On location of spectra of unbounded operators in the right half-plane
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On location of spectra of unbounded operators in the right half-plane Michael Gil’1 Received: 25 September 2018 / Accepted: 18 April 2020 © African Mathematical Union and Springer-Verlag GmbH Deutschland, ein Teil von Springer Nature 2020
Abstract Let T and T˜ be closed linear operators in a Hilbert space. Let the spectrum of T is in the right half-plane. We suggest the conditions, under which the spectrum of T˜ also lies in the right half-plane. Keywords Hilbert space · unbounded operators · spectrum perturbation Mathematics Subject Classification 47A10 · 47A55
1 Introduction and statement of the main result √ Let H be a separable complex Hilbert space with a scalar product ., ., the norm . = ., . and unit operator I . B(H) means the algebra of all bounded linear operators in H. For a linear operator A, Dom(A) is the domain, σ (A) denotes the spectrum, Rz (A) = (A − z I )−1 (z ∈ / σ (A)) is the resolvent, rs (A) is the spectral radius, A∗ is the adjoint operator. If A ∈ B(H), then A is its operator norm. Let T and T˜ be closed linear operators on H, and inf σ (T ) > γ
(1.1)
for a constant γ > 0. In this paper we suggest conditions that provide the inequality inf σ (T˜ ) > γ .
(1.2)
Conditions of the type (1.2) play an essential role in various applications. In particular, if the spectrum of an operator lies in the open left half-plane, then the corresponding differential equation is asymptotically stable for a wide class of operators, cf. [1–3]. The study of the perturbations and stability of the operator spectra is a well-developed subject. For the classical results see [4–6]. The recent investigations can be found in the papers [7–13] and references given therein. At the same time, to the best of our knowledge the conditions providing inequality (1.2) were not investigated in the available literature.
B 1
Michael Gil’ [email protected] Department of Mathematics, Ben Gurion University of the Negev, P.O. Box 653, 84105 Beer-Sheva, Israel
123
M. Gil’
From (1.1) it follows that T is boundedly invertible. Below we check that condition (1.1) is equivalent to the condition rs (2γ T −1 − I ) < 1.
(1.3)
So ξ(T , γ ) :=
∞
(2γ T −1 − I )k 2 < ∞.
k=0
Now we are in a position to formulate our main result. Theorem 1.1 Assume that T˜ is T -bounded, i.e. T˜ : Dom(T ) → H is continuous (where Dom(T ) is endowed with the graph norm). In addition, let the conditions (1.1), δ := (T˜ − T )T −1 < 1
(1.4)
and ξ(T , γ )
2γ T −1 δ 1−δ
2
4γ 2γ T −1 − I T −1 δ + 1−δ
γ and |2γ /μ − 1|2 = |2γ − μ|2 /|μ|2 = ((2γ − x)2 + y 2 )/|μ|2 = (4γ 2 − 4γ x + x 2 + y 2 )/|μ|2 < (x 2 + y 2 )/|μ|2 = 1.
This proves the lemma. For an A ∈ B(H) assume that rs (A) < 1,
(2.1)
and put ∞
χ(A) :=
Ak 2 .
k=0
Note that
2π
(I − Aeit )−1 h2 dt =
0
2π
∞
0
= 2π
Ak eitk h,
k=0 ∞
∞
A j eit j h dt
j=0
Ak h2 (h ∈ H).
k=0
Hence we easily have χ(A) =
1 2π
2π
(I − Aeit )−1 2 dt.
(2.2)
0
Lemma 2.2 Let A, A˜ ∈ B(H) and (2.1) hold. If, in addition, ˜ 2 + 2AA − A) ˜
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