Let $\{f_n\}$ in $C[0,1]$ satisfy $\sup_n \sup_{x\in[0,1]}|f(x)|<\infty$. Show that $$f_n \ \ \text{converges weakly to 0} \Longleftrightarrow \lim_{n}f_n(x)=0 \ \ \text{for all }x\in[0,1].$$
I have worked out the necessary part, how to prove the essential part? I have no idea.
Let $\mathbb{K}$ be the base field you're working with ($\mathbb{R}$ or $\mathbb{C}$).
Given $x\in[0,1]$, let $\widehat{x}:C[0,1]\to\mathbb{K}$, be the continuous linear functional given by $\widehat{x}(f)=f(x)$.
If $f_n\to 0$ weakly, then in particular $f_n(x)=\widehat{x}(f_n)\to 0$ in $\mathbb{K}$, that is $0=\lim_nf_n(x)$ for every $x\in[0,1]$.
Now, the only proof of the other implication that I know is not so trivial: Suppose $0=\lim_n f_n(x)$ for every $x$. Let $T:C[0,1]\to\mathbb{K}$ be any continuous linear functional. By Riesz-Markov-Kakutani Theorem, there exists a (unique) regular Borel measure $\mu$ on $[0,1]$ such that $T(g)=\int_{[0,1]}gd\mu$ for every $g\in C[0,1]$ (and $\mu$ must be finite). By hypothesis, $\sup_n\sup_x|f_n(x)|<\infty$ and $f_n\to 0$ pointwise, so by Dominated Convergence we have $0=\lim_n\int_{[0,1]}f_nd\mu=\lim_nT(f_n)$, and that means that $f_n\to 0$ weakly.