Proving every sequence in an infninite dimensional hilbert space with norm one contains a subsequence which converges weakly to f in H Let H be an infinite dimensional (separable) Hilbert space, $(f_k)$ a sequence in H, with $||f_k|| = 1$
for all k. Then there exists a subsequence $(f_{k_j})$ of $(f_k)$ and a unique f ∈ H such that $f_{k_j}→ f$weakly,  that is $⟨f_{k_j},g⟩ → ⟨f, g⟩ $for all $g \in H$.
I have seen a proof of this online which uses diagonalisation argument and I followed the proof until the end where they said let $f= \sum_{i=1}^{\infty} c_i e_i$ where $c_i$ were the limits of each subsequence $<f_{k_j},e_j>$  and $e_i $ orthonormal basis. Then followed by finishing the proof. What I dont get is, that kind of infinite sum may not be in H! How are they jusifying that it is in H. I will copy paste the proof if needed. Otherwise any other proof or the same proof without the infinite sum is welcome. I am self studying, found this problem in Stein and Shakarski measure theory book. Thanks
 A: Take a look at the sequence $\langle f_n, e_1 \rangle$. Because $\lVert{f_n \rVert} =1$, by the Cauchy-Schwarz inequality the sequence will be in a the compact set $\{ z \in \mathbb{C} \mid \lvert z \rvert \leq 1\}$. So $\langle f_n, e_1 \rangle$ has a convergent subsequence which we will call $f_{n,1}$. Now we can take a look at $\langle f_{n,1},e_2 \rangle$, by the same argument this also has  a convergent subsequence $f_{n,2}$. Now we can continue this inductively, such that for each $k \in \mathbb{N}: \langle f_{n,k},e_k \rangle$ is a convergent subsequence. Now set $g_n = f_{n,n}$. For some fixed $k \in \mathbb{N}$, we can view $(\langle g_n, e_k \rangle)_{n=k}^\infty$ as a subsequence of $\langle f_{n,k}, e_k \rangle$ so it will converge. Now we can define a linear functional $g:\text{span}\{ e_k \mid k \in \mathbb{N} \} \to \mathbb{C}$ by $g(e_k) = \lim_{n \to \infty} \langle e_k, g_n \rangle$. This function is clearly continuous and $\text{span}\{ e_k \mid k \in \mathbb{N}$ is dense in $H$ so it can be continuously extended to the entire of $H$. Now we have defined a continuous linear functional $g: H \to \mathbb{C}$ and by the Riesz-representation theorem there is a unique element $f \in H$ such that $g(x) = \langle x, f \rangle$ which is precisely the $f$ such that $g_n \to f$ weakly.
There is also a more general statement called the Banach-Alaoglu theorem which you might be interested in. It says that for any normed space $X$, the unit ball of the dual space of $X$ is weakly* compact. Since any Hilbert space is a Banach space that is also isometrically isomorphic to its dual space (Riesz representation theorem), the unit ball of any Hilbert space must be weakly compact. 
Another way to look at it is via Kakutani's theorem, this say that a Banach space $X$ is reflexive if and only if the unit ball of $X$ is weakly compact. Since a Hilbert space is reflexive (again by the Riesz representation theorem), the unit ball of the Hilbert space is weakly compact.
