Suppose $f_k \to 0$ in $L^2(U)$ where $U$ is an open and bounded domain in $\mathbb{R}^n$. Suppose further that $f_k \in H_0^1(U)$ where $H_0^1(U)$ is the Sobolev space whose trace being zeros.
I want to claim or disclaim that $f_k$ weakly converges to $0$ in $H_0^1(U)$. I think the above is true, but not entirely sure. Here is my argument.
Let $C^\infty_0(U)$ be the set of test functions whose support is compact in $U$. Then, for any $v \in C^\infty_0(U)$, we have $$ \langle f_k, v\rangle_{H^1(U)} = \langle f_k, v\rangle_{L^2(U)} + \langle Df_k, Dv\rangle_{L^2(U)}. $$ Since $f_k \to 0$ in $L^2(U)$, the first term in the right hand side vanishes as $k \to \infty$. The second term also vanishes since $$ \langle Df_k, Dv\rangle_{L^2(U)} = -\langle f_k, D^2v \rangle_{L^2(U)} \to 0. $$ Since $C^\infty_0(U)$ is dense in $H_0^1(U)$ with respect to $\|\cdot\|_{H_0^1(U)}$, we conclude that $f_k$ weakly converges to 0 in $H_0^1(U)$.
It seems that the above argument is true, however, I feel a bit uneasy about the conclusion. Normally, one cannot have the convergence of $\{f_k'\}$ from the convergence of $\{f_k\}$. This is also described in this Wikipedia article [link]. But, in this case where $f_k \to 0$, the above argument also shows that $f_k'$ weakly converges to 0 in $L^2(U)$. In a way, this makes sense as the limit is a constant function, however, I just wanted to make sure my argument is solid.
Any comments/suggestions/answers will be very appreciated.
What you claim is true if and only if you know that the sequence$(f_k)_k$ is bounded in $H_0^1$.
One direction is clear (weakly convergent sequences are bounded), and the other direction follows by approximating a given $v \in H_0^1$ by $v_\epsilon \in C_c^\infty$, and then bounding the error in the inner product. See also here: Is it sufficient to check weak convergence on a (weak* or strongly) dense subset of the dual?