Generalisation of the norm of bounded linear operators II

Question

Let $E$ be a complex Hilbert space, with inner product $\langle\cdot\;, \;\cdot\rangle$ and the norm $\|\cdot\|$ and let $\mathcal{L}(E)$ be the algebra of all bounded linear operators on $E$.

Let $M\in \mathcal{L}(E)^+$ (i.e. $\langle Mx\;, \;x\rangle \geq0,\;\forall x\in E$), we consider the following subspace of $\mathcal{L}(E)$: $$\mathcal{L}_M(E)=\left\{A\in \mathcal{L}(E):\,\,\exists c>0 \quad \mbox{such that}\quad\|Ax\|_M \leq c \|x\|_M ,\;\forall x \in \overline{\mbox{Im}(M)}\right\},$$ with $\|x\|_M:=\|M^{1/2}x\|,\;\forall x \in E$. If $A\in \mathcal{L}_M(E)$, the $M$-semi-norm of $A$ is defined us $$\|A\|_M:=\sup_{\substack{x\in \overline{\mbox{Im}(M)}\\ x\not=0}}\frac{\|Ax\|_M}{\|x\|_M}$$

According to this answer, for $A\in \mathcal{L}_M(E)$, we have $$\|A\|_M=\displaystyle\sup_{\|x\|_M\leq1}\|Ax\|_M=\displaystyle\sup_{\|x\|_M=1}\|Ax\|_M.$$

Let $A\in \mathcal{L}_M(E)$, I see in a paper that it is straightforward that $$\|A\|_M=\sup\left\{|\langle Ax, y\rangle_M|;\;x,y\in \overline{\mbox{Im}(M)} ,\;\|x\|_{M}\leq1,\|y\|_{M}\leq 1\right\},$$ where $\langle Ax, y\rangle_M=\langle MAx, y\rangle.$ How can I prove this result?

Thank you everyone !!!

Answer 1

Note first that for $A\in \mathcal{L}_M(E)$, we have $$\|A\|_M=\displaystyle\sup_{\|x\|_M\leq1,x \in \overline{\mbox{Im}(M)}}\|Ax\|_M,$$ and for all $x \in \overline{\mbox{Im}(M)}$ we have $$\|Ax\|_M\leq \|A\|_M\|x\|_M.$$

For any $x,y \in \overline{\mbox{Im}(M)}$ with $\|x\|_M, \|y\|_M \leq1$ we have $$|\langle Ax\mid y\rangle_M| \leq\|Ax\|_M\|y\|_M \leq \|A\|_M\|x\|_M\|y\|_M \leq\|A\|_M.$$ Now, for any $x \in \overline{\mbox{Im}(M)}$ with $Ax \in \overline{\mbox{Im}(M)}\setminus \{0\}$ we have: $$\left|\langle Ax\mid\frac{Ax}{\|Ax\|_M}\rangle_M\right| = \frac{\|Ax\|_M^2}{\|Ax\|_M} = \|Ax\|_M.$$ This implies that for all $x \in \overline{\mbox{Im}(M)}$ with $Ax \in \overline{\mbox{Im}(M)}\setminus \{0\}$ we have \begin{equation}\label{proofnorm1} \sup_{\|y\|_M\leq 1,y \in \overline{\mbox{Im}(M)}}|\langle Ax\mid y\rangle_M| \geq \left|\langle Ax\mid \frac{Ax}{\|Ax\|_M}\rangle_M\right| =\|Ax\|_M. \end{equation} So taking the supremum over $\|x\|_M\leq 1$ with $x \in \overline{\mbox{Im}(M)}$ gives: $$\sup_{\|x\|_M, \|y\|_M\leq 1, x,y \in \overline{\mbox{Im}(M)}}|\langle Ax\mid y\rangle_M| \geq \sup_{\|x\|_M\leq 1,x \in \overline{\mbox{Im}(M)}} \|Ax\|_M = \|A\|_M.$$

Answer 2

If you know that $$ \|A\|_M=\sup\{\|Ax\|_M:\ \|x\|_M\leq1\}, $$ you have $$ \|Ax\|_M=\frac{\langle Ax,Ax\rangle_M}{\|Ax\|_M}=\langle Ax,\frac{Ax}{\|Ax\|_M}\rangle_M, $$ so $$\tag1 \|A\|_M\leq \sup\left\{|\langle Ax, y\rangle_M|;\;x,y\in \overline{\mbox{Im}(M)} ,\;\|x\|_{M}\leq1,\|y\|_{M}\leq 1\right\}. $$ On the other hand, if $\|y\|_M\leq1$, then $$ \langle Ax,y\rangle_M\leq\|Ax\|_M\,\|y\|_M\leq\|Ax\|_M, $$ so $$\tag2 \sup\left\{|\langle Ax, y\rangle_M|;\;x,y\in \overline{\mbox{Im}(M)} ,\;\|x\|_{M}\leq1,\|y\|_{M}\leq 1\right\}\leq\|A\|_M. $$ Then $(1)$ and $(2)$ together give $$ \|A\|_M=\sup\left\{|\langle Ax, y\rangle_M|;\;x,y\in \overline{\mbox{Im}(M)} ,\;\|x\|_{M}\leq1,\|y\|_{M}\leq 1\right\}. $$