For the sake of completeness, taking Nate Eldredge's comment and elevating to an answer. This is known as Schauder's theorem; the statement (as Nate Eldredge says) is that an operator is compact iff its adjoint is compact. A proof sketch can be found in this MSE answer. In this context, $L^2(\Omega)$ will embed compactly into $H^{-1}(\Omega)$ exactly when $H^1_0(\Omega)$ embeds compactly into $L^2(\Omega)$, which will follow from Sobolev embedding theorems.

For the sake of completeness, taking Nate Eldredge's comment and elevating to an answer. This is known as Schauder's theorem; the statement (as Nate Eldredge says) is that an operator is compact iff its adjoint is compact. A proof sketch can be found in this MSE answer. In this context, $L^2(\Omega)$ will embed compactly into $H^{-1}(\Omega)$ exactly when $H^1_0(\Omega)$ embeds compactly into $L^2(\Omega)$, which is the case when $\Omega$ is bounded by the Rellich-Kondrachov theorem.

For the sake of completeness, taking Nate Eldredge's comment and elevating to an answer. This is known as Schauder's theorem; the statement (as Nate Eldredge says) is that an operator is compact iff its adjoint is compact. A proof sketch can be found in this MSE answer. In this context, $L^2(\Omega)$ will embed compactly into $H^{-1}(\Omega)$ exactly when $H^1_0(\Omega)$ embeds compactly into $L^2(\Omega)$, which will follow from Sobolev embedding theorems.

For the sake of completeness, taking Nate Eldredge's comment and elevating to an answer. This is known as Schauder's theorem; the statement (as Nate Eldredge says) is that an operator is compact iff its adjoint is compact. A proof sketch can be found in this MSE answer. In this context, $L^2(\Omega)$ will embed compactly into $H^{-1}(\Omega)$ exactly when $H^1_0(\Omega)$ embeds compactly into $L^2(\Omega)$, which will follow from Sobolev embedding theorems.