The relations between $ L^2 $ and $ H^{-1} $.

Question

Let $ \Omega $ be a bounded domain in $ \mathbb{R}^d $. Suppose that $ f\in L^2(\Omega) $. Define $ \phi\in H^{-1}(\Omega) $ such that \begin{align} \langle\phi,u\rangle_{H^{-1}\times H_0^1}=\int_{\Omega}f(x)u(x)dx \end{align} for all $ u\in H_0^1(\Omega) $. I want ask that what is the relations between $ \left\|f\right\|_{L^2(\Omega)} $ and $ \left\|\phi\right\|_{H^{-1}(\Omega)} $. Obviously, we have \begin{align} \left\|\phi\right\|_{H^{-1}(\Omega)}\leq \left\|f\right\|_{L^2(\Omega)}. \end{align} Can I obtain $ C>0 $ such that $ \left\|f\right\|_{L^2(\Omega)}\leq C\left\|\phi\right\|_{H^{-1}(\Omega)} $? Can you give me some references and hints?

Answer 1

The embedding $i:H^1_0(\Omega) \hookrightarrow L^2(\Omega)$ is compact. The Riesz isomorphism $j:L^2(\Omega) \to L^2(\Omega)^*$ is continuous. It is now easy to see that $\phi = i^*(j(f))$, where $i^*: L^2(\Omega)^* \to H^{-1}(\Omega)$. Then the mapping $f\mapsto \phi$ is compact from $L^2(\Omega)$ to $H^{-1}(\Omega)$, and the second bound cannot be true, i.e., there is no $c>0$ such that $\|f\|_{L^2} \le c \|\phi\|_{H^{-1}(\Omega)}$.

Answer 2

If the bound $\left\Vert f\right\Vert_{L^2}\leq C\left\Vert\phi\right\Vert_{H^{-1}}$ holds, it would imply the inclusion $L^2\hookrightarrow H^{-1}$ is invertible, which is obviously not true.