Generalisation of the norm of bounded linear operators 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 !!!
 A: 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 $A(\ker M) \subset \ker M$.
