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