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?

  • $\begingroup$ How do you define $L^2(X)$ and $H^{-1}(X)$ for a Banach space $X$? $\endgroup$ – 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.


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