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?

  • 2
    $\begingroup$ I don't think the norm bound is relevant. The claim to be proven is invariant under scaling of B. $\endgroup$
    – PhoemueX
    Dec 27, 2022 at 10:43
  • $\begingroup$ You are right, so it seems that we only need to assume that $B$ is a bounded operator. $\endgroup$
    – Robin
    Dec 27, 2022 at 11:05
  • 2
    $\begingroup$ 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 ${\cal B}(H,H)$. $\endgroup$
    – Gerd
    Jan 1, 2023 at 10:29
  • $\begingroup$ The case of finite dimension, solution to $AX+XA =B$ $\endgroup$
    – Apass.Jack
    Jan 4, 2023 at 12:38

1 Answer 1


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.

  • $\begingroup$ Thanks! How did you get the uniqueness at the very end (which is not needed to answer the question)? From what you showed, it only follows that your integral expression is a right inverse of the Ljapunov operator, which gives you existence, but not uniqueness. [I will award the bounty to you in any case] $\endgroup$
    – PhoemueX
    Jan 5, 2023 at 10:52
  • $\begingroup$ 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. $\endgroup$
    – Hanno
    Jan 6, 2023 at 11:29

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