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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.

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Let me elaborate on user3148's answer and comment.

There are two facts:

  1. A weak Cauchy sequence $(f_{n})$ is bounded.
  2. 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).

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    @Greg: I used it to make the question specific and the answer useful. I'm not assuming that there is a mysterious $f$ at all. I'm showing that if for all $g$ we have convergence $\langle f_{n}, g \rangle \to c(g)$, where $c(g)$ is a real number then $c(g) = \langle f, g \rangle$ for some $f$. This is more general than the asked question and actually has some practical use.2011-03-02
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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).

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    Well, a sequence is weakly converging, if $\langle f_n,g\rangle$ is a Cauchy sequence. Then one concludes that the sequence $f_n$ is bounded and due to lower semicontinuity its limit will have a bounded norm and hence, lies in the same space. Or probably I did not get the point of the question...2011-03-02