Estimation of commutator norms in Davidson's book “C star algebra by example” In ChapterII Davidson's book "$C^{*}$-algebra by example" he used a estimation of norms to prove Thm II.5.3, that is
$$
||A^{\frac{1}{2}}T - TA^{\frac{1}{2}}|| \le 2 ||T||^{\frac{1}{2}}||AT-TA||^{\frac{1}{2}},\quad \forall A \ge 0,A,T \in\mathcal{B}(\mathcal{H})
$$
where $\mathcal{H}$ is a separable Hilbert space, $\mathcal{B}(\mathcal{H})$ is set of bounded linear operators on $\mathcal{H}$. This is actually exII.9 of this book. But I was struggling to prove this from unitary cases to self-adjoint cases. In this book, Davidson hints like below

First prove when $T$ is unitary, $||A^{\frac{1}{2}}T - TA^{\frac{1}{2}}|| \le ||AT-TA||^{\frac{1}{2}},\quad \forall A \ge 0$, then for $T =T^*$ set
$$
U_{s} = (sT + \sqrt{-1})(sT - \sqrt{-1})^{-1} ,\quad s>0
$$
is unitary and apply the results before.

I do not get the usage of construction of $U_{s}$. I try to use
$s(AT-TA) = (AU_{s}-U_{s}A)(sT-\sqrt{-1}) + sU_{s}(AT-TA),s>0$ to control norms but it seems no use.
So how can one engage the construction to prove the inequality? I appreciate any help of this.
 A: Note that $1\not\in\sigma(U_s)$. Your equality can be written as
\begin{align}
AT-TA&=s^{-1}(I-U_s)^{-1}\,(AU_s-U_sA)\,(sT-i).\\[0.3cm]
\end{align}
One can also check that
$$
(I-U_s)^{-1}=\frac{i}2\,(sT-i).
$$
and
$$
\|(sT-i)^{-1}\|\leq1
$$
So
\begin{align}
AT-TA&=\frac{i}{2s}(sT-I)\,(AU_s-U_sA)\,(sT-i).\\[0.3cm]
\end{align}
Assume first that $\|T\|=1$.
Then
\begin{align}
\|A^{1/2}T-TA^{1/2}\|
&\leq\frac1{2s}\,\|sT-i\|^2\,\|A^{1/2}U_s-U_sA^{1/2}\|\\[0.3cm]
&\leq \frac1{2s}\,\|sT-i\|^2\,\|AU_s-U_sA\|^{1/2}\\[0.3cm]
&\leq \frac1{2s}\,\|sT-i\|^2\,s^{1/2}\|I-U_s\|^{1/2}\,\|(sT-i)^{-1}\|^{1/2}\,\|AT-TA\|^{1/2}\\[0.3cm]
&\leq \frac1{\sqrt{2s}}\,\|sT-i\|^2\,\|AT-TA\|^{1/2}\\[0.3cm]
&= \frac{1+s^2\|T\|^2}{\sqrt{2s}}\,\|AT-TA\|^{1/2}\\[0.3cm]
&= \frac{1+s^2}{\sqrt{2s}}\,\|AT-TA\|^{1/2}\\[0.3cm]
\end{align}
Taking $s=1$,
$$
\|A^{1/2}T-TA^{1/2}\|\leq \sqrt2\,\|AT-TA\|^{1/2}.
$$
For arbitrary selfadjoint $T$, applying the above to $T/\|T\|$, we get
$$
\|A^{1/2}T-TA^{1/2}\|\leq \sqrt2\,\|T\|^{1/2}\,\|AT-TA\|^{1/2}.
$$
