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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 $A\in \mathcal{L}_S(E)$, I want to give a necessary and sufficient conditions under which $\|A\|_S=0$.

I think that $$\|A\|_S=0\Leftrightarrow A(\overline{\mbox{Im}(S)})\subseteq \mbox{Ker}(S)$$

Thank you everyone !!!

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  • $\begingroup$ At a glance, it's equivalent to $S^{1/2}AS=0$. $\endgroup$ – tomasz Dec 24 '17 at 13:13
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Your implication $$\tag1\|A\|_S=0\Leftrightarrow A(\overline{\mbox{Im}(S)})\subseteq \mbox{Ker}(S)$$ holds true.

If $A(\overline{\mbox{Im}(S)})\subseteq \mbox{Ker}(S)$, then clearly $\|A\|_S=0$. Conversely, if $A(\overline{\mbox{Im}(S)})\subsetneq \mbox{Ker}(S)$, there exist $x\in A(\overline{\text{Im}\,S})$ such that $sx=0$; in particular, $s^{1/2}x\ne0$. So there exists $y\in \overline{\mbox{Im}(S)}$ such that $x=Ay$. Thus $S^{1/2}Ay=S^{1/2}x\ne0$. And $s^{1/2}y=0$, because $y\in(\ker S)^\perp$. In conclusion, $$ \frac{\|Ay\|_S}{\|y\|_S}>0, $$ and so $\|A\|_S>0$.

The condition $(1)$ is trivially equivalent with $SAS=0$. Indeed, if $SAS=0$ it follows trivially that $\|A\|_S=0$. And, conversely, if $A(\overline{\text{Im}\,S})\subset\ker S$, then $SAS=0$.

Thus, $$\tag2\|A\|_S=0\Leftrightarrow A(\overline{\mbox{Im}(S)})\subseteq \mbox{Ker}(S)\iff SAS=0 $$

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