Q:

Weak* continuity

Let $B$ be the open unit ball in $\mathbb{R}^2$ and $\mathcal{M}^+$ the set of nonnegative Radon measures on $B$ and $\mathcal{M}^2$ the set of $\mathbb{R}^2 \text{-valued}$ Radon measures on $B.$ I am struggling to see that the map
$I \colon \mathcal{M}^+ \times \mathcal{M}^2 \to \mathbb{R}, \quad 
(\alpha, \mu) \mapsto \int_B (1-f) \, d\alpha + \int_B g \cdot d\mu$
is weakly* continuous, where $f \in C_c(B)$ and $g \in [C_c(B)]^2.$
How do I approach that?
Or more general, how do I show that a function $h \colon X \to \mathbb{R}$ is weakly* continuous for a normed vectorspace $(X,\| \cdot \|)?$
Thanks for any kind of help

A:

First, you should note that weak-* continuity is only defined for functions, whose domain is a dual space, such as $\mathcal{M}(B) = C_0(B)'$.
For a general dual space $X'$, there are two typical weak-* continuous maps: $x^* \mapsto \|x^*\|$ and $x^* \mapsto x^*(x)$, where $x \in X$ is fixed.
Concerning your example, the map
$$(\alpha,\mu) \mapsto -\int_B f \, d\alpha + \int_B g \, d\mu$$
is weak-* continuous, since $C_c(B) \subset C_0(B)$.
It remains to check $\alpha \mapsto \int_B \,d\alpha = \alpha(B)$. This map, however, is not weak-* continuous:
Let $x_n \in B$ be a sequence such that $x_n \to x \in \partial B$.
Now, let $\alpha_n = \delta_{x_n}$ be the sequence of Dirac measures at $x_n$.
It is easy to check that $\delta_{x_n}$ converges weak-* to $0$, but
$$\delta_{x_n}(B) = 1 \not\to 0.$$
Hence, your map is not weak-* continuous.

