# Weak-* Convergence of functionals defined by probability measures $\{\delta_n\}$

Consider the sequence of Dirac delta probability measures $\{\delta_n\}_{n=1}^\infty$ on $\mathbb R$, i.e. the measures which assign probability 1 to the point $n\in\mathbb R$.

These give the associated linear functionals $\Lambda_n$ defined by $\Lambda_n(f) = \int f\,d\delta_n = f(n)$ for all bounded continuous functions on $\mathbb R$, i.e. all $f\in C_b(\mathbb R)$. All these $\Lambda_n$ therefore have norm $\leq 1$.

So while this sequence is not tight and does not have a subsequence converging to a probability measure, it is contained in the compact unit ball of the weak-* topology by the Banach-Alaoglu theorem. So it has a subsequence which converges in weak-* to some linear functional $\Lambda$.

I am having a hard time seeing how this can be so. What can this limit be? How can we have that $\lim_{k\to\infty} f(n_k) = \Lambda(f)$ even exists for every bounded function?

• While the closed unit ball is (topologically) compact in the weak-$*$ topology, this topology is not metrizable, and hence it does not necessarily follow that it is sequentially compact as well. – Joey Zou Sep 23 '16 at 16:20
• Oh wow, thank you! I now see more clearly why we had to use nets to prove the Banach-Alaoglu theorem. – Milind Sep 23 '16 at 16:37
• As a side note, the weak* topology on a dual space $X^*$ is metrizable on bounded sets if $X$ is a separable Banach space (even though not in the whole space, unless the dimension is finite), but this is not the case since $C_b(\mathbb R)$ is not separable. If you considered instead $C_0(\mathbb R)$, being this separable you would obtain a metrizable dual. In fact in this case the limit exists and it is the zero functional. – Del Sep 23 '16 at 17:08

The space $C_b(\mathbb R)$ is not separable. Thus, your sequence $\{\Lambda_n\}$ may not have a weak-* convergent subsequence (and indeed, it does not have any), but it does have a weak-* convergent subnet.