functional analysis – Exaction of convergent sub sequence in \$H^{-1}(Omega)\$

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?