# Generalisation of the norm of bounded linear operators II

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 !!!

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 $$$$\label{proofnorm1} \sup_{\|y\|_M\leq 1,y \in \overline{\mbox{Im}(M)}}|\langle Ax\mid y\rangle_M| \geq \left|\langle Ax\mid \frac{Ax}{\|Ax\|_M}\rangle_M\right| =\|Ax\|_M.$$$$ So taking the supremum over $$\|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.$$

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 $$\|A\|_M=\sup\left\{|\langle Ax, y\rangle_M|;\;x,y\in \overline{\mbox{Im}(M)} ,\;\|x\|_{M}\leq1,\|y\|_{M}\leq 1\right\}.$$

• I don't understand why the inequality $(1)$ holds? – Student Oct 26 '18 at 15:46
• Because $\|Ax\|_M=\langle Ax,y\rangle$ if you chose $y$ carefully, as the line above $(1)$ shows. – Martin Argerami Oct 26 '18 at 16:07