Decomposition of a bounded Hilbert space operator I am trying to prove the following homework problem:
Let $B$ be a positive operator on a Hilbert space $H$ with $\Vert B\Vert=1$ and $B$ is invertible. Try to prove that for each $\,T\in \mathfrak{B}(H,H)$, there exists an $S\in\mathfrak B(H,H)\,$ such that $\,T=\frac{1}{2}(BS+SB)\,$.
I have tried to apply the polar decomposition of $T$ but seemingly in vain. Actually, I am wondering how the condition $\Vert B\Vert=1$ is used.
Can somebody give me some hints on this problem?
 A: Let us begin within a more general setting. Fix an $\,A\in\mathfrak B=\mathfrak B(H,H)$,
and define the Lyapunov operator
$$L:\mathfrak B\to\mathfrak B,\quad X\mapsto A^*X+XA$$
(which is a special case of the Sylvester operator).
An operator $A\in\mathfrak B\,$ is said to be stable if its spectrum is contained in the open left half-plane of $\,\mathbb C$.
An $A\in\mathfrak B\,$ is said to be positively stable, if $-A\,$ is stable, i. e., each spectral value of $A$ has a strictly positive real part.
In particular, $A^*$ is positively stable iff $A$ is positively stable.
For a positively stable $A$ the expressions $\,\exp(-tA^*)\,$ and $\,\exp(-tA)\,$
where $0\leqslant t< \infty\,$ define norm-continuous one-parameter semigroups tending to zero for large $t$. They allow for each $X\in\mathfrak B\,$ to define the nicely convergent integral
$$S(X)\:=\:\int\limits_0^\infty e^{-tA^*}X\,e^{-tA}\,dt\,,$$
which is an explicit inverse of the Lyapunov operator.
Checking this involves the product rule for differentiation with respect to $t\,$:
$$L\big(S(X)\big)=A^*S(X)+S(X)A
=\underbrace{\int\limits_0^\infty e^{-tA^*}
(A^*X+XA)\:e^{-tA}\,dt}_{\qquad\quad=\,S(L(X))}
=-\int\limits_0^\infty \frac d{dt}\big(e^{-tA^*}X
\:e^{-tA}\big)dt = X$$
If $B\in\mathfrak B$ is a positive operator $($then $B^*=B\,)$ and invertible, then $\frac12 B\,$ is strictly positive, hence positively stable. It follows from the preceding that for each $T\in\mathfrak B$ there is a Lyapunov preimage $S\in\mathfrak B$, i.e.,
$\,L(S)=\frac12(BS+SB)\,=T$, and $\,S\,$ is unique.
