On the limits of weakly convergent subsequences Let $\{ f_n \}$ be a sequence in a Hilbert space $L^2(\mathbb{R}^d)$. We say that this sequence converges weakly to an element $f \in L^2$ if $\langle f_n, g \rangle \to \langle f,g \rangle$ for every $g \in L^2$ (where $\langle \cdot,\cdot \rangle$ denotes the inner product on $L^2$). By definition, we are given that the weak limit $f$ is in $L^2$. 
However, suppose we know that a sequence "formally" converges weakly to a limit $f$ (i.e. $\langle f_n, g \rangle \to \langle f,g \rangle$ for every $g \in L^2$ for some $f$ which we don't necessarily know yet to be in $L^2$) . 
Does this, purely by the characteristics of weak convergence, directly imply that $f \in L^2$? 
I think you could also generalize this question to any Hilbert space, provided that taking the inner product of an element possibly not in the Hilbert space makes sense.
 A: Let me elaborate on user3148's answer and comment.
There are two facts:


*

*A weak Cauchy sequence $(f_{n})$ is bounded.

*Every bounded sequence has a weakly convergent subsequence.


Combining these two facts it is easy to see that every weak Cauchy sequence converges. Recall that a weak Cauchy sequence is a sequence $(f_{n})$ such that $\langle f_{n}, g\rangle$ is Cauchy in $\mathbb{R}$ for all $g$. The condition you impose on the sequence $(f_{n})$ means in particular that it is a weak Cauchy sequence, so it necessarily converges to some $f \in L^2$.

Proof of 1. This follows immediately from the Banach-Steinhaus theorem applied to the operators $\langle f_{n}, \cdot \rangle: X^{\ast} \to \mathbb{R}$, see Sokal's recent paper for a neat proof of that theorem (without Baire!).
Proof of 2. This is immediate from the version of the Banach-Alaoğlu theorem saying that the unit ball in a separable reflexive space is compact metrizable in the weak topology (= weak$^{\ast}$-topology by reflexivity).
A: Well, there are answers to this on different levels. One answer is, that the norm in a Hilbert space is always weakly lower semicontinuous meaning that for a weakly converging sequence $(f_n)$ it holds that
$$
\lim\inf \|f_n\| \geq \|w-\lim f_n\|.
$$
Another answer is that due to the principle of uniform boundedness (or Banach-Steinhaus-Theorem) every weakly convergent sequence is bounded (see here).
