By Rellich compactness theorem the inclusion map $i: H_0^{1}(Omega) rightarrow L^2(Omega)$ compact. Which means if $f_n$ is a bounded sequence in $H^1(Omega)$ then there exists a subsequence $f_{n_k}$ such that $f_{n_k} rightarrow f,$ in $L^2(Omega).$

Since $L^2(Omega)$ and $H_0^{1}(Omega)$ are Banach spaces the map $i^*:(L^2(Omega))^* rightarrow H^{-1}(Omega)$ is also compact.

Suppose $T_f in (L^2(Omega))^*$ denotes the functional $langle T_f,g rangle =intlimits_{Omega}fg$ for all $gin L^2(Omega).$

Then $i^*(T_f)in H^{-1}(Omega)$ is the functional defined by $langle i^*(T_f),g rangle=intlimits_{Omega}fg$ for all $gin H_0^1(Omega) subset L^2(Omega).$

Which means that $i^*(T_f)$ is the restriction of the functional $T_f$ to the set $H^1_0(Omega)$. Furthermore $||T_f||_{(L^2(Omega))^*}=||f||_{L^2(Omega)}=||i^*(T_f)||_{H^{-1}}(Omega)$

Let $f_n$ be a bounded sequence in $L^2,$ then $T_{f_n}$ is a bounded sequnce in $H^{-1}(Omega)$, hence, there exists a subseqeunce $f_{n_k}$ such that $i^*(T_{f_{n_k}}) rightarrow T in H^{-1}(Omega).$

**Is this argument correct?**