# When do inner products of weakly convergent subsequences converge?

If we have 2 weakly convergent subsequences in $L^2(U)$ (for $U$ some bounded open domain with smooth boundary), $u_k\rightharpoonup u$ and $v_k\rightharpoonup v$, under which conditions do we have $$\langle u_k,v_k \rangle \to \langle u, v \rangle?$$

I can see that if $u_k \to u$ strongly and $\{v_k\}_{k=1}^{\infty}$ is bounded then the result follows but I don't know when it would be true if both convergences are only weak.

Thanks!

• Note that by the uniform boundness principle, any weakly convergent sequence is bounded, so it is sufficient that one of the sequences converges strongly. – Nate Eldredge Nov 22 '13 at 16:56
• Is there any reasonable condition that would make it true in which both sequences are strictly weakly convergent? In the book I am reading (Evans: Weak convergence methods for nonlinear PDES) in the proof of the div-curl lemma, I don't think we can't be sure of strong convergence! – David Nov 22 '13 at 17:00

To better illustrate the issue, subtract off the weak limits, so that you have two weakly null sequences $u_k$, $v_k$. Suppose they do not converge strongly. Then neither $\|u_k\|$ nor $\|v_k\|$ tend to $0$. But you want to have $\langle u_k,v_k\rangle \to 0$, which means the vectors $u_k$ and $v_k$ must become nearly orthogonal when $k$ is large. There is no intrinsic property of $u_k$ or of $v_k$ that will make that happen. Two vectors are (nearly) orthogonal when they are (nearly) orthogonal; to say something less tautological, you really need some specific information about how these sequences are built.
In general this is false. Let $e_k$ be an orthonormal basis for your Hilbert space, and take $u_k = v_k = e_k$. By Bessel's inequality, $e_k \rightharpoonup 0$ but $\langle e_k, e_k \rangle = 1$ for all $k$.
The Div-Curl lemma gives that $\int u_n v_n \rightarrow \int uv$ if each sequence converges weakly, and the divergence of one sequence and the curl of the other are precompact in $H^{-1}$.