# Under which conditions $\mathcal{L}^{S}(E)=\mathcal{L}(E)$?

Let $E$ be a complex Hilbert space, $\mathcal{L}(E)$ be the algebra of all bounded linear operators on $E$. For $S\in \mathcal{L}(E)^+$, we set

$$\mathcal{L}^{S}(E)=\left\{A\in \mathcal{L}(E);\;\exists \,M> 0;\;\|S^{1/2}Ay\| \leq M\|S^{1/2}y\|,\;\forall\,y\in E\right\}.$$ If $SA=AS$ and $S^{1/2}$ is surjective then $\mathcal{L}^{S}(E)=\mathcal{L}(E)$.

I want to get a more weaker conditions under which $\mathcal{L}^{S}(E)=\mathcal{L}(E)$. Do you think if $S$ is injective operator with a closed range, the above equality holds?

Thank you!

• Your condition $SA=AS$ for all $A\in\mathcal L(E)$ means that $S$ has to be proportional to $\Bbb 1$. – s.harp Dec 24 '17 at 15:14

## 1 Answer

Suppose there is a $c>0$ so that $S≥c\Bbb1$, ie $S-c\Bbb1$ is a positive operator. If $S$ is positive then this is equivalent to $S$ being bounded from below.

Then $S^{1/2}$ is bounded from below and (if $A\neq0$): $$\|S^{1/2}y\|≥c^{1/2}\|y\|≥\frac{c^{1/2}}{\|S^{1/2}\|\,\|A\|}\|S^{1/2}Ay\|.$$ So for such $S$ we have that $\mathcal L^S(E)=\mathcal L(E)$.

On the other hand every positive $S$ so that $\mathcal L^S(E)=\mathcal L(E)$ must be bounded from below (or $0$). There are two possible situations:

1. $S^{1/2}$ has non-zero kernel.
2. $S^{1/2}$ doesn't have non-zero kernel.

In the first case let $y$ be in the kernel and let $x$ be so that $S^{1/2}(x)\neq0$. Then with $A= |x\rangle\langle y|$ you have $S^{1/2}Ay=S^{1/2}x\neq0$. On the other hand $S^{1/2}y=0$. This means the condition $A\in \mathcal L^S(E)$ becomes impossible.

In the second case let $y_n$ be a sequence of norm $1$ vectors so that $S^{1/2}y_n\to0$ (since $S^{1/2}$ is not bounded from below this exists). We can assume that the $y_n$ are orthogonal and that $\|S^{1/2}y_n\|≤2^{-n}$. Let $x$ be as above and $A=\sum_n \frac1{n^2}|x\rangle\langle y_n|$. Then if $A\in\mathcal L^S(E)$ we must have: $$\|S^{1/2}Ay_n\|=\frac1{n^2}\|S^{1/2}x\|\overset?≤M\|S^{1/2}y_n\|≤M2^{-n}.$$ This is impossible.

• Thank you for your answer. But what do you mean by $$A= |x\rangle\langle y|?$$ – Student Dec 26 '17 at 6:36
• @Student, this is physicists notation. $|x\rangle\langle y|$ means $x\otimes y^*$, ie it is the operator $A(z)=\langle y,z\rangle \, x$ (if the left part of the scalar product is the anti-linear one). – s.harp Dec 27 '17 at 8:56
• Thank you very much – Student Dec 27 '17 at 9:08