In a vector lattice $V$, is it the case that $\sup(0,v) = P_{V^+}v$? 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?
 A: 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.
A: 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 $Pv = v^+$. $\square$
