Dual cone of 'positive' Bochner integrable functions

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...

• Why is the reverse inclusion tricky? Take a separating hyperplane, that leads to a contradiction. Mar 31, 2016 at 7:47

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 $$g\in L^\infty([0,1],Y_+^*).$$