Weak convergence in $C[0,1]$ For a uniformly bounded sequence $(f_n)$ in $C[0,1]$, show that 
$f_n$ converges weakly to $0$ $\iff $ $\lim \limits_{n \to \infty} f_n(y) =0$ for all $y \in [0,1]$ 
Is the equivalence true if we do not assume that $(f_n)$is uniformly bounded,explain?
My first confusion, is $C[0,1]$ Hilbert space? I think it is not because this space is dense in $L^2[0,1]$ so it not closed subspace $L^2$ so not Hilbert space then how could the concept of weak convergence comes here? I'm confused, could anybody plz help me. I'm preparing for qualifying exam in august so I'm writing many questions frequently. I, really appreciate any kind of help..
 A: If $(E,\lVert\cdot\rVert)$ is a normed space, then we say that the sequence $(x_n)_{n\geqslant 1}$ converges weakly to $x$ if for each linear continuous functional $f$ on $E$, $f(x_n)\to f(x)$. 
There is no need to consider a Hilbert space in this context and actually $C[0,1]$ endowed with the uniform norm is not a Hilbert space (the parallelogram identity does not hold). 
To check the equivalence, for one direction, consider the linear continuous functional $f_t\colon x\in C[0,1]\mapsto x(t)$. The other is more difficult unless we know a characterization of the topological dual space of $C[0,1]$ with Radon measures. Here, since $[0,1]$ is compact, we know that a continuous linear functional can be represented by a Borel (signed) measure on the unit interval (no need to consider Radon measures). The crucial fact is that if $\mu$ is a positive Borel measure on the unit interval and $\varepsilon\gt 0$, then there exists an integer $N$ and $a_1,\dots,a_N\in\mathbf R$, $c_1,\dots,c_N\in [0,1]$ such that 
$$\forall h\in C[0,1],\quad \left|\int_{[0,1]}h(x)\mu(dx)-\sum_{j=1}^Nc_jh(a_j)\right|\lt\varepsilon.$$
Recall a that weakly convergent sequence in a normed space is necessarily bounded.
