I'm a little embarrassed to be asking this, but surely there is a simple argument that I didn't see?
Let $(f_\lambda)$ be a net in $l^\infty$ which converges weak* to $f \in l^\infty$. We do not assume the net is bounded. Does the net $(f_\lambda^+)$ converge weak* to $f^+$, where $f^+ = \max(f,0)$ is the positive part of $f$?
It's false in $L^\infty[0,1]$.
Given a finite set $\cal F$ of functions in $\ell_1$, choose a function $z_{\cal F}$ in $\ell_\infty$ s.t. $\langle z_{\cal F}, x \rangle =0$ for all $x$ in $\cal F$ s.t. $z_{\cal F}$ has at least one positive coordinate, and normalized s.t. $\langle z^+_{\cal F}, u \rangle = 1$, where $u := \sum_{n=1}^\infty 2^{-n} e_n$ and $e_n$ is the $n$th unit vector in $\ell_1$. The net $(z_{\cal F})$ converges weak$^*$ to zero when the finite subsets of $\ell_1$ are directed by inclusion.
$\def\bosy{\boldsymbol}\def\conc{\kern.6mm\boldsymbol{\hat{\phantom{.}}}\kern.4mm}\def\seq#1{\langle\kern.6mm{#1}\kern.6mm\rangle}\def\sp{\kern.4mm}$Since it is not clear from Bill Johnson's answer that the axiom of choice is not needed to construct a net $\bosy n$ giving a negative answer to Nik Weaver's question, I wish to point out it here. Let $\bosy n=(D,n)$ where $D$ is the set of all pairs $(\alpha,\beta)$ such that $\alpha$ and $\beta$ are nonempty finite sets included in $\ell^{\,1}$ with $\alpha\subseteq\beta$. Note that $\ell^{\,1}\subset\ell^{\,2}\subset\ell^{\,+\infty}$, and let $n$ be the set of all pairs $(\alpha,y)$ where $\alpha\in{\rm dom\,}D$ and $y$ is obtained as follows. Order $\alpha$ lexicographically, and successively delete possible elements contained in the linear span of the preceding ones. Let $\bosy a$ be the finite sequence obtained in this way, and let $e$ be the first standard unit vector not in the linear span of ${\rm rng\kern.8mm}\bosy a\,$. Apply Gramm−Schmidt to the concatenated finite sequence $\bosy a\conc\seq{e}\sp$, and let $y_0$ be the last member of the obtained orthonormal finite sequence. Let $i\in\mathbb N_0$ be the first one with $y_0i\not=0$. Put $y_1=y_0$ if $y_0i>0$ and $y_1=-y_0$ otherwise. Let $t=\sum_{\,k\,\in\sp N\,}2^{-1-k}\,(y_1k)$ where $N$ is the set of $k\in\mathbb N_0$ with $y_1k\ge 0$, and put $y=t^{-1}\,y_1$.
