Let $G$ be a locally compact group. A mean $M$ on $L^{\infty}(G)$ is a continuous linear functional on $L^{\infty}(G)$ such that $1 = \langle 1 , M\rangle = \|M\|$.
My Exercise: Show that the set $\{f\in L^{1}(G): f\geq 0, \|f\|_{1}=1\}$ is weak-$*$ dense in the set of means on $L^{\infty}(G)$.
The text has already proven that if we assume that for a mean $M$ on $L^{\infty}(G)$, we have $\langle \phi, M\rangle \geq 0$ for every $\phi\geq 0$.
By Goldstine's theorem, $(L^{1}(G))_{1}:=\{f\in L^{1}(G): \|f\|_{1}\leq 1\}$ is weak-$*$ dense in $(L^{\infty}(G)^{*})_{1}$.
Therefore I can choose a net $(f_{\alpha})$ in $L^{1}(G)$, with $\|f_{\alpha}\|_{1}\leq 1$ such that $\langle \phi, f_{\alpha}\rangle\to \langle\phi, M\rangle$.
I'm trying to show that the $f_{\alpha}$ can be chosen to be non-negative of norm $1$.
Intuitively, I just replace $f_{\alpha}$ with $g_{\alpha}(y):= \begin{cases}f_{\alpha}(y) & \text{ if }f_{\alpha}(y)\geq 0\\0 &\text{ otherwise}\end{cases}$, and then normalize them by setting $h_{\alpha} := \begin{cases}\frac{1}{\|g_{\alpha}\|}g_{\alpha} &\text{ if }g_{\alpha}\neq 0\\0&\text{ otherwise.}\end{cases}$.
It is immediate that $h_{\alpha}$'s are non-negative of norm $1$ (if we throw away the ones which are $0$). But it feels like I've lost weak-$*$ convergence to $M$.
I try to compute, for any $\phi\in L^{\infty}(G)$, \begin{align*} \left|\lim_{\alpha} \langle \phi,h_{\alpha}\rangle - \langle \phi, M\rangle\right| &= \left|\lim_{\alpha} \langle \phi,h_{\alpha}\rangle - \lim_{\alpha}\langle \phi, f_{\alpha}\rangle\right|\\ &= \lim_{\alpha}\left| \langle \phi,h_{\alpha}\rangle - \langle \phi, f_{\alpha}\rangle\right|\\ &\leq \lim_{\alpha}\|\phi\|\int_{G}|f_{\alpha} - h_{\alpha}|d\mu\\ &= \|\phi\|\cdot\lim_{\alpha}\|f_{\alpha} - h_{\alpha}\|_{1} \end{align*} where $\mu$ is a left Haar measure on $G$.
I would be done if I could show $\lim_{\alpha}\|h_{\alpha} - f_{\alpha}\|_{1} = 0$, but all of my attempts failed, and I suspect it may not be true.
I realize now it is best to view $f_{\alpha}$ as a linear combination of four elements of the positive cone of $(L^{\infty}(G))^{*}$, and replace each $f_{\alpha}$ with its positive real part, which must only converge faster to $M$ since $M$ is positive. Then normalizing each element doesn't harm anything and the problem is solved.
I noticed a problem with my earlier logic: I assumed each $f_{\alpha}$ was real valued without even asking them to be.