Corollary

In $L^{\infty}(G)$ endowed with the weak topology, the set Pos(G) is closed and the set Pos$_{\leqslant 1}(G)$ is compact.

The first assertion follows from the proposition, and the second then follows from EVT, III, loc. cit.

In general, Pos$_1(G)$ is not compact for the weak topology, as is shown by the example of the group $\mathbf{R}$ (exercise 10 of V, p. 497).