Let
- $U_i,H$ 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 $\alpha=1$?