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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 $E$.

For $S\in \mathcal{L}(E)^+$, we consider the following subspace of $\mathcal{L}(E)$: $$\mathcal{L}_S(E)=\left\{A\in \mathcal{L}(E):\,\,\exists M>0 \quad \mbox{such that}\quad\|Ay\|_S \leq M \|y\|_S ,\;\forall y \in \overline{\mbox{Im}(S)}\right\},$$ with $\|y\|_S:=\|S^{1/2}y\|,\;\forall y \in E$. If $A\in \mathcal{L}_S(E)$, the $S$-semi-norm of $A$ is defined us $$\|A\|_S:=\sup_{\substack{y\in \overline{\mbox{Im}(S)}\\ y\not=0}}\frac{\|Ay\|_S}{\|y\|_S}$$

If $A\in \mathcal{L}_S(E)$, is it true that $$\|Ay\|_S\leq \|A\|_S\|y\|_S,\;\forall y\in E\;?$$

Clearly we have $$\|Ay\|_S\leq \|A\|_S\|y\|_S,\;\forall y\in \overline{\mbox{Im}(S)}.$$ So the problem when $y\in \mbox{Ker}(S)$. Is it true that $$\|Ay\|_S\leq \|A\|_S\|y\|_S,\;\forall y\in \mbox{Ker}(S)\;?$$

Thank you everyone !!!

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  • $\begingroup$ What is $L(E)^+$? $\endgroup$ – mathworker21 Dec 25 '17 at 6:41
  • $\begingroup$ Thank you. It is the set of all positive operators $\endgroup$ – Student Dec 25 '17 at 6:47
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If $y\in\ker(S)$ you have that $\|y\|_S=0$. For the inequality to hold you must then have $\|Ay\|_S=0$ for such $y$. This need not be given for an $A\in\mathcal L_S(E)$, as the example in the answer below shows.

However for it holds for any $A\in\mathcal L^S(E)=\{A\in\mathcal L(E)\mid \exists M>0:\ \|Ay\|_S≤M\|y\|_S\ \forall y\in E \}$, since you now you must have an $M$ so that: $$0≤\|Ay\|_S≤M\|y\|_S =0.$$ And $\|Ay\|_S=0$ follows, ie $A$ sends $\ker(S)$ to $\ker(S)$.

In general if $\pi$ is the orthogonal projection onto $\overline{\mathrm{im}(S)}$ then $\|y\|_S=\|\pi(y)\|_S$. Because $A$ sends $\ker(S)$ to $\ker(S)$ we have $\pi A=\pi A\pi.$ Thus $$\|Ay\|_S=\|\pi Ay\|_S=\|\pi A\pi y\|_S=\|A\pi y\|_S$$ and you can actually for $A\in\mathcal L^S(E)$ reduce to the case $y\in\overline{\mathrm{im}(S)}$. For such $A$ you have the inequality on the entire Hilbert space.

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It is not true. Consider for example $$ S= \begin{pmatrix}1 & 0 \\ 0 & 0 \end{pmatrix},\; A= \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} . $$ If $y=(0,1)$, then $\|y\|_S=0$, but $\|Ay\|_S=1$.

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  • $\begingroup$ Ah, I misread the question. In a previous question the OP was asking about a space $\mathcal L^S(E)$, where $\|Ay\|_S≤M\|y\|_S$ had to hold for all $y\in E$. Here he is asking about a bigger space $\mathcal L_S(E)$, where the inequality only needs to hold for $y\in \overline{\mathrm{im}(S)}$. $\endgroup$ – s.harp Dec 27 '17 at 9:02

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