a question on weak * convergence

Question

It is written in an article that $\rho_{n_k}*\mu_k$, where $\mu_{k}=k^{-1}|\Omega|\sum_{i=1}^{k}\delta_{a_i}$, converges to $1$ in the weak $*$ notion, i.e., $\int_{\Omega}\phi(\rho_{n_k}*\mu_k) dx\rightarrow\int_{\Omega}\phi dx$ for every $\phi\in C_0(\overline{\Omega})$ as $k\rightarrow\infty$, where $[C_0(\overline{\Omega})]^{*}=\mathfrak{M}(\Omega)$, $\mathfrak{M}(\Omega)$ being the space of measures. $\rho_{n_k}$ refers to a mollifier with support $B(0,1/n)$. $(a_i)$ are uniformly distributed points in $\Omega$ a bounded domain in $\mathbb{R}^N$. How do we show that in the weak * notion this holds?.

Answer 1

First, note that $$ (\rho_n*\mu_k)(x)=k^{-1}\lvert\Omega\rvert\sum_{i=1}^k \rho_n(x-a_i), $$ and therefore $$ \begin{split} \int_\Omega \varphi(x)(\rho_n*\mu_k)(x)\,dx &= k^{-1}\lvert\Omega\rvert\sum_{i=1}^k \int_\Omega \varphi(x) \rho_n(x-a_i)\,dx \\&= k^{-1}\lvert\Omega\rvert\sum_{i=1}^k(\tilde\rho_n*\varphi)(a_i) \\&= \color{blue}{k^{-1}\lvert\Omega\rvert\sum_{i=1}^k\varphi(a_i)}+ \color{red}{k^{-1}\lvert\Omega\rvert\sum_{i=1}^k\bigl((\tilde\rho_n*\varphi)(a_i)-\varphi(a_i)\bigr)} \end{split} $$ where $\tilde\rho_n(x)=\rho_n(-x)$.

Here, I identify $C_0(\overline\Omega)$ with the set of continuous functions on $\mathbb{R}^N$ vanishing outside $\Omega$. It follows from the equidistribution of the sequence $(a_i)$ that $$ \color{blue}{\lim_{k\to\infty}k^{-1}\lvert\Omega\rvert\sum_{i=1}^k\varphi(a_i)=\int_\Omega \varphi\,dx}. $$

To finish the proof, you only need to notice that $\tilde\rho_n*\varphi\to\varphi$ uniformly as $n\to\infty$, by the uniform continuity of $\varphi$, so the red term above vanishes in the limit as $n\to\infty$ (uniformly in $k$).