Weakly convergence in C(X) for metric compact spaces Let $X$ be a compact metric space.
Let $f_n$ be a sequence of functions in $C(X)$. Show that if $f_n$ converges to $0$ weakly then $|f_n|$ converges to $0$ weakly.
Actually, I need to show then $f_n$ converges to $0$ weakly iff $\sup ||f_n||$ is bounded (uniform norm) and $f_n$ converges to $0$ pointwise.
But I can show the direction: weakly convergence implies $\sup$ norm is bounded by the first argument.
I'm familiar with Reisz Representation Theorem.

Thank you in advance.
Assume that $f_n \to 0$ weakly. It can be shown that every weakly boundes sequence is norm bounded in any Banach space. So $\sup ||f_n ||&lt;\infty,$ Moreover for any $x\in X$ define regular Borel measure $$\mu_x (A) =\begin{cases} 1 \mbox{ iff } x\in A \\ 0 \mbox{ iff } x\notin A\end{cases}. $$
Then by the weak convergence we have $$f_n (x) =\int_X f(u) \mu_x (du) \to 0$$
hence $f_n (x) \to 0 $ pointwise.
On the other hand if $\sup ||f_n ||$ is finite and $f_n \to 0 $ pointwise then by the Lebesgu'e dominated convergence theorem we get $$\int_X f_n (u) \mu (du )\to 0 $$ for any regular borel measure on $X.$ Hence $f_n \to 0$ weakly.
Assume that $f_n \to 0$ weakly. It can be shown that every weakly boundes sequence is norm bounded in any Banach space. So $\sup ||f_n ||&lt;\infty,$ Moreover for any $x\in X$ define regular Borel measure $$\mu_x (A) =\begin{cases} 1 \mbox{ iff } x\in A \\ 0 \mbox{ iff } x\notin A\end{cases}. $$
Then by the weak convergence we have $$f_n (x) =\int_X f(u) \mu_x (du) \to 0$$
hence $f_n (x) \to 0 $ pointwise.
On the other hand if $\sup ||f_n ||$ is finite and $f_n \to 0 $ pointwise then by the Lebesgu'e dominated convergence theorem we get $$\int_X f_n (u) \mu (du )\to 0 $$ for any regular borel measure on $X.$ Hence $f_n \to 0$ weakly.