# 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.

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).

• 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).

• 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