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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?

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

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  • $\begingroup$ Thank you! Do you have any idea which conditions I'd need to ensure that it is the projection? $\endgroup$ – soup Apr 9 at 11:30
  • $\begingroup$ @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. $\endgroup$ – Theo Bendit Apr 9 at 11:33
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    $\begingroup$ @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. $\endgroup$ – Theo Bendit Apr 9 at 13:29
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    $\begingroup$ @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 $\|v^-\|$.) $\endgroup$ – Jochen Glueck Apr 9 at 17:18
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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 $Pv = v^+$. $\square$

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