0
$\begingroup$

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?

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

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.