# 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.

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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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What exactly is the Cauchy sequence you have in mind? The functionals $\langle f_n,\cdot\rangle$ with the operator norm? Why is that sequence Cauchy? – Greg Graviton Mar 2 '11 at 12:55
@Greg: I'm not sure I understand your question. The sequence of operators $(\langle f_{n},\cdot\rangle)_{n}$ is pointwise bounded (Cauchy sequences in $\mathbb{R}$ are bounded) by hypothesis, hence it is uniformly bounded by Banach-Steinhaus. The operator norm of $\langle f_{n}, \cdot \rangle$ is $\|f_{n}\|$ by Hahn-Banach. But it certainly isn't a Cauchy sequence with respect to the operator norm, take e.g. an orthonormal system $(f_{n})$ (such a system converges weakly to $0$ by Parseval's identity). I've added some clarifying remarks, I hope it's understandable now. – t.b. Mar 2 '11 at 13:13
Ah that makes sense. And weak limits are unique, so the $f$ that is the weak subsequential limit must be the same as the $f$ that is the limit that in my hypothesis right? Thanks! – user1736 Mar 2 '11 at 16:52
Ah, ok, you use the term "weak Cauchy". For this explanation, I would avoid it completely, however, because it just replaces a problem with a definition. – Greg Graviton Mar 2 '11 at 16:54
@user1736: Yes, exactly. Note that you don't even have to pass to a subsequence since the sequence was weakly Cauchy in the first place. The subsequence argument is only needed for finding an accumulation point, but a Cauchy sequence can have at most one accumulation point. – t.b. Mar 2 '11 at 16:58

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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I don't understand what this answer has to do with the question. All the things you're saying are correct but don't seem to address the question asked, as far as I can tell. – t.b. Mar 2 '11 at 8:32
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... – Dirk Mar 2 '11 at 11:22