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

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?

• How do you define $L^2(X)$ and $H^{-1}(X)$ for a Banach space $X$? – gerw Mar 25 at 7:05

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.