Suppose $U \subset \mathbb{R}^d$ is a smooth bounded domain. Let $(u_n)_{n\geq 1}$ be a sequence in the Sobolev space $W^{1,2}(U)$ such that it weakly converges to zero: $u_n \rightharpoonup 0$ as $n \to \infty$. Also assume that for any bounded sequence of test functions
$$(\phi_n)_{n\geq 1} \subset W^{1,2}_0(U)$$
we have the convergence
$$\int_U \nabla u_n(x) \cdot \nabla \phi_n(x) dx \to 0$$
Can I conclude that $\nabla u_n \to 0$ in $L^2(U)$? Clearly I cannot take that test function $\phi_n = u_n$ because of the zero boundary condition.
Best Answer
Assume $d \ge 2$. Let $K \subset H^1(U)$ be the subspace of weakly harmonic functions, i.e., all $u$ satisfying $$ \int_U\nabla u \cdot \nabla \varphi \, \mathrm d x = 0 $$ for all $\varphi \in H_0^1(U)$. Then, $K$ is an infinite-dimensional Hilbert space. Then, any orthonormal sequence $(u_n)$ satisfies your assertions but does not converge strongly to zero.
[I believe that your assertion is true in case $d = 1$, since the dimension of $K$ is $2$ in this case.]