A weakly but not strongly convergent sequence in $H_0^1$

Question

I am looking for a sequence that weakly converges in $H_0^1$ but not strongly. I am not sure how to define a weakly convergent sequence in $H_0^1$ in the first place.

Answer 1

Since $H_0^1$ is infinite dimensional (I assume that the support of your measure is not a finite set), you may find $\{e_1, e_2, \dots\} \subset H_0^1$ linearly independent. Using the Gram-Schmidt procedure you may obtain an orthonormal subset $\{f_1, f_2, \dots\}$.

If the sequence $(f_n)$ were convergent (i.e. strongly), it would also be Cauchy. But

$$\| f_n - f_m \|^2 = \| f_n \|^2 - \langle f_n, f_m \rangle - \langle f_m, f_n \rangle + \| f_m \|^2 = 1 - 0 - 0 + 1 = 2 ,$$

which shows that the distance between any two terms of that sequence does not decrease, staying constant $\sqrt 2$, so the sequence is not Cauchy, therefore not convergent.

On the other hand, it is a consequence of Alaoglu's theorem that the unit ball in any Hilbert space is weakly compact. Since $\| f_n \| = 1$ for all $n$, it follows that the sequence $(f_n)$ lives inside the unit ball of $H_0^1$. Since this one is weakly compact, it follows that $(f_n)$ has a weakly-convergent subsequence, call it $(g_n)$. This $(g_n)$ is the sequence that you are looking for (it is not convergent for the same reason for which $(f_n)$ is not.)

Answer 2

Consider the sequence $f_n(x)=\frac1n\sin(2\pi nx)$ for $x\in (-1,1)$. Note that $f_n\to 0$ uniformly. Also, $f^{\prime}_n(x)=2\pi\cos(2\pi nx)$. Using the Riemann-Lebesgue lemma, you have that $$\int_{-1}^1 2\pi\cos(2\pi nx)g(x)\,dx\to 0$$ for every function $g\in L^2(-1,1)$. This shows that $f_n$ converges to zero weakly in $H^1_0$. However, $f^{\prime}_n$ does not converge to $0$ in $L^2$, since $\int_{-1}^1 (2\pi\cos(2\pi nx))^2\,dx$ does not go zero, so you don't have strong convergence in $H^1_0$.