• $U_i,H$ be $\mathbb R$-Hilbert spaces
  • $A_i\in\mathfrak L(U_i,H)$ with $$\left\|A_1^\ast h\right\|_{U_1}\le C\left\|A_2^\ast h\right\|_{U_2}\;\;\;\text{for all }h\in H\tag1$$ for some $C\ge0$
  • $B_1:=\left\{u_1\in U_1:\left\|u_1\right\|_{U_1}\le1\right\}$ and $B_2(C):=\left\{u_2\in U_2:\left\|u_2\right\|_{U_2}\le C\right\}$

Assume there is a $u_1\in B_1$ with $$A_1u_1\not\in A_2B_2(C)\tag2.$$ Since $H$ is reflexive and $A_2B_2(C)$ is closed and convex, we can apply the Hahn-Banach separation theorem and obtain the existence of a $x\in H\setminus\left\{0\right\}$ and a $\alpha\in\mathbb R$ with $$\langle x,A_2u_2\rangle_H\le\alpha\;\;\;\text{for all }u_2\in B_2(C)\tag3$$ and $$\langle x,A_1u_1\rangle_H>\alpha.\tag4$$

Why do $(2)$ and $(3)$ actually hold for $\alpha=1$?

  • $\begingroup$ What is $u_1$ in (3) and where does it come from? $\endgroup$ – Frederik vom Ende Mar 25 '18 at 21:23
  • $\begingroup$ @FrederikvomEnde Sorry, I've missed an important part of the question. Edited now. $\endgroup$ – 0xbadf00d Mar 25 '18 at 21:34

Your Answer

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

Browse other questions tagged or ask your own question.