on the closedness of the sum of ranges of operators with almost compact products

12
J. Math. Anal. Appl. 416 (2014) 24–35 Contents lists available at ScienceDirect Journal of Mathematical Analysis and Applications www.elsevier.com/locate/jmaa On the closedness of the sum of ranges of operators A k with almost compact products A i A j Ivan S. Feshchenko Taras Shevchenko National University of Kyiv, Faculty of Mechanics and Mathematics, Kyiv, Ukraine article info abstract Article history: Received 1 July 2013 Available online 19 February 2014 Submitted by Richard M. Aron Keywords: Sum of operator ranges Sum of subspaces Closedness Essential norm Essential reduced minimum modulus Let H 1 ,..., H n , H be complex Hilbert spaces and A k : H k →H be a bounded linear operator with closed range Ran(A k ), k =1,...,n. It is known that if A i A j is compact for all i = j , then n k=1 Ran(A k ) is closed. We show that if all the products A i A j , i = j , are “almost” compact (in a certain sense), then the subspaces Ran(A 1 ),..., Ran(A n ) are essentially linearly independent and their sum is closed. © 2014 Elsevier Inc. All rights reserved. 1. Introduction 1.1. On the closedness of the sum of subspaces Let X be a complex Hilbert space, and X 1 ,..., X n be subspaces of X (by a subspace we always mean a closed linear set). Define the sum of X 1 ,..., X n in the natural way, namely, n k=1 X k = n k=1 x k x k ∈X k ,k =1,...,n . We are interested in the following question: when is n k=1 X k closed? If X if finite dimensional, then, clearly, n k=1 X k is closed. However, in infinite dimensional space X this is not true (generally speaking). E-mail address: [email protected]. http://dx.doi.org/10.1016/j.jmaa.2014.02.022 0022-247X/© 2014 Elsevier Inc. All rights reserved.

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J. Math. Anal. Appl. 416 (2014) 24–35

Contents lists available at ScienceDirect

Journal of Mathematical Analysis and Applications

www.elsevier.com/locate/jmaa

On the closedness of the sum of ranges of operators Ak

with almost compact products A∗iAj

Ivan S. FeshchenkoTaras Shevchenko National University of Kyiv, Faculty of Mechanics and Mathematics, Kyiv, Ukraine

a r t i c l e i n f o a b s t r a c t

Article history:Received 1 July 2013Available online 19 February 2014Submitted by Richard M. Aron

Keywords:Sum of operator rangesSum of subspacesClosednessEssential normEssential reduced minimum modulus

Let H1, . . . ,Hn,H be complex Hilbert spaces and Ak : Hk → H be a boundedlinear operator with closed range Ran(Ak), k = 1, . . . , n. It is known that if A∗

iAj

is compact for all i �= j, then∑n

k=1 Ran(Ak) is closed. We show that if all theproducts A∗

iAj , i �= j, are “almost” compact (in a certain sense), then the subspacesRan(A1), . . . ,Ran(An) are essentially linearly independent and their sum is closed.

© 2014 Elsevier Inc. All rights reserved.

1. Introduction

1.1. On the closedness of the sum of subspaces

Let X be a complex Hilbert space, and X1, . . . ,Xn be subspaces of X (by a subspace we always mean aclosed linear set). Define the sum of X1, . . . ,Xn in the natural way, namely,

n∑k=1

Xk ={

n∑k=1

xk

∣∣∣ xk ∈ Xk, k = 1, . . . , n}.

We are interested in the following question:

when isn∑

k=1

Xk closed?

If X if finite dimensional, then, clearly,∑n

k=1 Xk is closed. However, in infinite dimensional space X this isnot true (generally speaking).

E-mail address: [email protected].

http://dx.doi.org/10.1016/j.jmaa.2014.02.0220022-247X/© 2014 Elsevier Inc. All rights reserved.

I.S. Feshchenko / J. Math. Anal. Appl. 416 (2014) 24–35 25

Example 1.1. Let Y,Z be Hilbert spaces, and A : Y → Z be a bounded linear operator with nonclosedRan(A). Set X = Y ⊕ Z. Define the subspaces

X1 = Y ⊕ 0 ={(y, 0)

∣∣ y ∈ Y}, X2 = Graph(A) =

{(y,Ay)

∣∣ y ∈ Y}.

Then X1 + X2 = Y ⊕ Ran(A) is not closed.

Systems of subspaces X1, . . . ,Xn for which the question

isn∑

k=1

Xk closed?

is very important arise in various branches of mathematics, for example, in

(1) theoretical tomography and theory of ridge functions (plane waves). See, e.g., [5,17,21,18], where theproblem on closedness of the sum of spaces of ridge functions (plane waves) is studied;

(2) theory of wavelets and multiresolution analysis. Here the problem on the closedness of the sum ofshift-invariant subspaces of L2(Rd) is studied. See, e.g., [16] and references therein.

(3) statistics. See, e.g., [3], where the closedness of the sum of two marginal subspaces is important forconstructing an efficient estimation of linear functionals of a probability measure with known marginaldistributions;

(4) projection algorithms for solving convex feasibility problems (problems of finding a point in thenonempty intersection of n closed convex sets). See, e.g., [2, Theorem 5.19], [1, Theorem 4.1], [19]and the bibliography therein;

(5) a problem of finding an element of a Hilbert space with prescribed best approximations from a finitenumber of subspaces. This problem is a common problem in applied mathematics, it arises in harmonicanalysis, optics, and signal theory. See, e.g., [7] and references therein;

(6) theory of Banach algebras. See, e.g., [8,9];(7) theory of operator algebras. See, e.g., [13], where the closedness of finite sums of full Fock spaces over

subspaces of Cd plays a crucial role for construction of a topological isomorphism between universal

operator algebras;(8) quadratic programming. See, e.g., [20];

and others.

1.2. On the closedness of the sum of ranges of operators Ak with almost compact products A∗iAj

Let H1, . . . ,Hn,H be complex Hilbert spaces and Ak : Hk → H be a bounded linear operator with closedrange Ran(Ak), k = 1, . . . , n. We study a question on the closedness of

n∑k=1

Ran(Ak) ={

n∑k=1

yk

∣∣∣ yk ∈ Ran(Ak)}

={

n∑k=1

Akxk

∣∣∣ xk ∈ Hk

}

under additional conditions imposed on Ak, k = 1, . . . , n. The following result follows immediately from[10, Theorem 2].

Theorem A. If A∗iAj is compact for all i �= j, then

∑n Ran(Ak) is closed.

k=1

26 I.S. Feshchenko / J. Math. Anal. Appl. 416 (2014) 24–35

We will show that if the essential norms of A∗iAj , i �= j, are “small enough” (in some sense), then

the subspaces Ran(A1), . . . ,Ran(An) are essentially linearly independent and their sum∑n

k=1 Ran(Ak) isclosed.

To formulate our main result we need

(1) to introduce a notion of the essential linear independence of a system of subspaces;(2) to introduce a notion of the essential reduced minimum modulus of an operator.

It is by using of the essential reduced minimum modulus of Ak, k = 1, . . . , n that we will specify the meaningof the fuzzy words “small enough”.

1.3. Structure of the paper

In Section 2 we present the notation used in this paper and recall some facts on operators in a Hilbertspace.

In Sections 3, 4 we introduce auxiliary notions, namely, the essential reduced minimum modulus of anoperator and the essential linear independence of a system of subspaces.

In Section 5 we formulate our main result (see the Main Theorem). Using the Main Theorem, we providesufficient conditions under which Ran(A1), . . . ,Ran(An) are essentially linearly independent and their sumis closed. As a consequence, we provide sufficient conditions under which n subspaces are essentially linearlyindependent and their sum is closed.

In Section 6 we prove two auxiliary lemmas.Finally, in Section 7 we prove the Main Theorem.

2. Notation and basic facts on operators in a Hilbert space

1. In this paper we consider only complex Hilbert spaces usually denoted by the letters X ,H,K. The scalarproduct in a Hilbert space is denoted by 〈·,·〉, and ‖ · ‖ stands for the corresponding norm, ‖x‖ =

√〈x, x〉.

2. By a subspace of a Hilbert space we always mean a closed linear set. If K is a subspace of H, thenHK denotes the orthogonal complement to K in H. This means that HK is the set of all elements of Hwhich are orthogonal to K. It is well-known that HK is a subspace of H, and H is the orthogonal directsum of K and HK.

3. The identity operator on H is denoted by IH or simply I if it is clear which Hilbert space is beingconsidered.

4. For a bounded linear operator A : H → H, σ(A) denotes the spectrum of the operator A. Recall thatσ(A) = C\ρ(A), where ρ(A) (the resolvent set of A) is the set of all complex λ such that A−λI is invertible.

5. A bounded self-adjoint operator A in H is said to be nonnegative (and we write A � 0) if 〈Ax, x〉 � 0for every x ∈ H. It is well-known that A is nonnegative if and only if σ(A) ⊂ [0,+∞). For two boundedself-adjoint operators A,B in H we write A � B if A−B is nonnegative.

6. Let A be a bounded self-adjoint operator in H. By the definition, the essential spectrum of A, σe(A),consists of all λ ∈ σ(A) such that either λ is a limit point of σ(A) or Ker(A−λI) is infinite-dimensional. Thereare characterizations of σe(A) in the terms of the spectral measure of A and singular sequences of A (see, e.g.,[4, Chapter 9.1]). By the Weyl theorem, the essential spectrum is stable under compact perturbations, i.e., ifK is a compact self-adjoint operator in H, then σe(A+K) = σe(A) (see, e.g., [4, Chapter 9.1, Theorem 3]).

7. For a bounded linear operator A : H1 → H2,

‖A‖e = inf{‖A + K‖

∣∣ K : H1 → H2 is compact}

is the essential norm of A.

I.S. Feshchenko / J. Math. Anal. Appl. 416 (2014) 24–35 27

3. An auxiliary notion: the essential reduced minimum modulus

To make the definition of the essential reduced minimum modulus of an operator more clear and natural,we first recall the well-known and widely used notions of the minimum modulus and the reduced minimummodulus.

Let H1,H2 be Hilbert spaces, A : H1 → H2 be a bounded linear operator.

3.1. The minimum modulus

The minimum modulus of A, γ(A), was introduced by H.A. Gindler and A.E. Taylor in [11]. By thedefinition,

γ(A) = inf{‖Ax‖

∣∣ x ∈ H1, ‖x‖ = 1}.

In other words, γ(A) is the maximum of all γ � 0 such that

‖Ax‖ � γ‖x‖

for all x ∈ H1.

3.2. The reduced minimum modulus

It is natural to define the reduced minimum modulus of a nonzero operator A, γr(A), as follows. (See,e.g., [15, Section 5.1] and [12, Definition 4.1.3]. Note that in [12] the reduced minimum modulus is calledthe minimum modulus.)

Set

γr(A) = inf{‖Ax‖

∣∣ x ∈ H1 Ker(A), ‖x‖ = 1}.

In other words, γr(A) is the maximum of all γ � 0 such that

‖Ax‖ � γ‖x‖ (3.1)

for all x ∈ H1 Ker(A).

Remark 3.1. For the zero operator we set γr(0) = +∞.

The reduced minimum modulus possesses the following important property (see, e.g., [15, Theorem 5.2]and [12, Theorem 4.1.6]): Ran(A) is closed if and only if γr(A) > 0.

Indeed, if γr(A) > 0, then A(H1 Ker(A)) = A(H1) = Ran(A) is closed. Conversely, suppose Ran(A) isclosed. Define an operator

A′ = A �H1�Ker(A): H1 Ker(A) → Ran(A).

Then A′ is an invertible operator between two Hilbert spaces. Consequently, there exists γ > 0 such that‖A′x‖ � γ‖x‖, x ∈ H1 Ker(A). This means that ‖Ax‖ � γ‖x‖, x ∈ H1 Ker(A) ⇒ γr(A) � γ > 0.

Note that (3.1) is equivalent to ‖Ax‖2 � γ2‖x‖2. This inequality can be rewritten as

⟨A∗Ax, x

⟩� γ2‖x‖2. (3.2)

28 I.S. Feshchenko / J. Math. Anal. Appl. 416 (2014) 24–35

Define a self-adjoint operator

B = A∗A �H1�Ker(A): H1 Ker(A) → H1 Ker(A).

(Thus, A∗A = B ⊕ 0 with respect to the orthogonal decomposition H1 = (H1 Ker(A)) ⊕ Ker(A).) Now,from (3.2) we see that γr(A) is the maximum of all γ � 0 such that

B � γ2I.

3.3. The essential reduced minimum modulus

Now we are ready to introduce, in a natural way, the essential reduced minimum modulus of an operator.The essential reduced minimum modulus of a nonzero operator A, γer(A), is the supremum of all γ � 0 forwhich there exists a compact self-adjoint operator K : H1 Ker(A) → H1 Ker(A) such that

B + K � γ2I.

Remark 3.2. For the zero operator we set γer(0) = +∞.

Since the notion of the essential reduced minimum modulus plays a crucial role in this paper, we presentsome properties and formulas for γer(A).

1. Clearly, γer(A) � γr(A). Consequently, if Ran(A) is closed, then γer(A) > 0.2.1. If H1 Ker(A) is finite dimensional, then γer(A) = +∞.2.2. Suppose H1 Ker(A) is infinite dimensional. Then

γer(A) =(min

{λ∣∣ λ ∈ σe(B)

})1/2. (3.3)

To see this, we need the following simple proposition, which seems to be a part of mathematical folklore.

Proposition 3.1. Let C : H → H be a bounded self-adjoint operator in an infinite dimensional Hilbertspace H. Define me(C) to be the supremum of all m for which there exists a compact self-adjoint operatorK : H → H such that

C + K � mI.

Then me(C) = min{λ | λ ∈ σe(C)}.

The proof is trivial. Indeed, denote by λ0 the minimal point of σe(C). Then, by the Weyl theorem,me(C) � λ0. Moreover, from the definition of the essential spectrum it follows that there exists a compactself-adjoint operator K such that C + K � λ0I.

Using the proposition, we have

(γer(A)

)2 = me(B) = min{λ∣∣ λ ∈ σe(B)

};

it follows (3.3).Now suppose that Ran(A) is closed. Then γer(A) > 0 and σe(B) ⊂ [(γer(A))2,+∞). Hence, from (3.3)

it follows that

γer(A) =(min

{λ∣∣ λ ∈ σe

(A∗A

)\ {0}

})1/2.

I.S. Feshchenko / J. Math. Anal. Appl. 416 (2014) 24–35 29

3. If for γ � 0 there exists a compact operator T : H1 Ker(A) → H2 such that

‖Ax‖2 + ‖Tx‖2 � γ2‖x‖2 (3.4)

for all x ∈ H1 Ker(A), then γer(A) � γ. Indeed, (3.4) can be rewritten as⟨(A∗A + T ∗T

)x, x

⟩� γ2‖x‖2;

hence,

B + T ∗T � γ2I.

Since T ∗T is compact and self-adjoint, we conclude that γer(A) � γ.

Remark 3.3. Suppose that dimH1 � dimH2. (Here dimH is the Hilbert dimension of H, i.e., dimH is thecardinality of an orthonormal basis of H. This inequality holds in the most interesting case when H1,H2are separable and infinite dimensional.) Then γer(A) is the supremum of all γ � 0 for which there exists acompact operator T : H1 Ker(A) → H2 such that (3.4) holds for all x ∈ H1 Ker(A).

This follows from the arguments above and the following observations:

(1) since |K| � K for any self-adjoint operator K, we see that in the definition of the essential reducedminimum modulus one can consider only nonnegative compact self-adjoint operators K;

(2) since dim(H1 Ker(A)) � dimH2, we conclude that every nonnegative compact self-adjoint operatorK : H1 Ker(A) → H1 Ker(A) can be represented as K = T ∗T for some compact operator T :H1 Ker(A) → H2.

4. An auxiliary notion: the essential linear independence

Let X be a Hilbert space, X1, . . . ,Xn be its subspaces. We say that X1, . . . ,Xn are linearly independentif the equality

n∑i=1

xi = 0,

where xi ∈ Xi, i = 1, . . . , n, implies that xi = 0, i = 1, . . . , n. Clearly, X1, . . . ,Xn are linearly independentif and only if

Xi ∩∑j �=i

Xj = {0}

for i = 1, . . . , n.Now, it is natural to say that X1, . . . ,Xn are essentially linearly independent if the linear set Xi∩

∑j �=i Xj

is finite dimensional for i = 1, . . . , n.

Remark 4.1. It is useful to reformulate the properties of linear independence and essential linear indepen-dence it terms of properties of the operator S : X1 ⊕ · · · ⊕ Xn → X defined by

S(x1, . . . , xn) =n∑

i=1xi, xi ∈ Xi, i = 1, . . . , n.

Clearly,

30 I.S. Feshchenko / J. Math. Anal. Appl. 416 (2014) 24–35

(1) X1, . . . ,Xn are linearly independent ⇔ Ker(S) = {0};(2) X1, . . . ,Xn are essentially linearly independent ⇔ Ker(S) is finite dimensional.

Remark 4.2. Suppose that n-tuple of subspaces X1, . . . ,Xn is such that∑n

i=1 Xi is closed. Then this n-tuple isessentially linearly independent if and only if this n-tuple is lower semi-Fredholm in the sense of [6, Section 5].Unfortunately, the notion “lower semi-Fredholm” is not convenient, it should be replaced by “upper semi-Fredholm”. Let us explain this.

Clearly, X1, . . . ,Xn are essentially linearly independent iff Ker(S) is finite dimensional. Recall that weassumed that

∑ni=1 Xi is closed, this means that Ran(S) is closed. Conclusion: the n-tuple X1, . . . ,Xn is es-

sentially linearly independent iff S is upper semi-Fredholm (an operator is said to be upper semi-Fredholm ifits range is closed and its kernel is finite dimensional). So, it is natural to call essentially linearly independentn-tuples of subspaces with closed sum by “upper semi-Fredholm”.

The essentially linearly independent systems of subspaces can be regarded as a finite dimensional per-turbation of the linearly independent systems of subspaces. More precisely, we have

(1) if Xi = Yi + Zi, i = 1, . . . , n, where subspaces Y1, . . . ,Yn are linearly independent and subspacesZ1, . . . ,Zn are finite dimensional, then X1, . . . ,Xn are essentially linearly independent;

(2) if X1, . . . ,Xn are essentially linearly independent, then there exists a representation Xi = Yi ⊕ Zi,i = 1, . . . , n, where the subspaces Y1, . . . ,Yn are linearly independent and the subspaces Z1, . . . ,Zn arefinite dimensional.

To prove (1), note that

dim(Xi ∩

∑j �=i

Xj

)� dimZi + dim

∑j �=i

Zj .

To prove (2), it is sufficient to define

Zi = Xi ∩∑j �=i

Xj , Yi = Xi Zi

for i = 1, . . . , n.

5. The main result and an example of its application

5.1. The main result

Let H1, . . . ,Hn,H be complex Hilbert spaces and Ak : Hk → H be a bounded linear operator with closedrange Ran(Ak), k = 1, . . . , n.

Suppose that

(1) γer(Ak) � γk > 0, k = 1, . . . , n;(2) ‖A∗

iAj‖e � εi,j for i �= j.

In what follows, we assume that εj,i = εi,j for all i �= j (note that, since A∗jAi = (A∗

iAj)∗, we have‖A∗

jAi‖e = ‖A∗iAj‖e).

I.S. Feshchenko / J. Math. Anal. Appl. 416 (2014) 24–35 31

Define a real symmetric n× n matrix M = (mi,j) by

mi,j ={γ2i , if i = j;

−εi,j , if i �= j.

Main Theorem. If M is positive definite, then Ran(A1), . . . ,Ran(An) are essentially linearly independentand their sum is closed.

Now we mention some consequences of this theorem; while the proof will be given in Section 7, by usingauxiliary lemmas in Section 6.

From the Main Theorem it follows that if the numbers εi,j are “sufficiently small” in comparison withthe numbers γk (that is, if the operators A∗

iAj are “almost compact”), then Ran(A1), . . . ,Ran(An) areessentially linearly independent and their sum is closed. But what is the precise meaning of the words“sufficiently small”? We will specify this meaning by using the notion of a strictly diagonally dominantmatrix.

Let Λ = (λi,j) be a real symmetric n × n matrix with positive diagonal entries. Λ is said to be strictlydiagonally dominant if

∑j �=i

|λi,j | < λi,i

for i = 1, . . . , n. The notion of a strictly diagonally dominant matrix is well-known (see, e.g., [14, Sec-tion 6.1]).

Strictly diagonally dominant matrices are important for us for the following reason: it is well-known (andone can easily check) that if Λ is strictly diagonally dominant, then Λ is positive definite.

Hence, if

∑j �=i

εi,j < γ2i (5.1)

for i = 1, . . . , n, then M is a strictly diagonally dominant matrix; consequently, M is positive definite. Thus,we get the following corollary of the Main Theorem.

Corollary 5.1. If (5.1) holds for i = 1, . . . , n, then Ran(A1), . . . ,Ran(An) are essentially linearly independentand their sum is closed.

5.2. Example

Let X be a Hilbert space, and X1, . . . ,Xn be its subspaces. Using the Main Theorem, we will obtainsufficient conditions under which X1, . . . ,Xn are essentially linearly independent and their sum is closed.

For a subspace Y of X , define PY to be the orthogonal projection onto Y. Clearly, Ran(PY) = Y and

γer(PY) ={

1, if Y is infinite dimensional;+∞, if Y is finite dimensional.

The last equality follows from the formulas for the essential reduced minimum modulus given in Section 3.3.Indeed, Ker(PY) = Y⊥ and with respect to the orthogonal decomposition X = Y ⊕Y⊥ we have PY = I⊕0.Hence, P ∗

YPY = I ⊕ 0. This means that the restriction of P ∗YPY to Y, regarded as an operator from Y to Y,

equals I. Using (3.3), we get the required formula for γer(PY).

32 I.S. Feshchenko / J. Math. Anal. Appl. 416 (2014) 24–35

We apply the Main Theorem to the operators PXk: X → X . Suppose that numbers εi,j = εj,i, i �= j, are

such that ‖PXiPXj

‖e � εi,j for i �= j. Define a real symmetric n× n matrix M = (mi,j) by

mi,j ={

1, if i = j;−εi,j , if i �= j.

By the Main Theorem, if M is positive definite, then X1, . . . ,Xn are essentially linearly independentand their sum is closed. In particular, if

∑j �=i εi,j < 1 for any i = 1, . . . , n, then X1, . . . ,Xn are essentially

linearly independent and their sum is closed.

6. Auxiliary lemmas

6.1. On the closedness of the sum of operator ranges

Let H1, . . . ,Hn,H be Hilbert spaces, Ak : Hk → H be a bounded linear operator, k = 1, . . . , n. Definean operator B : H → H by

B =n∑

k=1

AkA∗k.

Lemma 6.1. If σ(B) ∩ (0, ε) = ∅ for some ε > 0, then∑n

k=1 Ran(Ak) is closed.

Remark 6.1. The converse is also true. However, this fact is not needed in the paper.

Proof of Lemma 6.1. Define R =∑n

k=1 Ran(Ak) and consider the orthogonal decomposition

H = R⊕ (HR).

With respect to this orthogonal decomposition

Ak =(A′

k

0

),

where A′k : Hk → R, A′

k is Ak considered as the operator from Hk to R. Define an operator B′ : R → R by

B′ =n∑

k=1

A′k

(A′

k

)∗.

We have

Ker(B′) =

n⋂k=1

Ker((A′

k

)∗)

=n⋂

k=1

{x ∈ R

∣∣ x⊥Ran(A′

k

)}=

n⋂k=1

{x ∈ R

∣∣ x⊥Ran(Ak)}

= {x ∈ R∣∣ x⊥R} = {0}.

It is easily seen that B = B′ ⊕ 0. Hence, σ(B′) ∩ (0, ε) = ∅. Recall the following simple fact from thespectral theory of self-adjoint operators (see, e.g., [4]): every isolated point of spectrum of a self-adjointoperator is its eigenvalue. Using this fact, we immediately see that 0 /∈ σ(B′), that is, B′ is invertible.

I.S. Feshchenko / J. Math. Anal. Appl. 416 (2014) 24–35 33

Consequently, Ran(B′) = R. Since Ran(B′) ⊂∑n

k=1 Ran(A′k) ⊂ R, we see that

∑nk=1 Ran(A′

k) = R.Hence,

∑nk=1 Ran(Ak) = R is closed as required. �

6.2. On a lower bound for the essential spectrum of block operators

Let H1, . . . ,Hn be Hilbert spaces, A : H1⊕· · ·⊕Hn → H1⊕· · ·⊕Hn be a bounded self-adjoint operator.Let

A = (Ai,j | i, j = 1, . . . , n)

be the block decomposition of A. Suppose that reals ai, i = 1, . . . , n, and aj,i = ai,j , i �= j, satisfy thefollowing conditions:

(1) me(Ai,i) � ai for i = 1, . . . , n;(2) ‖Ai,j‖e � ai,j for any i �= j.

(Recall that for a bounded self-adjoint operator C, me(C) is the supremum of all m for which there existsa compact self-adjoint operator K such that C + K � mI.)

Define a real symmetric n× n matrix M = (mi,j) by

mi,j ={ai, if i = j;−ai,j , if i �= j.

Lemma 6.2. σe(A) ⊂ [λmin(M),+∞), where λmin(M) is the minimum eigenvalue of M .

Proof. First, let us show that for any ε > 0 there exists a compact self-adjoint operator K such that

A + K �(λmin(M) − ε

)I. (6.1)

Set δ = ε/n. There exist compact self-adjoint operators Ki,i, i = 1, . . . , n, and compact operators Ki,j ,i �= j, such that

(1) Ai,i + Ki,i � (ai − δ)I for i = 1, . . . , n;(2) ‖Ai,j + Ki,j‖ � ai,j + δ for i �= j.

Since Aj,i = A∗i,j , we can assume that Kj,i = K∗

i,j for i �= j. Set

K = (Ki,j | i, j = 1, . . . , n).

For any v = (v1, . . . , vn) ∈ H1 ⊕ · · · ⊕ Hn we have

⟨(A + K)v, v

⟩=

n∑i=1

⟨(Ai,i + Ki,i)vi, vi

⟩+ 2

∑i<j

Re⟨(Ai,j + Ki,j)vj , vi

�n∑

i=1(ai − δ)‖vi‖2 − 2

∑i<j

(ai,j + δ)‖vj‖‖vi‖

=(

n∑ai‖vi‖2 − 2

∑ai,j‖vi‖‖vj‖

)− δ

(n∑

‖vi‖2 + 2∑

‖vi‖‖vj‖)

i=1 i<j i=1 i<j

34 I.S. Feshchenko / J. Math. Anal. Appl. 416 (2014) 24–35

=⟨M

(‖v1‖, . . . , ‖vn‖

)T,(‖v1‖, . . . , ‖vn‖

)T ⟩− δ

(n∑

i=1‖vi‖

)2

� λmin(M)n∑

i=1‖vi‖2 − nδ

n∑i=1

‖vi‖2 =(λmin(M) − nδ

)‖v‖2

=(λmin(M) − ε

)‖v‖2.

It follows (6.1).Now we are ready to prove the assertion of the lemma. Consider any ε > 0. There exists a compact

self-adjoint operator K such that (6.1) holds. By the Weyl theorem, σe(A) = σe(A+K) ⊂ [λmin(M)−ε,+∞).Since ε > 0 is arbitrary, we conclude that σe(A) ⊂ [λmin(M),+∞). �7. Proof of the Main Theorem

Set Ki = Hi Ker(Ai), i = 1, . . . , n. Define an operator Γ : K1 ⊕ · · · ⊕ Kn → H by

Γ (x1, . . . , xn) =n∑

i=1Aixi, xi ∈ Ki, i = 1, . . . , n.

Then Γ ∗ : H → K1 ⊕ · · · ⊕ Kn and

Γ ∗x =(A∗

1x, . . . , A∗nx

), x ∈ H.

Hence, ΓΓ ∗ =∑n

i=1 AiA∗i . Consider the operator

G = Γ ∗Γ : K1 ⊕ · · · ⊕ Kn → K1 ⊕ · · · ⊕ Kn.

Its block decomposition is the following:

G = (Gi,j | i, j = 1, . . . , n), Gi,j = A∗iAj �Kj

: Kj → Ki.

By Lemma 6.2 and the positive definiteness of M , we have 0 /∈ σe(G) (note that me(Gi,i) = (γer(Ai))2 forall i, and ‖Gi,j‖e = ‖A∗

iAj‖e for i �= j).Let us show that Ran(A1), . . . ,Ran(An) are essentially linearly independent. Since 0 /∈ σe(G), we conclude

that Ker(G) is finite dimensional. We have Ker(G) = Ker(Γ ∗Γ ) = Ker(Γ ). Hence,

Ker(Γ ) ={

(x1, . . . , xn) ∈ K1 ⊕ · · · ⊕ Kn

∣∣∣ n∑i=1

Aixi = 0}

is finite dimensional, whence the linear set{(y1, . . . , yn) ∈ Ran(A1) ⊕ · · · ⊕ Ran(An)

∣∣∣ n∑i=1

yi = 0}

is also finite dimensional.It follows that Ran(A1), . . . ,Ran(An) are essentially linearly independent.Let us show that

∑ni=1 Ran(Ai) is closed. Since 0 /∈ σe(G), we conclude that σ(G) ∩ (0, ε) = ∅ for

some ε > 0. Since σ(ΓΓ ∗) \ {0} = σ(Γ ∗Γ ) \ {0}, we see that σ(∑n

i=1 AiA∗i ) ∩ (0, ε) = ∅. By Lemma 6.1,∑n

i=1 Ran(Ai) is closed.The proof is complete.

I.S. Feshchenko / J. Math. Anal. Appl. 416 (2014) 24–35 35

Acknowledgments

I am grateful to the referee for a careful reading and valuable remarks that have improved the paper.

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