Strong convergence in $L^1 \implies$ weak convergence in $H^{-1}$?

Let $U$ be a bounded open set in $\mathbb{R}^3$. Prove or disprove:Strong convergence in $L^1(U) \implies$ weak convergence in $H^{-1}(U).$

Attempt: Suppose $f_n \to f$ in $L^1(U)$ strongly. As the space is reflexive, we need to show for $g \in H_0^1(U)$, $$\big<f_n,g\big> \to \big<f,g\big> $$

where $\big<.,.\big> $ is the pairing between $H^{-1}(U)$ and $H_0^1(U)$.

Since $ L^1(U) \subset L^2 (U) \subset H^{-1}(U)$ and $L^{\infty}(U)\cap H_0^1(U)$ is dense in $H_0^1(U)$, if the pairing is given by $$\big<u,v\big>=\int_U uv$$ then $$\Big|\big<f_n-f,g\big>\Big|\leq ||f_n-f||_{L^1}||g||_{L^{\infty}} \to 0$$

Is there a way to extend this for any arbitrary pairing or is this result false in general?