If we consider the space of integrable functions $L^1([0,1];\mathbb{R})$, it can be ordered by the convex cone of positive integrable functions $L^1([0,1];\mathbb{R}_+)$. It is known that the corresponding dual cone is $L^\infty ([0,1];\mathbb{R}_+)$. I wonder if the same kind of statement holds when replacing $\mathbb{R}$ by an arbitrary Banach space $Y$?

More formally :

Let $L^1([0,1];Y)$ be the Banach space of Bochner integrable functions from $[0,1]$ to $Y$ (identifying the functions a.e. equal on $[0,1]$). Its topological dual space is known to be $L^\infty_{w^*}([0,1];Y^*)$, the space of $w^*$-measurable functions from $[0,1]$ to $Y^*$ (with also an identification there).

Suppose now that $Y$ is ordered by a closed convex cone, with non-empty interior, that we note $Y_+$. We note $Y_+^*$ the correspoding dual cone, defined by :

$$Y_+^* := \{ y^* \in Y^* \ | \ \langle y^*,y\rangle \geq 0 \ \forall y \in Y_+ \}.$$

Can we prove that the dual cone of $L^1([0,1];Y_+)$ is $L_{w^*}^\infty([0,1];Y^*_+)$ ?

By just applying the definitions, we directly see that $L_{w^*}^\infty([0,1];Y^*_+)$ is included in the dual cone of $L^1([0,1];Y_+)$. But the reverse inclusion seems tricky...