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Let $B$ be a Banach space and $K$ a closed proper cone in $B$ such that the induced partial order makes $B$ a vector lattice. Let $K'=\{x^*\in B':\langle x^*, x\rangle\geq 0\ \forall x\in K\}$ the positive cone in the topological dual $B'$. Is it true that any (topologically) continuous linear functional which is order-bounded belongs to $K'-K'$?

In other words, let $x^*:B\to \mathbb{R}$ be a linear order bounded functional. Then, being $B$ a lattice, there exists $x^*\lor 0$, which is a linear order bounded functional on $B$. Is it true that $x^*\lor 0$ is topologically continuous if $x^*$ is so?

The main concern is when $K$ is not normal, such as $B=H^1([0,1])$ with the natural norm and ordering given by $K=\{u\geq 0\}$, but I am still not able to find a counterexample in this space.

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I found out in Peressini's book "Ordered topological vector spaces" that any positive linear functional on a Banach space ordered by a closed generating cone $K$ is actually topologically continuous, regardless of the normality of $K$. The result seems due to Kakutani.

Regarding my question: since $(B, K)$ is a lattice $K$ is generating, and thus $x^*\lor 0$ is topologically continuous for any order bounded, linear $x^*$. Any order bounded functional $x^*$ can be written as $x^*=x^*\lor 0+x^*\land 0$ (because the order dual of a lattice is a lattice). Thus any order bounded linear functional is topologically continuous.

So the answer is yes.

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