$\sup_{n \in \mathbb{N}} \|f\|_{\infty} < \infty$ and $f_n(x) \to f(x) \Rightarrow f_n \to f$ weakly

Question

Let $X$ be a compact Hausdorff space, $(f_n)_{n \in \mathbb{N}} \subset C(X)$ and $f \subset C(X)$. Show that if $\sup_{n \in N} \|f_n\| < \infty$ and $f_n(x) \to f(x) \ \forall x \in X$ then $f_n \to f$ weakly.

$f_n, f \in C(X) \Rightarrow$ measurable, $f_n \to f$ pointwise $\forall x \in X$. Continuous functions on compact interval $\Rightarrow$ max exists. Let $g(x) = \max_{n \in N} f_n(x),$ then $|f_n(x)| < g(x) \ \forall x \in X, n \in \mathbb{N}$. Now by Riesz there exists a finite regular Borel measure s.t $\phi (f) = \int_X f(x) d \mu (x),$ $\phi \in C(X)^*$. Now by DCT

$$ \lim_{n \to \infty} \phi (f_n) = \lim_{n \to \infty} \int f_n d \mu = \int f d \mu = \phi (f). $$ Thus $\phi (f_n) \to \phi (f)$, and $f_n \to f$ weakly.

Is this correct?

Answer 1

A couple of comments:

$(1)$ Your definition of $g$ doesn't imply that $|f_n(x)|\leq g(x)$ for all $x\in X$ and $n\in\mathbb N$ (what if $f_n(x)<0$ for all $x$ and $n$?). Instead, since $\sup_n\|f_n\|<\infty$, let $g$ be the constant function $g(x)=\sup_n\|f_n\|$ for all $x$. Then $|f_n|\leq g$ for all $n$, and the hypotheses of the dominated convergence theorem are satisfied.

$(2)$ The dominated convergence theorem applies to positive measures, so you really have to split the measure into four measures (positive/negative, real/imaginary) and apply it on each.

Other than these two items, everything seems to be in order. Nice work.