# 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?

• I don't think the norm bound is relevant. The claim to be proven is invariant under scaling of B. Dec 27, 2022 at 10:43
• You are right, so it seems that we only need to assume that $B$ is a bounded operator. Dec 27, 2022 at 11:05
• Using a result on the spectrum of inner automorphisms of Banach algebras (see Bonsall F.F., Duncan J.: Complete Normed Algebras, p.88, Proposition 9(ii)) it follows that the spectrum of $S \mapsto B^{-1}SB$ is a subset of $\{\mu/\lambda: \mu,\lambda \in \sigma(B)\} \subseteq (0,\infty)$. Hence $S \mapsto S +B^{-1}SB$ is invertible in the algebra of bounded operators on ${\cal B}(H,H)$.
– Gerd
Jan 1, 2023 at 10:29
• The case of finite dimension, solution to $AX+XA =B$ Jan 4, 2023 at 12:38

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.
• Thank you @PhoemueX for the bounty $-$ very appreciated! I shall edit the answer to respond to your comment, including that $S(T)$ is also a left inverse. Jan 6, 2023 at 11:29