# In a vector lattice $V$, is it the case that $\sup(0,v) = P_{V^+}v$? [closed]

We have a reflexive Banach space which is strictly convex. Let $$V$$ be a vector lattice with a partial ordering and write $$V^+ := \{v \in V : v \geq 0\}$$ to be the positive cone. Suppose that it is a closed set. We have the projection operator $$P\colon V \to V^+$$ defined as the closed point in the usual way.

Is it the case that $$P(v) = \sup(0,v)$$? Does the projection agree with the lattice structure?

• What have you tried? Have you tried finding a counterexample, or a proof? – supinf Apr 9 at 11:15
• What is the meaning of $\sup(0,u)$? What is $u$? – mathcounterexamples.net Apr 9 at 11:16
• @mathcounterexamples.net It's defined via the lattice structure. $u$ was supposed to be $v$, was a typo. – soup Apr 9 at 11:29

## 2 Answers

No, not necessarily. Equip $$V = \Bbb{R}^2$$ with the partial order $$(a, b) \le (c, d) \iff (a \le c \text{ and } a + b \le c + d).$$ Then, $$V^+ = \{(x, y) \in \Bbb{R}^2 : x \ge 0 \text{ and } x + y \ge 0\}.$$ Consider the point $$(1, -2)$$. Note that $$(x, y) \ge (1, -2)$$ and $$(x, y) \ge (0, 0)$$ if and only if $$x \ge 0$$, $$x \ge 1$$, $$x + y \ge -1$$ and $$x + y \ge 0$$. Together, this simplifies to $$x \ge 1$$ and $$x + y \ge 0$$, i.e. $$(x, y) \ge (1, -1)$$. Thus, $$(1, -1)$$ is the positive part of $$(1, -2)$$.

But, $$(1, -1)$$ is not the projection of $$(1, -2)$$ onto $$V^+$$; if it were, then the horizontal line line $$y = -1$$ would be tangent to the cone $$V^+$$ at the point $$(1, -2)$$, which is quite clearly not the case.

• Thank you! Do you have any idea which conditions I'd need to ensure that it is the projection? – soup Apr 9 at 11:30
• @soup I'm not sure, but off the top of my head, I would guess that such a property is actually quite rare. I think you'd want the faces of your cone to be orthogonal to each other? Most lattice orders would fail this. – Theo Bendit Apr 9 at 11:33
• @soup Jochen's answer gives a very good sufficient condition that I can't believe I didn't think of. It's been too long since I've looked at Banach lattices. – Theo Bendit Apr 9 at 13:29
• @TheoBendit: Thanks for the advertisement! (There was a stupid mistake in my answer, though, which I have now corrected: the distance of $v$ to the cone is, of course, not $\|v^+\|$ but $\|v^-\|$.) – Jochen Glueck Apr 9 at 17:18

The answer is yes in the important special case where $$V$$ is a Banach lattice. This means that the Banach space $$V$$ is lattice ordered and that the following compatibility property between order and norm is satisfied:

For all $$v, w \in V$$ such that $$\lvert v \rvert \le \lvert w \rvert$$, we have $$\|v\| \le \|w\|$$.

(Equivalently, we have $$\|v\| \le \|w\|$$ for all $$0 \le v \le w$$ in $$V$$ and $$\|\lvert v \rvert\| = \|v\|$$ for all $$v \in V$$.)

Important examples of Banach lattices are (with their standard orders, respectively):

• $$L^p(\Omega,\mu)$$ for every measure space $$(\Omega, \mu)$$ and every $$p \in [1,\infty]$$.

• The space $$C(K)$$ of continuous real-valued functions on a compact Hausdorff space $$K$$, endowed with the supremum norm.

Here is a general result about the distance to the cone in Banach lattices:

Proposition. If $$V$$ is a Banach lattice and $$v \in V$$, then the norm $$\|v^-\|$$ of the negative part $$v^-$$ of $$v$$ coincides with the distance $$\operatorname{d}(v,V_+)$$ of $$v$$ to the positive cone $$V_+$$.

Proof. The inequality $$\operatorname{d}(v,V_+) \le \|v^-\|$$ follows from $$\|v^-\| = \|v - v^+\|$$.

For the converse inequality, let $$w \in V_+$$. Then $$\lvert v - w\rvert \ge (v - w)^- \ge v^-,$$ so $$\|v-w\| \ge \|v^-\|$$. $$\square$$

Corollary. If the Banach lattice $$V$$ is uniformly convex and $$P: V \to V_+$$ denotes the proximum projection onto $$V_+$$, then $$Pv = v^+$$ for each $$v \in V$$.

Proof. Let $$v \in V$$. We have $$\|Pv - v\| = \operatorname{d}(v,V_+) = \|v^-\| = \|v^+ - v\|,$$ so by the uniqueness of the proximum it follows that $$Pv = v^+$$. $$\square$$