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. $M^*=M$ and $\langle Mx\;, \;x\rangle \geq0,\;\forall x\in E$), 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 $$\|A\|_M:=\sup_{\substack{x\in \overline{\mbox{Im}(M)}\\ x\not=0}}\frac{\|Ax\|_M}{\|x\|_M}$$

It is true that, if $A\in \mathcal{L}_M(E)$, we have $$\|A\|_M=\displaystyle\sup_{\|x\|_M\leq1}\|Ax\|_M=\displaystyle\sup_{\|x\|_M=1}\|Ax\|_M\,?$$

Thank you everyone !!!


We have

$$\lVert A\rVert_M = \sup_{\substack{x \in \overline{\operatorname{Im} M} \\ \lVert x\rVert_M \leqslant 1}} \lVert Ax\rVert_M = \sup_{\substack{x \in \overline{\operatorname{Im} M} \\ \lVert x\rVert_M = 1}} \lVert Ax\rVert_M\,$$

(provided we interpret $\sup \varnothing = 0$ - since we're looking at a set of non-negative values - in the case $M = 0$), but we cannot in general replace $\overline{\operatorname{Im} M}$ with $E$.

For example, if $M\neq 0$ is not injective and its image is not dense, choose $\xi \in \ker M$ and $\eta \in \operatorname{Im} M$ with $\lVert\xi\rVert_E = \lVert\eta\rVert_E = 1$ and define

$$Ax = \langle x,\xi\rangle\cdot \eta\,.$$

Then clearly $A \in \mathcal{B}(E)$, and since $M$ is normal we have $\overline{\operatorname{Im} M} = (\ker M)^{\perp} \subset \ker A$, so $A \in \mathcal{L}_M(E)$ with $\lVert A\rVert_M = 0$. But we have $M^{1/2}\xi = 0$, so $\lVert\xi\rVert_M = 0$ and

$$\sup_{\lVert x\rVert_M \leqslant 1} \lVert Ax\rVert_M \geqslant \lVert A\xi\rVert_M = \lVert \langle \xi,\xi\rangle\eta\rVert_M = \lVert \eta\rVert_M > 0.$$

Since $t\xi \in \ker M$ for all $t \in \mathbb{C}$, it follows that in fact

$$\sup_{\lVert x\rVert_M \leqslant 1} \lVert Ax\rVert_M = +\infty$$

in this situation.

Since $\lVert y\rVert_M = 0$ for all $y \in \ker M$, and consequently $\lVert x+y\rVert_M = \lVert x\rVert_M$ for $x\in E$ and $y\in \ker M$, we have

$$\lVert A\rVert_M = \sup_{\lVert x\rVert_M \leqslant 1} \lVert Ax\rVert_M$$

for an $A \in \mathcal{L}_M(E)$ if and only if $A(\ker M) \subset \ker M$.

  • $\begingroup$ Thank you. According to your answer, It is not true that, if $A\in \mathcal{L}_M(E)$, we have $$\|A\|_M=\displaystyle\sup_{\substack{\|x\|_M \leq 1,\\ \|y\|_M \leq 1}}|\langle M^{1/2}Ax\;,\;M^{1/2}y\rangle|\,$$, yes or no? $\endgroup$ – Student Dec 15 '17 at 18:51
  • $\begingroup$ In general, no. Or "yes, it is not true in general". $\lVert A\rVert_M$ cares only about what happens on $\overline{\operatorname{Im} M}$, and if $A$ doesn't vanish on $\ker M = (\overline{\operatorname{Im} M})^{\perp}$, $$\sup_{\substack{\lVert x\rVert_M \leqslant 1 \\ \lVert y\rVert_M \leqslant y}} \lvert\langle M^{1/2} Ax, M^{1/2} y\rangle\rvert$$ does. But I made a mistake, the condition is not $\ker M \subset \ker A$ but $A(\ker M) \subset \ker M$, since $\lVert\cdot\rVert_M$ only measures the component in $\overline{\operatorname{Im} M}$. Fixing that. $\endgroup$ – Daniel Fischer Dec 15 '17 at 20:01
  • $\begingroup$ I think there is a wrong in the paper. please see this question :math.stackexchange.com/questions/2568966/… $\endgroup$ – Student Dec 16 '17 at 18:25
  • $\begingroup$ Yes, there's a mistake in the paper. Martin Argerami's example leaves no doubt about that, I presume. $\endgroup$ – Daniel Fischer Dec 17 '17 at 19:26
  • $\begingroup$ I want to show that the inclusion between $$\mathcal{L}_M(E)=\left\{A\in \mathcal{L}(E):\,\,\exists c>0 \quad \mbox{such that}\quad\|Ax\|_M \leq c \|x\|_M ,\;\forall x \in \overline{\mbox{Im}(M)}\right\}$$ and $\mathcal{L}(E)$ is strict. And thank you. $\endgroup$ – Student Jan 21 '18 at 16:37

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