Is it true that $\|Ay\|_S\leq \|A\|_S\|y\|_S,\;\forall y\in E\;?$ 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 !!!
 A: 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.
A: 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 $\|Ay\|_S=1$.
