Does the order induced by a self-dual cone produce a Riesz space?

Question

Let $X$ be a Hilbert space, $K$ a closed, convex, and self-dual cone. The latter property means $$ K = \{ y\in X : \ \langle x,y\rangle \ge0 \ \forall x\in K\}. $$ Then $K$ induces an order on $X$ by $$ x \le y \ \Leftrightarrow\ y-x \in K, $$ which makes $(X,\le)$ an ordered vector space.

An particular consequence is that any increasing sequence with upper bound is (strongly) converging, i.e., $x_n\le x_{n+1}\le \bar x$ for all $n$ implies convergence of $(x_n)$.

My question is: Is $(X,\le)$ a Riesz space? That is, given two points $x,y\in X$ then $\{x,y\}$ has a least upper and a greatest lower bound.

If yes, then a result by Penney implies that $(X,\le)$ is isomorph to some $(L^2,\le)$, where the order is induced by non-negative functions.

Answer 1

The answer is no. The Lorentz cone is a counter-example.

To proof this, let me show an implication of the existence of $\sup$. Let me note a simple equivalence $$ z\ge x \ \Leftrightarrow \ z \in x+K. $$

Lemma: Let $(X,K$) be a Riesz space. Then for $z:=\sup(x,y)$ we have $z+K = (x+K)\cap (y+K)$.

Proof: Let $k\in K$. Then $z+k \ge x$ and $z+k \ge y$, which implies $z+k \in (x+K)\cap (y+K)$.

Now let $z+k \in (x+K) \cap (y+K)$. This implies $z+k \ge x$, $z+k\ge y$. Since $z$ is the least upper bound of $x,y$, it follows $z+k\ge z$, $k\ge0$, and $k\in K$. $\square$

The Lorentz cone $$ K = \{ x\in \mathbb R^3: x_3 \ge \sqrt{ x_1^2 + x_2^2}\} $$ is self-dual. But clearly, the intersection property from the lemma above fails to hold. So $(\mathbb R^3,K)$ is not a Riesz space.