where $E$ is conditional expectation. Let $h^p(\mathbb{R}_-, (\mathcal A_n), Y^*)$ denote the space of such martingales with finite norm $\|(M_n)\| = \sup_n \|M_n\|_p$.

Proposition 2.20 in Pisier's book states that the dual of $L^p(\mathbb{R}_-; Y)$ is isometrically isomorphic to $h^{p'}(\mathbb{R}_-, (\mathcal A_n); Y^*)$, for every $1\le p<\infty$, and for every Banach space $Y$.

Sketch of the proof: Every continuous linear functional on $L^p(\mathbb{R}_-; Y)$, restricted to the subspace $L^p((\mathbb{R}_-, \mathcal A_n); Y)$, is identified with a function $M_n \in L^{p'}((\mathbb{R}_-, \mathcal A_n); Y^*)$ and it's easy to check that the sequence $(M_n)$ is an $L^{p'}$-bounded martingale.

Conversely, every martingale in $h^{p'}(\mathbb{R}_-, (\mathcal A_n); Y^*)$ induces a sequence of linear functionals on the subspaces $L^p((\mathbb{R}_-, \mathcal A_n); Y^*)$ that extend each other and have uniformly bounded norms; and since the union of subspaces $L^p((\mathbb{R}_-, \mathcal A_n); Y)$ is dense in $L^p(\mathbb{R}_-; Y)$, we get an element of $L^p(\mathbb{R}_-; Y)^*$ from this martingale. $\quad\Box$

The space $L^{p'}(\mathbb{R}_-; Y^*)$ is isometrically isomorphic to a subspace of $h^{p'}(\mathbb{R}_-, (\mathcal A_n); Y^*)$, since every $L^{p'}$ function induces an $L^{p'}$-bounded martingale.

The final piece is the result I will not prove, referring to Pisier's book: these two spaces are the same (i.e., all $L^{p'}$ bounded martingales come from $L^{p'}$ functions) if and only if $Y^*$ has the Radon–Nikodym property. And the space $L^1(\mathbb{R})^* = L^\infty(\mathbb{R})$ fails the Radon–Nikodym property, as does $L^1(\mathbb{R})$ itself.