# Positive cone of Banach lattice algebra

From the literature on Banach lattice algebras (and that on ordered Banach algebras) there does not appear to be a consensus on a definition.

What is agreed is that one should be a Banach lattice, be an associative algebra with a sub-multiplicative norm and that the product of positive elements should be positive. To standardise terminology we propose that a Banach lattice algebra simply is at the same time a Banach lattice, an algebra with the sub-multiplicative norm, and with the product of positive elements being positive.

Question: For $$X$$ being a Banach lattice algebra endowed with an ordering $$\leq$$, let $$X_+:=\{x\in X:x>0\}$$ be the positive cone. This cone would generally be open or closed? Please provide proof or hints to prove it with references.

• How do you define $x > 0$? Is it satisfied iff $x \ne 0$ and $x \ge 0$?
– gerw
Apr 24 '17 at 9:49
• actually yes. we say $x>y$ iff $x-y>0$ or $x-y\in X_+$ for $x,y\in X$. Apr 24 '17 at 13:17

Actually, I think the set you defined as $X_+$ is neither closed nor open. Let me change the notation slightly: $X$ is the normed space, $P=\{x\in X: x\geq 0\}$ its positive cone and $P_+=P\setminus \{0\}$ the set of non zero positive elements. Your question regards the topological properties of $P_+$ (although there is also a chance that you mistook $P$ for $P_+$).

First of all, in every Banach lattice, the set $P$ is a closed proper subset. Since normed spaces are connected, it can not be open.

Concerning $P_+$, to see that it is not closed, just pick a nonzero $x_0\in P$ and consider the sequence $(\tfrac{1}{n}x_0)_{n\in \mathbb{N}}$ which is a sequence in $P_+$ that converges to $0\notin P_+$.

Now to show that $P_+$ is not open, you need a little extra work.

1. If $P$ is not a half line, we just need to find a nonzero boundary point of $P$. Lets pick an $x_0\in P_+$ and a $y_0\in X\setminus P$ such that $0$ does not belong in the line segment that connects $x_0$ to $y_0$. It should be easy to find such a pair $(x_0, y_0)$: Just pick a $y_0\in X\setminus P$, set $L=\{\lambda y_0: \lambda \in \mathbb{R}\}$ to be the line passing through $0$ and $y_0$ and pick an $x_0\in P\setminus L$. This set is nonempty since $P$ is not a half line.

Now define $f: [0,1]\rightarrow X$, with $f(\lambda)=\lambda x_0+(1-\lambda)y_0$, for $\lambda\in [0,1]$. Clearly, $f(0)=y_0\in X\setminus P$ and $f(1)=x_0\in P$. Set $\lambda_0=\inf\{\lambda\in [0,1]: f(\lambda)\in P\}$. Since $P$ is closed, $f(\lambda_0) \in P$. Additionally, $f(\lambda) \in X\setminus P$, for every $\lambda<\lambda_0$, since $P$ is convex.

This imples that $f(\lambda_0)\in \partial P$, $f(\lambda)\neq 0 \forall \lambda$, so $f(\lambda_0)$ can't be an interior point of $P_+$.

2. If $P$ is a half line, then your original space can't be a lattice space: Since $P$ is a half line, $P=\{\lambda x_0: \lambda\geq 0\}$, for some nonzero $x_0\in X$. Then for $x, y\in X$,

$$x\leq y \iff y-x=\lambda x_0,$$ for some $\lambda \geq 0$. Now pick $y_0, z_0\in X$ such that the set $\{x_0, y_0, z_0\}$ is linearly independent. Claim: The set of common upper bounds of $y_0$ and $z_0$ is empty: Suppose that $w\in X$ is a common upper bound. Then $$w-z_0=\lambda x_0 \ \ \text{ and }\ \ w-y_0=\lambda' x_0,$$ for some $\lambda, \lambda'\geq 0$. Then $y_0-z_0+(\lambda'-\lambda)x_0=0$, which contradicts the linear independency of these three vectors.

• thank you so much for your effort. Apr 25 '17 at 15:08
• You are welcome. I hope I didn't miss anything Apr 25 '17 at 17:57
• @treedetective What would be the interior of $P?$ Feb 3 '18 at 20:37
• @SahibaArora It's hard to give a concrete description in general. To begin with, there exist Banach lattices for which the positive cone $P$ has an empty interior (for example the $\ell_p$ spaces for $1<p<\infty$. The same holds for the $L_p$ spaces and they provide examples of Banach lattice algebras with an empty cone interior). However, if the cone is nonempty, then its interior coincides with the set of strictly positive elements of $P$. An $x\in P$ is called strictly positive, if $f(x)>0$ for every $f\in P^*\setminus\{0\}$. This is often a very useful characterization. Feb 4 '18 at 16:42
• @treedetective Thank you. Feb 4 '18 at 17:25