Proving $f_n$ is weakly convergent to $f$ if and only if following holds

Question

Weakly convergent definition (from Wikipedia): A sequence of points $(x_n)$ in a Hilbert space $H$ is said to converge weakly to a point $x$ in $H$ if $\langle x_n, x \rangle \rightarrow \langle x,y \rangle$ for all $y$ in $H.$

Let $X=C[a,b]$ and $\{f_n\}_{n=1}^{\infty}\in X$. Then $f_n$ is weakly convergent to $f$ if and only if there exists a constant $M$ such that \begin{equation*} \|f_n\|\leq M \end{equation*} for all $n$ and \begin{equation*} \lim_{n\rightarrow\infty}f_n(x)=f(x) \end{equation*} for any $x\in[a,b]$.

I don't really see how to approach this problem.

Answer 1

By weak-convergence, do you mean the following?

Let $X$ be a Banach space and $X^{\ast}$ its dual space (i.e., the Banch space of all bounded linear functionals on $X$). Let $(x_{\alpha})$ be a net in $X$, $x\in X$. We say that the net $(x_{\alpha})$ converges to $x$ iff for each $f\in X^{\ast}$, $\langle f,x_{\alpha}\rangle\rightarrow\langle f,x\rangle$. In general, the weak-topology $\sigma(X,X^{\ast})$ on $X$ is not first countable, so it is inappropriate to describe it using sequences. (One should use nets or "generalized sequences"). However, we do have: If a sequence $(x_{n})$ converges weakly, it is bounded. For, recall that $X\hookrightarrow X^{\ast\ast}$, $x\mapsto\hat{x}$, where $\langle\hat{x},f\rangle=f(x)$, $f\in X^{\ast}$ is an isometric embedding. Now, let $(x_{n})$ be a sequence in $X$ and suppose that $x_{n}\rightarrow x$ weakly for some $x\in X$. For each $f\in X^{\ast}$, $\{\langle\hat{x}_{n},f\rangle\mid n\in\mathbb{N}\}$ is bounded because $(\langle\hat{x}_{n},f\rangle)_{n}$ is a convergent sequence of complex numbers. By uniform boundedness principle, $\sup_{n}||\hat{x}_{n}||<\infty$. Note that $||x_{n}||=||\hat{x}_{n}||$...

For your question. $X=C([a,b])$ is a Banach space with respect to the supremum norm (in fact, it has far more structures: It is a unital commutative $C^{\ast}$-algebra). It is well-known that $X^{\ast}$ can be identified with the space of all complex-valued, regular Borel measures on $[a,b]$, with total variation norm. (Riesz representation theorem). Let $(f_{n})$ be a sequence in $X$ and $f\in X$. Suppose that $f_{n}\rightarrow f$ weakly. By the previous discussion, $\sup_{n}||f_{n}||<\infty$. For each $x\in[a,b]$, let $\delta_{x}$ be the Dirac measure at $x$, i.e., $\delta_{x}(A)=\begin{cases} 1, & \mbox{ if }x\in A\\ 0, & \mbox{ if }x\notin A \end{cases}$, $A\in\mathcal{B}([a,b])$. Note that $\delta_{x}\in X^{\ast}$. Then $\langle\delta_{x},f_{n}\rangle\rightarrow\langle\delta_{x},f\rangle$. Recall that $\langle\delta_{x},f\rangle=\int fd\delta_{x}=f(x)$. Hence, $f_{n}(x)\rightarrow f(x)$.

Conversely, suppose that $\sup_{n}||f_{n}||=M<\infty$, and for each $x\in X$, $f_{n}(x)\rightarrow f(x)$. Let $\mu\in X^{\ast}$. By Riesz representation theorem, $\mu$ is a complex measure and the duality $\langle\mu,f\rangle$ is actually integral: $\langle\mu,f\rangle=\int f\,d\mu$. Note that $|f_{n}|\leq M$ for all $n$ and the constant function $M$ is $\mu$-integrable. By Lebesgue dominated convergence theorem, we have $\int f_{n}\,d\mu\rightarrow\int f\,d\mu$. That is, $\langle\mu,f_{n}\rangle\rightarrow\langle\mu,f\rangle$. In another word, $f_{n}\rightarrow f$ weakly.