AN be bounded self-adjoint operators in H such that AiAj is compact for any i 6= j

B

J

M

A

www.emis.de/journals/BJMA/

ON THE ESSENTIAL SPECTRUM OF THE SUM OF SELF-ADJOINT OPERATORS AND THE CLOSEDNESS OF

THE SUM OF OPERATOR RANGES

IVAN S. FESHCHENKO Communicated by C. Badea

Abstract. Let Hbe a complex Hilbert space, and A₁, . . . , A_N be bounded self-adjoint operators in H such that AiAj is compact for any i 6= j. It is well-known thatσe(PN

i=1Ai)\ {0}= (∪^N_i=1σe(Ai))\ {0}, whereσe(B) stands for the essential spectrum of a bounded self-adjoint operatorB.

In this paper we get necessary and sufficient conditions for 0∈σe(PN i=1Ai).

This conditions are formulated in terms of the projection valued spectral mea- sures ofA_i, i= 1, . . . , N. Using this result, we obtain necessary and sufficient conditions for the sum of ranges ofA_i,i= 1, . . . , N to be closed.

1. Introduction

Let H be a complex Hilbert space, and A₁, . . . , A_N be N > 2 bounded self- adjoint operators inH such thatA_iA_j is compact for i6=j. Define

A=

N

X

i=1

A_i.

1.1. On the essential spectrum of the sum of self-adjoint operators.

The following result on the essential spectrum ofA,σ_e(A), is well-known and has applications in scattering theory and spectral analysis of Hankel operators (see, e.g., [7, Proposition 3.1],[6, Chapter 10, Lemma 1.5]).

Date: Received: 26 December 2012; Accepted: 14 April 2013.

2010Mathematics Subject Classification. Primary 47B15; Secondary 46C07.

Key words and phrases. Self-adjoint operator, compact operator, essential spectrum, sum of operator ranges, closedness.

55

Theorem A. We have

σe(A)\ {0}=

N

[

i=1

σe(Ai)

!

\ {0}.

For convenience of the reader, we will prove Theorem A (following [7]) in Sec- tion 3.

What one can say about the point 0?

If H is infinite dimensional, then 0 ∈ ∪^N_i=1σ_e(A_i). Indeed, it is easily checked, that ifB₁, B₂ are bounded self-adjoint operators inHsuch thatB₁B₂ is compact, then 0 ∈/ σ_e(B₁) ⇒ B₂ is compact. Hence, if 0 ∈/ σ_e(A₁), then A₂ is compact, and, consequently,σ_e(A₂) ={0}.

Thus, the following question arises naturally: when 0 ∈ σ_e(A)? We will give an answer to this question in terms of the projection valued spectral measures of Ai, EAi(·),i= 1, . . . , N. Define the subspace

Hε=Hε(A1, . . . , AN) =

N

\

i=1

EAi([−ε, ε])H, ε>0. (1.1) Main Theorem. 0∈σ_e(A) if and only if the subspace H_εis infinite dimensional for any ε >0.

1.2. On the closedness of the sum of operator ranges. The problem on the closedness of the sum of operator ranges (criteria for the sum of operator ranges to be closed, properties of collections of operatorsX_i such thatPn

i=1Ran(X_i) is closed), in particular, the problem on the closedness of the sum ofnsubspaces of a Hilbert space (criteria for the sum ofnsubspaces of a Hilbert space to be closed, properties of collections of subspaces with closed sum), is an important problem of functional analysis. This problem was studied in numerous publications and has many applications in various branches of mathematics, see, for example, [4]

and the bibliography therein (unfortunately, English translation of the Russian original of this paper in some places is bad. For example, the word ”closeness”

must be replaced by ”closedness”), [3], [1].

We consider the following question: when the sum of ranges ofA_i,

N

X

i=1

Ran(A_i) = ( _N

X

i=1

y_i |y_i ∈Ran(A_i) )

= ( _N

X

i=1

A_ix_i |x_i ∈ H )

,

is closed? This question arises naturally in connection with the following re- sult having applications in theoretical tomography (see [5] and the bibliography therein).

Theorem B. Let P1, . . . , PN be orthogonal projections in H. Suppose PiPj is compact for i6=j. ThenPN

i=1Ran(P_i) is closed.

Remark 1.1. Theorem B is stated only for Hilbert spaces, while it was proved by Lars Svensson for reflexive Banach spaces [8], by Harald Lang for Frechet spaces [5], and by Lars Svensson for Hausdorff locally compact topological vector spaces [9].

Using Main Theorem, we prove a criterion for PN

i=1Ran(A_i) to be closed.

Recall that the subspacesH_ε, ε>0, are defined by (1.1). Clearly, H₀ =

N

\

i=1

Ker(A_i).

Note thatH0 ⊂ Hε for any ε >0.

Theorem 1.2. PN

i=1Ran(A_i) is closed if and only if H_ε =H₀ for some ε >0.

Remark 1.3. We will prove that if H_ε H₀ is finite dimensional for someε > 0, then PN

i=1Ran(A_i) is closed.

Remark 1.4. If PN

i=1Ran(A_i) is closed, then PN

i=1Ran(A_i) = H H₀. Indeed, suppose PN

i=1Ran(Ai) is closed. We have H

N

X

i=1

Ran(Ai)

!

=

N

\

i=1

(H Ran(Ai)) =

N

\

i=1

Ker(Ai) =H0. Hence, PN

i=1Ran(A_i) = H H₀. Using this criterion forPN

i=1Ran(A_i) to be closed, we get the following gener- alization of Theorem B.

Theorem 1.5. Let Ap,i : Kp,i → H, i = 1, . . . , Np, p = 1, . . . , m be bounded linear operators (hereK_p,i are Hilbert spaces,m, N1, . . . , Nm are natural numbers) such that A^∗_p,iA_q,j is compact for any p 6= q, i = 1, . . . , N_p, j = 1, . . . , N_q. If R_p =PNp

i=1Ran(A_p,i) is closed for any p= 1, . . . , m, then Pm

p=1R_p is closed.

Corollary 1.6. If Ran(Ai) is closed for i = 1, . . . , N, then PN

i=1Ran(Ai) is closed.

Note that our proof of Theorem 1.5 is not a generalization of the proof of Theorem B (and we don’t use Theorem B in the proof of Theorem1.5).

1.3. Notation. In this paper we consider only complex Hilbert spaces usually denoted by the lettersH,K. The scalar product inHis denoted byh·,·i, andk · k stands for the corresponding norm, kxk² =hx, xi. The identity operator onH is denoted by IH or simply I if it is clear which Hilbert space is being considered.

For a bounded linear operator X : H → H, σ(X) denotes the spectrum of the operator X.

2. Auxiliary results and notions

2.1. The essential spectrum of a self-adjoint operator. Following [2, Chap- ter 9], we recall the definition and some properties of the essential spectrum of a self-adjoint operator.

Let A be a bounded self-adjoint operator in a complex Hilbert space H. The essential spectrum of A, σ_e(A), is the set of all λ ∈ σ(A) such that either λ is a limit point of σ(A) or Ker(A−λI) is infinite-dimensional. The set σ_d(A) =

σ(A)\σ_e(A) is called the discrete spectrum of A. Clearly, λ∈σ_d(A) if and only if λ is an isolated point ofσ(A) and Ker(A−λI) is finite-dimensional.

It terms of the spectral measure E_A(·) the essential spectrum of A can be characterized as follows: λ∈σ_e(A) if and only if E_A((λ−ε, λ+ε))H is infinite- dimensional for anyε >0.

More convenient for us is a description ofσ_e(A) in terms of singular sequences.

Recall that a sequence{xk |k >1}, xk ∈ H, k>1, is called a singular sequence for A at the pointλ if

(1) x_k →0 weakly as k → ∞;

(2) x_k does not converge to 0 as k → ∞;

(3) (A−λI)x_k→0 as k → ∞.

Let us denote by sing(A, λ) the set of all sequences {x_k | k > 1} which are singular for A at λ.

Proposition 2.1. λ∈σ_e(A) if and only if sing(A, λ)6=∅. 2.2. On the sum of operator ranges.

Proposition 2.2. Let K1, . . . ,Kn and H be Hilbert spaces, Bi : Ki → H a bounded linear operator, i= 1, . . . , n. If Pn

i=1Ran(B_i) =H, then Pn

i=1B_iB_i^∗ >

εI for some ε >0.

Proof. Define an operator B :K₁⊕. . .⊕ K_n→ H by B(x₁, . . . , x_n) =

n

X

i=1

B_ix_i, x_i ∈ K_i, i= 1, . . . , n.

Then B^∗ :H → K₁⊕. . .⊕ K_n and

B^∗x= (B₁^∗x, . . . , B^∗_nx), x∈ H.

Since Ran(B) = Pn

i=1Ran(B_i) = H, we conclude that B^∗ is an isomorphic embedding, that is, there existsc > 0 such that kB^∗xk>ckxk, x∈ H. We have

kB^∗xk² =

n

X

i=1

kB_i^∗xk² =

n

X

i=1

hB_iB_i^∗x, xi=

* _n X

i=1

B_iB_i^∗

! x, x

+ . Hence, Pn

i=1B_iB_i^∗ >c²I.

We will need a simple corollary of Proposition 2.2.

Corollary 2.3. Let K₁, . . . ,K_n and H be Hilbert spaces, B_i :K_i → H a bounded linear operator, i= 1, . . . , n. If Pn

i=1Ran(B_i) is closed, then

n

X

i=1

Ran(Bi) =Ran

n

X

i=1

BiB_i^∗

! . The following proposition is well-known.

Proposition 2.4. Let B be a bounded self-adjoint operator in H. Ran(B) is closed if and only if σ(B)∩((−ε,0)∪(0, ε)) =∅ for some ε >0.

The proof is trivial and is omitted.

3. Proof of Theorem A

To prove Theorem A, we need the following simple lemma which shows relation between singular sequences of two bounded self-adjoint operators whose product is compact.

Lemma 3.1. Let B₁, B₂ be bounded self-adjoint operators in H. Suppose B₁B₂ is compact. If {x_k | k > 1} ∈ sing(B₂, λ), where λ 6= 0, then {x_k | k > 1} ∈ sing(B₁,0).

Proof. We have (B₂−λI)x_k → 0, k → ∞. It follows that B₁(B₂−λI)x_k → 0, k → ∞, that is, B₁B₂x_k −λB₁x_k → 0, k → ∞. Since B₁B₂ is compact and x_k → 0 weakly, we conclude that B₁B₂x_k → 0. Hence, λB₁x_k → 0, k → ∞, whence, B₁x_k→0, k→ ∞. Therefore, {x_k |k >1} ∈sing(B₁,0).

Proof of Theorem A. 1. Suppose that λ∈ σe(Ai) for some i, and λ 6= 0. Let us show that λ ∈ σe(A). There exists a sequence {xk | k > 1} ∈ sing(Ai, λ). By Lemma3.1, {x_k|k >1} ∈sing(A_j,0) for j 6=i. We have

(A−λI)x_k= (A_i−λI)x_k+X

j6=i

A_jx_k →0, k → ∞.

Hence, {x_k|k >1} ∈sing(A, λ), whence λ∈σ_e(A).

2. Suppose that λ ∈ σ_e(A), λ 6= 0. Let us prove that λ ∈ σ_e(A_i) for some i.

There exists {x_k |k>1} ∈sing(A, λ). We have (A−λI)xk=

N

X

j=1

Aj −λI

!

xk →0, k → ∞. (3.1)

Leti∈ {1, . . . , n}. ThenA_i(PN

j=1A_j−λI)x_k →0,k → ∞, that is,PN

j=1A_iA_jx_k− λA_ix_k → 0, k → ∞. Since A_iA_j is compact for j 6= i and x_k → 0 weakly, we conclude thatA_iA_jx_k→0,k → ∞forj 6=i. Hence,A²_ix_k−λA_ix_k →0,k → ∞, i.e., (A_i−λI)A_ix_k →0,k → ∞.

Suppose that there exists i such that the sequence A_ix_k does not converge to 0 as k → ∞. Then{A_ix_k |k >1} ∈sing(A_i, λ), whence λ∈σ_e(A_i).

Now assume thatA_ix_k →0,k→ ∞for anyi= 1, . . . , N. From (3.1) it follows that λx_k →0,k → ∞. Hence,x_k →0, k→ ∞, a contradiction.

The proof is complete.

4. Proof of Main Theorem

To prove Main Theorem, we need the following two lemmas.

Lemma 4.1. Let B, C be bounded self-adjoint operators in H. If BC is compact, then

E_B(R\[−ε, ε])E_C(R\[−δ, δ]) is compact for any ε, δ >0.

Proof. Set P =E_B(R\[−ε, ε]),Q=E_C(R\[−δ, δ]). ThenB² >ε²P,C² >δ²Q.

SinceBCis compact, we see thatBC²B is compact. Clearly,BC²B >δ²BQB.

Hence, BQB is compact. But BQB = (BQ)(BQ)^∗. Consequently, BQ is com- pact. Then QB²Q is compact. Clearly, QB²Q > ε²QP Q. Hence, QP Q is compact. But QP Q = (QP)(QP)^∗. Consequently, QP is compact. It follows

that P Q= (QP)^∗ is compact.

Lemma 4.2. Let P1, . . . , Pn be orthogonal projections in H. Suppose that PiPj

is compact for any i6=j. Then there exists ε >0 such that σ(

n

X

i=1

P_i)∩(0, ε) =∅.

Proof. Set K_i =Ran(P_i),i= 1, . . . , n. Define an operator Γ :H → K₁⊕. . .⊕ K_n by

Γx= (P₁x, . . . , P_nx), x∈ H.

Then Γ^∗ :K₁⊕. . .⊕ K_n → H and

Γ^∗(x₁, . . . , x_n) = x₁+. . .+x_n, x_i ∈ K_i, i= 1, . . . , n.

Hence, Γ^∗Γ =Pn

i=1P_i. Clearly, the operator ΓΓ^∗ :K₁⊕. . .⊕ K_n→ K₁⊕. . .⊕ K_n and its block decomposition is equal to

ΓΓ^∗ = (P_i ^Kj:K_j → K_i |16i, j 6n).

Since P_iP_j is compact for any i6=j, we conclude that (ΓΓ^∗)_i,j =P_i ^Kj=P_iP_j ^Kj

is compact for i 6=j. Consequently, ΓΓ^∗−I is compact. By the Weyl theorem, σ_e(ΓΓ^∗) = σ_e(I)⊂ {1}. Hence, 0∈/ σ_e(ΓΓ^∗), whenceσ(ΓΓ^∗)∩(0, ε) =∅for some ε >0. Sinceσ(Γ^∗Γ)\ {0}=σ(ΓΓ^∗)\ {0}, we conclude thatσ(Pn

i=1P_i)∩(0, ε) =

∅.

Proof of Main Theorem. 1. Suppose the subspace H_ε is infinite dimensional for any ε >0. We claim that 0∈σ_e(A). To prove this, we construct {x_k| k>1} ∈ sing(A,0) as follows. Take

x₁ ∈ H₁, kx₁k= 1.

Suppose x₁, . . . , x_k have already been defined. Clearly, we can choose x_k+1 ∈ H_1/(k+1), kx_k+1k= 1

such that x_k+1 is orthogonal to x_i, i = 1, . . . , k. Hence, we obtain the sequence {xk | k > 1}. By the construction, {xk | k > 1} is orthonormal. Moreover, kAixkk61/k, i= 1, . . . , N. It follows that kAxkk6 N/k,k >1. Consequently, Ax_k →0 as k→ ∞. Hence, {x_k |k>1} ∈sing(A,0), whence 0∈σ_e(A).

2. SupposeH_εis finite dimensional for someε >0. Let us show that 0∈/ σ_e(A).

Define P_i = E_Ai(R\[−ε, ε]), i = 1, . . . , N. By Lemma 4.1, P_iP_j is compact for any i 6=j. By Lemma 4.2, there exists δ >0 such that σ(PN

i=1P_i)∩(0, δ) = ∅. It follows that

N

X

i=1

P_i +δQ>δI,

where Qis the orthogonal projection onto Ker(PN

i=1P_i). Clearly, Ker

N

X

i=1

Pi

!

=

N

\

i=1

Ker(Pi) =

N

\

i=1

EAi([−ε, ε])H =Hε.

Hence, Q is a finite dimensional orthogonal projection. Since A²_i > ε²P_i, i = 1, . . . , N, we have

N

X

i=1

A²_i +ε²δQ >

N

X

i=1

ε²P_i+ε²δQ>ε²δI.

Hence,

N

X

i=1

A²_i +µQ>µI, (4.1)

where µ=ε²δ.

Now we are in a position to prove that 0 ∈/ σ_e(A). Suppose that 0 ∈ σ_e(A).

There exists a sequence {x_k | k > 1} ∈ sing(A,0). We have (PN

j=1A_j)x_k → 0 as k → ∞. Let i ∈ {1, . . . , N}. Then A_i(PN

j=1A_j)x_k → 0 as k → ∞, that is, PN

j=1A_iA_jx_k →0 ask → ∞. SinceA_iA_j is compact forj 6=iandx_k →0 weakly, we conclude that A_iA_jx_k → 0 as k → ∞ for j 6= i. Consequently, A²_ix_k → 0, k → ∞. Since{x_k|k >1}is bounded, we conclude thathA²_ix_k, x_ki →0,k → ∞.

Hence, kA_ix_kk² → 0, k → ∞, whence A_ix_k → 0, k → ∞. Since Q is a finite dimensional operator and x_k → 0 weakly, we conclude that Qx_k → 0, k → ∞.

From (4.1) it follows that

N

X

i=1

hA²_ixk, xki+µhQxk, xki>µkxkk², that is,

N

X

i=1

kA_ix_kk² +µkQx_kk² >µkx_kk². It follows that x_k →0 as k → ∞, a contradiction.

Hence, 0∈/σ_e(A).

5. Proof of Theorem 1.2

Proof of Theorem 1.2. LetK=H H₀, then H=H₀⊕ K. With respect to this orthogonal decomposition A_i = 0⊕B_i,i= 1, . . . , N, where B_i is a bounded self- adjoint operator inK. SinceH₀ =∩^N_i=1Ker(A_i), we see that∩^N_i=1Ker(B_i) = {0}.

Since A_iA_j is compact for i6=j, we conclude that B_iB_j is compact for i6=j. 1. SupposePN

i=1Ran(A_i) is closed. Let us show thatH_δ =H₀ for someδ >0.

PN

i=1Ran(B_i) = PN

i=1Ran(A_i) is closed. Since K

N

X

i=1

Ran(B_i)

!

=

N

\

i=1

(K Ran(B_i)) =

N

\

i=1

Ker(B_i) = {0},

we conclude that PN

i=1Ran(B_i) = K. By Proposition 2.2, PN

i=1B_i² > εI for some ε > 0. Set δ = p

ε/(2N). We claim that Hδ = H0. Let us prove this.

We haveE_Ai(·) = E₀(·)⊕E_Bi(·). Hence,E_Ai([−δ, δ]) =I⊕E_Bi([−δ, δ]), whence E_Ai([−δ, δ])H=H₀⊕E_Bi([−δ, δ])K. Thus

H_δ =H₀⊕ ∩^N_i=1E_Bi([−δ, δ])K.

But∩^N_i=1E_Bi([−δ, δ])K={0}. Indeed, supposex∈ ∩^N_i=1E_Bi([−δ, δ])Kand x6= 0.

Then kB_ixk6δkxk, i= 1, . . . , N. Hence,

* _N X

i=1

B_i²

! x, x

+

=

N

X

i=1

kB_ixk² 6N δ²kxk² = ε 2kxk². Buth(PN

i=1B_i²)x, xi>εkxk². We get a contradiction. Hence,∩^N_i=1E_Bi([−δ, δ])K= {0}, and, consequently, H_δ =H₀.

2. Let us prove that if H_ε =H₀ for some ε >0, then PN

i=1Ran(A_i) is closed.

We will prove a more stronger fact: if H_ε H₀ is finite dimensional for some ε >0, thenPN

i=1Ran(Ai) is closed.

SinceA²_i = 0⊕B_i², we see thatE_A²

i(·) =E0(·)⊕E_B²

i(·). Hence,E_A²

i([−ε², ε²]) = I ⊕E_B²

i([−ε², ε²]). But E_A²

i([−ε², ε²]) = E_Ai([−ε, ε]). Hence, E_Ai([−ε, ε]) = I⊕E_B²

i([−ε², ε²]), whence E_Ai([−ε, ε])H=H₀⊕E_B²

i([−ε², ε²])K. Thus H_ε =H₀⊕ ∩^N_i=1E_B²

i([−ε², ε²])K.

Thus, ∩^N_i=1E_B²

i([−ε², ε²])K is finite dimensional. By the Main Theorem, 0 ∈/ σe(PN

i=1B_i²). Moreover, since Ker

N

X

i=1

B_i²

!

=

N

\

i=1

Ker(B_i) ={0}, we conclude that 0 ∈/ σ_d(PN

i=1B_i²). Hence, 0 ∈/ σ(PN

i=1B_i²), that is, the oper- ator PN

i=1B_i² is invertible. It follows that PN

i=1Ran(B_i) = K. Consequently, PN

i=1Ran(A_i) = K is closed.

6. Proof of Theorem 1.5 Proof of Theorem 1.5. Define

B_p =

Np

X

i=1

A_p,iA^∗_p,i, p= 1, . . . , m.

Clearly,B_p is a bounded self-adjoint operator in H. Moreover, B_pB_q is compact for p6=q.

From Corollary 2.3 it follows that Ran(B_p) = R_p, p = 1, . . . , m. By Propo- sition 2.4, there exists ε > 0 such that σ(B_p) ∩ ((−ε,0) ∪ (0, ε)) = ∅ for p = 1, . . . , m. It follows that E_Bp([−ε/2, ε/2]) = E_Bp({0}), and, consequently,

E_Bp([−ε/2, ε/2])H =Ker(B_p), p= 1, . . . , m. Hence, H_ε/2(B₁, . . . , B_m) =

m

\

p=1

E_Bp([−ε/2, ε/2])H=

m

\

p=1

Ker(B_p) =H₀(B₁, . . . , B_m).

By Theorem1.2, Pm

p=1Ran(Bp) = Pm

p=1Rp is closed.

Acknowledgement. The author is grateful to Alexei Yu. Konstantinov for helpful suggestions.

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Taras Shevchenko National University of Kyiv, Faculty of Mechanics and Mathematics, Kyiv, Ukraine.

E-mail address: [email protected]