2
$\begingroup$

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 !!!

$\endgroup$
3
$\begingroup$

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 $A(\ker M) \subset \ker M$.

$\endgroup$
  • $\begingroup$ 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

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.