Isometric embedding of $L^2$ onto $H^{-1}$

Question

Let $X$ be a Banach space. Many sources in the literature identify $L^2(X)$ with $H^{-1}(X)$ through the identification $$ \varphi: L^2(X) \to H^{-1}(X); \quad \quad \varphi(u)(v) := (u,v)_{L^2}, \quad \quad v \in H^{1}_0(X).$$ Clearly, by the Cauchy-Schwarz inequality, $$ \| \varphi(u) \|_{H^{-1}} = \sup_{ \|v\|_{H^{1}_0} \leq 1 } | \varphi(u)(v)| = \sup_{ \|v\|_{H^{1}_0} \leq 1 } \big| (u,v)_{L^2} \big| \leq \|u\|_{L^2}. $$ However, to show that it is an isometry, one also needs to show that

$$ \| \varphi(u) \|_{H^{-1}} \geq \|u\|_{L^2}.$$

I have trouble seeing this. Any ideas?

Answer 1

This is not true, $\varphi$ is usually not an isometry. By the Rellich-Kondrachov compactness theorem, $\varphi$ is a compact map. Since the involved spaces are infinite-dimensional, $\varphi$ cannot be an isometry.