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...



Dual cone of 'positive' Bochner integrable functions
Take some element in the dual cone $g\in L^1([0,1],Y\_+)^\*\subset L^\infty\_{w^\*}([0,1],Y^\*)$. 
Then, by definition, for every $f\in L^1([0,1],Y\_+)$


$$\int\_0^1 \langle f(t),g(t)\rangle\mathbb{d}t\geq0$$


In particular, for any $y\in Y\_+$ and $[a,b]\subset [0,1]$, we can take $f \equiv y\cdot \chi\_{[a,b]}$ so that


$$\int\_a^b \langle y,g(t)\rangle \mathbb{d}t\geq 0.$$


Hence, for every $y\in Y\_+$, and every $t\in[0,1]$ we have $\langle y,g(t) \rangle\geq 0$.
 Thus the image of $g$ is in the positive cone which means that
$$g\in L^\infty([0,1],Y\_+^\*).$$


