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...

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

1 Answer 1


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

  • 1
    $\begingroup$ Nice! A minor detail (which does not invalidate the proof): in the last part the inequality holds for almost every t (not every t). $\endgroup$
    – Guillaume
    Nov 20, 2018 at 11:49

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