Let G be an amenable locally compact group.
Does there exist a left invariant mean $m \colon L^\infty(G) \to \mathbb{C}$ on $L^\infty(G)$ which is in addition weak*-continuous ?
Recall that the von Neumann algebra $L^\infty(G)$ admits $L^1(G)$ as unique predual.
No, except in the trivial case where $G$ is compact.
If $m$ is weak-* continuous then, being a linear functional, it's given by a function $g\in L^1$: $$m(f)=\int_G f(x)g(x)\,dx,$$
where $dx$ is left-invariant Haar measure. (Or maybe it's right-invariant; I'm never sure which is which... Take $dx$ to be the flavor of Haar measure such that $\int\phi(cx)\,dx=\int\phi(x)\,dx$.)
Since $m$ is left-invariant it follows that $g$ is left invariant. So $g$ is constant, and since $m(1)=0$ it follows that $g\ne0$. So there exists $c\ne0$ with $\int_G c=1$, hence $G$ has finite measure.