On operator equation AX − XB = Y

Dragoljub Kečkić

Abstract In cases when the equation has a unique solution for all Y, i.e. when the mapping X → AX − XB is invertible, the integral representations of the solution are given, as well as the norm estimates of the solution in terms of Y. Results are, in fact, the extensions to infinitedimensional case, of those given by Bhatia and Rosenthal.

It is well known that if σ(A) ∩ σ(B) = ϕ the equation

AX − XB = Y  (1)

has a unique solution for all Y. It is a consequence of the fact that for an operator Δ_{A,B} : B(H) → B(H) Δ_{A,B}(X) = AX − XB, there holds σ(Δ_{A,B}) ⊂ σ(A) − σ(B), [4]. Also it is possible to find an explicit formula for this solution in terms of Y.

Proposition 1 Let Γ be a union of closed curve such that

Ind_{Γ}(a) = 1 for all a ∈ σ(A)  and  Ind_{Γ}(b) = 0 for all b ∈ σ(B).

(2)

Then X = \frac{1}{2πi} ∫_{Γ}(A − λI)^{−1}Y (B − λI)^{−1}dλ

Proof. The equation (1) is equivalent to (A − λI)X − X(B − λI) = Y for λ ∈ Γ. Multiplying the last equation by corresponding resolvents, and integrating over Γ, we easy obtain the result, taking into account that ∫_{Γ}(B − λI)^{−1}dλ = 0 and ∫_{Γ}(A − λI)^{−1}dλ = −2πiI.

In some cases when the spectra of A and B are, in a certain sense, regularly distributed, we can get some other formulae.

Proposition 2 a) If σ(A) ⊂ {z|Rez > 0} and σ(B) ⊂ {z|Rez < 0} then X = ∫_{0}^{+∞} e^{−tA}Y e^{tB}dt; b) If σ(A) ⊂ {z| |z| > ρ} and σ(B) ⊂ {z| |z| < ρ} then X = ∑_{n=0}^{+∞} A^{−n−1}Y B^{n}.

1Mathematics Subject Classification 2000 47B47, 47A30. Key words and Phrases: Generalized derivation

Proof. The spectra are compact sets. So σ(A) and σ(B) are separated by a strip of positive width d, in the first case. In the second case, σ(B) is contained in some disk of radius ρ_{1}, and σ(A) is contained in the complement of some disk of radius ρ_{2}. From this we conclude that for all t > 0 there holds ||e^{−tA}Y e^{tB}|| ≤ e^{−td}, and also that for all n ≥ 0 there holds ||A^{−n−1}Y B^{n}|| ≤ (ρ_{1}/ρ_{2})^{n}. From these inequalities it follows that the above integral and the series converge. Further, we can directly check: a) AX − XB = ∫_{0}^{+∞}(Ae^{−tA}Y e^{tB} − e^{−tA}Y e^{tB}Bdt = ∫_{0}^{+∞} \frac{d}{dt} (−e^{−tA}Y e^{tB}) dt = Y; b) AX − XB = ∑_{n=0}^{+∞} A^{−n}Y B^{n} − ∑_{n=0}^{+∞} A^{−n−1}Y B^{n+1} = Y.

Considering the previous proposition, part a), one can conjecture that for selfadjoint A and B there holds X = ∫_{−∞}^{+∞} e^{−itA}Y e^{itB}dt. Unfortunately, this integral does not necessarily converge. Nevertheless the following theorem holds.

Theorem 1 If A = A^{∗} and B = B^{∗}, and σ(A) ∩ σ(B) = ∅ then solution of equation (1) is given by X = ∫_{−∞}^{+∞} e^{−itA}Y e^{itB}f(t)dt, where f ∈ L^{1}(R) such that its Fourier transform f̃(t) = ∫_{−∞}^{+∞} e^{−tτ} f(τ)dτ = 1/t in some neighbourhood of σ(A) − σ(B).

For the proof of this theorem we need a lemma.

Lemma 1 Let X be solution of the equation (1), (with σ(A) ∩ σ(B) = ∅), and let X_{n} be solution of equation A_{n}X_{n} − X_{n}B = Y. If A_{n} → A in B(H) then X_{n} → X in B(H) as well.

Proof. Let Γ be a union of closed curves satisfying (2). For n large there holds σ(A) ⊂ int Γ, [3]. Now by proposition 1, we have

||X_{n} − X|| ≤ \frac{1}{2π} ∫_{Γ} ||(A_{n} − λI)^{−1} − (A − λI)^{−1}|| ||Y || ||(B − λI)^{−1}||d|λ|

However (A_{n} − λI)^{−1} − (A − λI)^{−1} = (A_{n} − λI)^{−1}(A − A_{n})(A − λI)^{−1}, and so

||X_{n} − X|| ≤ \frac{1}{2π} M_{n}MN||Y || ||A_{n} − A||length(Γ),

where M_{n} = max_{Γ}||(A_{n} − λI)^{−1}||, M = max_{Γ}||(A − λI)^{−1}||, N = max_{Γ}||(B − λI)^{−1}||. The proof is complete.

Remark 1. The previous proof is based on Proposition 1. So, one can arise the question whether the union of closed curves Γ (from Proposition 1.) always exists or not, i.e. whether the previous proof can fail in some occasions or not. However, the properties of compact sets on real line ensure that such a union of closed curves always exists, under the assumptions of Lemma 1. Namely, if σ(A), σ(B) are contained in some interval [a, b], and d = dist(σ(A), σ(B)) then we can choose Γ with total length less or equal to 2(b − a) + [(b − a)/d]ϵ (ϵ arbitrarily small), since (σ(A) ∪ σ(B))^{c} is an open set with no connected components of length less then d.

Proof of theorem. Suppose that A and B are diagonal operators, i.e. operators with pure point spectra. Let Au = αu and Bv = βv. We easy obtain

⟨u, Ae^{−itA}Y e^{itB}v⟩ = e^{it(β−α)}α ⟨u, Y v⟩ and
⟨u, e^{−itA}Y e^{itB}Bv⟩ = e^{it(β−α)}β ⟨u, Y v⟩ and also
⟨u, AX − XBv⟩ = (α − β) ⟨u, Y v⟩ ,

since α − β ∈ σ(A) − σ(B). The eigenvectors of a diagonal operators span the whole space and the result in this case follows.

Suppose now, that the operator B is diagonal, and the operator A is not diagonal. For all ϵ > 0 there exist operators A_{0} and P such that A = A_{0} + P, A_{0} is diagonal and ||P ||_{C_{2}} < ϵ (and clearly ||P ||_{B(H)} < ϵ). Let X_{0} satisfies the equation A_{0}X_{0} −X_{0}B = Y, and let X be the solution of the equation (1). Then for suitable choice of ϵ, the expression ||X_{0} − X|| can be made arbitrarily small, by lemma. Thus X = lim_{||P ||→0} X_{0} = lim_{||P ||→0} ∫_{−∞}^{+∞} e^{−itA_{0}}Y e^{itB}f(t)dt.

But also

||X_{0} − ∫_{−∞}^{+∞} e^{−itA}Y e^{itB}f(t)dt|| ≤ ∫_{−∞}^{+∞} ||e^{−itA_{0}} − e^{−itA}|| ||Y || ||e^{itB}|| |f(t)|dt.

The last integrand is less or equal to 2||Y || |f(t)|, and taking into account that ||e^{−itA_{0}} − e^{−itA}|| → 0, pointwise, the dominance convergence theorem gives us

lim_{||P ||→0} ∫_{−∞}^{+∞} e^{−itA_{0}}Y e^{itB}f(t)dt = ∫_{−∞}^{+∞} e^{−itA}Y e^{itB}f(t)dt,

as well. Hence X = ∫_{−∞}^{+∞} e^{−itA}Y e^{itB}f(t)dt. In a similar way the assumption that B is diagonal can be omitted. The proof is complete.

The integral representations obtained here, allow us to estimate the norm of the solution in terms of Y and δ = dist(σ(A), σ(B)).

Corollary 1 a) If the spectra of normal operators A and B are separated by a strip which width is greater or equal to δ, then the inequality

|||X||| ≤ \frac{1}{δ}|||Y |||  (3)

holds.

b) If the spectrum of B lies in a circle with center a and radius ρ, and if the spectrum of A lies outside a circle with center a and radius ρ + δ then inequality (3) holds.

Proof. Norms of X and Y in equation (1) will not be changed if A and B are replaced by e^{iθ}A − a and e^{iθ}B − a. Now the integral representations from Proposition 2 give us the result.

Corollary 2 Under the assumptions of the theorem 1. there holds |||X||| ≤ (c/δ)|||Y ||| in every unitarily invariant norm, where c = inf_{f∈Φ}||f||_{L^{1}(R)}, and Φ = {f ∈ L^{1}(R)|f̃(s) = 1/s in a neighbourhood of (−∞, −1] ∪ [1, +∞)}.

Proof. Let f_{δ} be a function from L^{1}(R) such that f̃_{δ}(t) = 1/t on the set {t| |z| > δ} ⊃ σ(A) − σ(B). Then

|||X||| ≤ ∫_{−∞}^{+∞} |||Y ||| |f_{δ}(t)|dt = |||Y ||| ∫_{−∞}^{+∞} |f_{δ}(τ/δ)| /δdτ.

Let us denote f(t) = f_{δ}(t/δ). Then f̃(s) = δf̃_{δ}(sδ) = δ \frac{1}{sδ} = 1/s for all s such that |sδ| > δ, i.e. |s| > 1. Thus

|||X||| ≤ |||Y |||\frac{||f||_{L^{1}}}{δ} ,

for any f ∈ Φ. Taking the infimum, we obtain the result.

Remark 2. The constant c in the previous corollary is equal to π/2, as shown in [6].

Remark 3. All results in this note were stated in [1] and [2] and also proved in finite dimensional case. However the infinite dimensional part of the proof is omitted in both works [1] and [2].

References

[1] Bhatia R., Matrix Analysis Springer, 1997

[2] Bhatia R. and Rosenthal P., How and why to solve the equation AX −XB = Y Bull. London Math. Soc. 29(1997)

[3] Gohberg I.C. and Krein M.G., Introduction to the theory of linear nonselfadjoint operators Transl. Math. Monographs Providence R.I. 1969

[4] Rudin W., Functional Analysis (in Russian) ”Mir” Moscow 1975

[5] von Neumann J., Charakterisierung des Spectrums eines Integraloperators, Actualites Sci. Ind., N^{0}229, Hermann, Paris 1935

[6] Nagy B.Sz., Ueber die Ungleichung von H. Bohr Math. Nachr. 9(1953)255-259