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- Open Access
General note on the theorem of Stampfli
- Cheniti Bensalloua^{1} and
- Mostefa Nadir^{1}Email author
https://doi.org/10.1186/s13660-016-1002-7
© Bensalloua and Nadir 2016
- Received: 25 September 2015
- Accepted: 3 February 2016
- Published: 12 February 2016
Abstract
It is well known that for every bounded operator A in \(L(H)\), there exists a compact operator K in \(K(H)\) such that the Weyl spectrum \(\sigma_{W}(A)\) of the operator A coincides with the spectrum \(\sigma(A+K)\) of the perturbed operator A. In this work, we show the extension of this relation by the use of Kato’s decomposition to the set of semi-Fredholm operators.
Keywords
- semi-Fredholm operator
- index of semi-Fredholm operator
- Weyl spectrum
MSC
- 05C38
- 15A15
- 05A15
- 15A18
1 Introduction
In 1965 Schechter demonstrated in [1] this result that for every bounded operator A in \(L(H)\), there exists a compact operator K in \(K(H)\) such that the Weyl spectrum \(\sigma_{W}(A)\) of the operator A coincides with the spectrum \(\sigma(A+K)\) of the perturbed operator A for a compact perturbation of Fredholm operators of index zero. In 1973 Stampfli showed in [2] that the minimum of this perturbation is achieved for a certain compact operator. This work is intended to extend this result to semi-Fredholm operators of any index using the decomposition of Kato, an extension already done by Apostol in [3] and Herrero in [4].
Let H be a complex, separable, infinite dimensional Hilbert space, and let \(L(H)\) denote the algebra of all linear bounded operators on H, \(C(H)\) the set of linear operators A with domain \(D(A)\) dense in H and range \(R(A)\) contained in H and a graph \(G(A)\) closed in \(H\times H\). \(K(H)\) is the set of compact elements of \(L(H)\).
For \(A\in C(H)\), we let \(\sigma(A)\), \(\rho(A)\), and \(N(A)\) denote the spectrum, the resolvent set, and the null space of A, respectively. The nullity \(\alpha(A)\) of A is defined as the dimension of \(N(A)\) and the deficiency \(\beta(A)\) of A is defined as the codimension of \(R(A)\) in H.
Proposition 1
(See [5])
- (1)
\(R(A)\) is a closed subspace in H if and only if \(c(A)>0\),
- (2)
\(c(A)=c(A^{\ast})\),
Definition 1
- (1)
\(R(A)\) is a closed subspace in H,
- (2)
\(\min \operatorname{ind}(A)= \{ \dim N(A),\dim N(A^{\ast}) \} <\infty\).
Theorem 1
(See [6])
- (1)
M and N are invariant by A,
- (2)
\(A/_{M}\) is regular,
- (3)
\(N\subseteq D(A)\), \(\dim N<\infty\), and \(A/_{N}\) is nilpotent.
This decomposition is known as the decomposition of Kato [6]. Operators admitting such decomposition were characterized in 1976 by Labrousse in [7] and are called quasi-Fredholm operators, also generalized by Mbekhta to the operators called pseudo-Fredholm [8].
Definition 2
Lemma 1
Definition 3
- (1)
\(R(A-\mu I)\) is a closed subspace in H,
- (2)
\(N[(A-\mu I)^{n}]\subseteq R(A-\mu I)\), \(\forall n\in\mathbb{N}\).
Remark 1
The set \(\operatorname{reg}(A)\) of regular points of A is open in \(\mathbb{C}\).
Theorem 2
(See [9])
\(A\in SF(H)\) if and only if \(A^{\ast}\in SF(H)\) and \(\operatorname{ind}(A)=-\operatorname{ind}(A^{\ast})\).
Theorem 3
(See [10])
For any \(A\in L(H)\cap SF(H)\) and \(B\in L(H)\) such that \((A-B)\in K(H)\) we get \(B\in SF(H)\) and \(\operatorname{ind}(A)=\operatorname{ind}(B)\).
Theorem 4
(See [11])
\(\operatorname{ind}(A-\mu I)\) is constant on each connected component of \(\rho_{e}(A)\).
Definition 4
Remark 2
\(g(A,B)\) defines a metric on \(C(H)\).
Proposition 2
(See [5])
Let M, N be two closed subspaces of H with \(\dim N<\infty\). Then if \(\delta(M,N) <1\) we have \(\dim M\leq\dim N\).
Definition 5
Let \(A\in L(H)\). A is called quasi-nilpotent and will be noted \(A\in QN(H)\) if and only if \(\lim_{n\rightarrow\infty} \Vert A^{n}\Vert ^{\frac{1}{n}}=0\).
Definition 6
\(A\in L(H)\) is called a Riesz operator if and only if \(\sigma_{e}(A)=0\).
Theorem 5
(See [1])
Passing to the complement in the expression \(\sigma_{W}(A)=\sigma (A+K)=\sigma(A)\cap [ \Phi_{0}(A) ] ^{c}\) we obtain \(\rho (A)\cup\Phi_{0}(A)=\rho(A+K)\). Hence \(\rho(A)\subseteq\Phi_{0}(A)\); then the result of Stampfli is equivalent to \(\Phi_{0}(A)=\rho(A+K)\).
2 Main results
Theorem 6
2.1 Correction of the operator A
2.1.1 First correction
Lemma 2
- (1)
\(\delta(N(B),N(A))\leq\frac{\Vert K_{1}\Vert }{c(A)}\),
- (2)
B is invertible on the left or on the right.
Proof
If A is one-to-one, then \(\dim N(A)=0\Rightarrow\dim N(B)=0\). Therefore \(N(B)=\{0\}\), hence B is one-to-one. As \(R(B)\) is closed (compact perturbation of semi-Fredholm operator), we deduce that B is left invertible.
If \(N(A)\neq \{ 0 \} \), then A is not left invertible. Hence A is right invertible. Then \(N(A^{\ast})= \{ 0 \} \). \(A,B\in L(H)\Rightarrow A^{\ast},B^{\ast}\in L(H)\), \(A^{\ast}-B^{\ast}=A-B\), and \(c(A)=c(A^{\ast})\).
Then \(N(B^{\ast})=\{0\}\) and \(B^{\ast}\) is one-to-one. \(B=(A+K_{1})\in SF(H)\), in particular \(R(B)\) is a closed subspace of H, \(R(B)=N(B^{\ast })^{\perp}=\{0\}^{\perp}=H\), then B is onto. Thus it is right invertible. □
Lemma 3
Proof
Lemma 4
- (1)
\(T(A,\beta)\) is of finite rank,
- (2)
\(\Vert T(A,\beta)\Vert \leq1\),
- (3)
\(\beta\in M_{e}(A+\alpha T(A,\beta))\).
Proof
Remark 3
If \(\operatorname{ind}(A-\beta I)=0\), then \(A+\alpha T(A,\beta)-\beta I\) is onto and one-to-one.
Remark 4
If \(\min \operatorname{ind}(A-\beta I)=0\), we take \(T=0\). Indeed in this case \(A+\alpha T(A,\beta)-\beta I=A-\beta I\) is invertible on the left or right.
Lemma 5
- (1)
\(\Vert K_{1}\Vert \leq\varepsilon\),
- (2)
\(\rho_{e}(A+K_{1})\cap \operatorname{reg}(A+K_{1})\subseteq M_{e}(A+K_{1})\).
Proof
- (1)
\(\alpha_{1}=\alpha\), \(A_{0}=A\),
- (2)
\(A_{i}=A_{i-1}+\alpha_{i}T_{i}\), \(i=1,2,3,\ldots,n\), where \(T_{i}=T(A_{i-1},\beta_{i})\),
- (2)
\(\alpha_{i+1}=\frac{1}{3}\min \{ \alpha_{i},c(A_{i}-\beta _{i}I), i=1,2,3,\ldots,n-1 \} \).
2.1.2 Second correction
Let \(B\in L(H)\) such that \(\rho_{e}(B)\cap \operatorname{reg}(B)\subseteq M_{e}(B)\). We will build a second compact perturbation \(K_{2}\) such that \(\rho _{e}(B)\subseteq M_{e}(B+K_{2})\). If \(B\in L(H)\) such that \(\rho _{e}(B)\cap \operatorname{reg}(B)\subseteq M_{e}(B)\), then \(\rho_{e}(B)\backslash M_{e}(B)\) is a finite or countable set denoted \(\mu_{j}\). Note that \(\sigma _{e}(B)=\emptyset\), because in [6], there exist \(n\in\mathbb{N}\) and \(\mu_{1},\mu_{2},\mu_{3},\ldots,\mu_{n}\in\mathbb{C}\) such that \(\sigma (B)=\{\mu_{1},\mu_{2},\mu_{3},\ldots,\mu_{n}\}\). Hence \(\dim H<\infty\), this is contrary to our hypotheses. As \(\sigma_{e}(B)\) is a closed set of \(\sigma(B)\) (which is compact), \(\sigma_{e}(B)\) is also compact. Since the distance function is continuous on \(\sigma_{e}(B)\), there is a sequence of points \((\mu_{l}^{\prime})_{j\in\mathbb{N}^{\ast}}\in \sigma _{e}(B)\) such that \(d(\mu_{j},\sigma_{e}(B))=|\mu_{j}-\mu _{j}^{\prime }|\). Finally if the sequence \((\mu_{n})\) is infinite, then \(\lim_{j\rightarrow\infty}|\mu_{j}-\mu_{j}^{\prime}|=0\), because otherwise \(\mu_{n}\) would have an accumulation point in \(\rho_{e}(B)\), which is contradictory. For every \(\mu_{j}\), let \(M_{j}\), \(N_{j}\) be the two subspaces of H corresponding to the decomposition of Kato of the operator \(B-\mu_{j}I\) and let \(P_{j}\) be the associated projection of the kernel \(M_{j}\) and the range \(N_{j}\).
Construction of subspaces \(M_{j}^{\prime}\) and \(N_{j}^{\prime}\)
For \(j\in\mathbb{N}\), let \(M _{0}^{\prime}=H\), \(M_{j+1}^{\prime }=M_{j}^{\prime}\cap M_{j+1}\), and \(N_{1}^{\prime}=M_{1}^{\perp}\), \(N_{j+1}^{\prime}=M_{j}^{\prime}\cap M_{j+1}^{\prime\perp}\). Finally, we denote by \(P_{j}^{\prime}\) the orthogonal projection onto \(N_{j}^{\prime}\).
Lemma 6
- (1)
If \(1\leq i\leq n\), \(1\leq j\leq n\), and \(i\neq j\) then \(N_{i}^{\prime }\perp N_{j}^{\prime}\) if and only if \(P_{i}^{\prime }P_{j}^{\prime}=0\).
- (2)
If \(1\leq j\leq n\), then \(N_{j}^{\prime}\oplus M_{j}^{\prime}=N_{j} \oplus M_{j}^{\prime}=M_{j-1}^{\prime}\) and \(\dim N_{j}^{\prime}=\dim N_{j}<\infty\).
- (3)
If \(1\leq j\leq n\), then \(\bigcap_{1\leq i\leq j}N_{i}^{\prime}=M_{j}^{\prime}\) and \(\operatorname{co}\dim M_{j}^{\prime}=\sum_{1\leq i\leq j}\dim N_{i}\).
Proof
For any \(i>j\), we have \(N_{i}^{\prime}\perp N_{j}^{\prime}\). Indeed \(N_{i}^{\prime}\subseteq M _{i-1}^{\prime}\subseteq M _{j}^{\prime}\); and \(N _{j}^{\prime}\subseteq M_{j}^{\prime\perp}\), then \(M_{j}^{\prime}\subseteq N_{j}^{\prime\perp}\) and we deduce \(N_{i}^{\prime}\subseteq N_{j}^{\prime\perp}\). Suppose \(P_{i}^{\prime }P_{j}^{\prime}=0\) and let \(x\in N_{i}^{\prime}\). Then \(\langle x,y \rangle= \langle P_{i}^{\prime}x,P_{j}^{\prime }y \rangle= \langle x,P_{i}^{\prime}P_{j}^{\prime}y \rangle=0\). \(N_{i}^{\prime}\) is orthogonal to \(N_{j}^{\prime}\). Reciprocally if \(N_{i}^{\prime}\) is orthogonal to \(N_{j}^{\prime}\) and \(u\in H\), then \(P_{j}^{\prime}u \in N_{j}^{\prime}\) and \(P_{j}^{\prime}u\in N_{j}^{\prime\perp}\) then \(P_{i}^{\prime }P_{j}^{\prime}u=0\) and \(P_{i}^{\prime}P_{j}^{\prime}=0\).
\(N_{j}^{\prime}\cap M_{j}^{\prime}\subseteq M_{j}^{\prime\perp}\cap M_{j}^{\prime}=\{0\}\) and \(N_{j}^{\prime}\cap M_{j}^{\prime}\subseteq N_{j}^{\prime}\cap M_{j}^{\prime}=\{0\}\). \(M_{j}^{\prime}+N_{j}^{\prime }=M_{j-1}^{\prime}\) (because \(N_{j}^{\prime}=M_{j-1}^{\prime}\cap M_{j}^{\prime\perp}\)).
For any \(i>j\), \(N_{i}^{\prime}\subseteq M_{j-1}^{\prime}=N_{j}^{\prime }\oplus M_{j}^{\prime}\), then \(M_{j}^{\prime}\oplus N_{j}^{\prime }\subseteq M_{j}^{\prime}\oplus N_{j}^{\prime}\); \(N_{j}^{\prime }\subseteq M_{j-1}^{\prime}=M_{j}^{\prime}\oplus N_{j}\Rightarrow M_{j}^{\prime}\oplus N_{j}^{\prime}\subseteq M_{j}^{\prime}\oplus N_{j}\).
It follows that for any \(j\in\mathbb{N}\), \(1\leq j\leq n\); \(M_{j+1}^{\prime \perp}=M_{j}^{\prime\perp}\oplus N_{j}^{\prime}\) It is easy to see that \(\operatorname{co}\dim M_{j}^{\prime}=\sum_{1\leq i\leq j}\dim N_{i}^{\prime}<\infty\). □
Remark 5
If \((P_{i})\) is a sequence of mutually orthogonal projections in H, then \(\sum_{1\leq i\leq n}P_{i}\) converges to P, which is an orthogonal projection (see [13]).
Suppose first that the sequence \((\mu_{n})\) is finite, \(1\leq j\leq n\). We put \(H_{1}=\bigcap_{1\leq j\leq n}M_{j}\) and \(K_{2}=\sum_{1\leq j\leq n}(\mu _{j}^{\prime}-\mu_{j})P_{j}\). It is obvious that \(K_{2}\) is normal.
Lemma 7
If \(\mu\in\rho_{e}(B)\), then \(N(B+K_{2}-\mu I)\subseteq N(B-\mu I)\cap H_{1}\).
Proof
Therefore \(\Vert \sum_{k\geq n}( \mu _{k}^{\prime}- \mu_{k})P_{k}^{\prime}u \Vert \leq\varepsilon\), then \(\sum_{1\leq j\leq \infty}( \mu _{j}^{\prime}- \mu _{j})P_{j}^{\prime}\) is normally convergent.
The rest of the proof is similar to the case of finite \(\mu_{j}\). □
Lemma 8
If \(\mu\in\rho_{e}(B)\), then \(H_{1}^{\perp}\subseteq R(B+K_{2}-\mu I)+H_{1}\).
Proof
Uniqueness. Suppose that \(w_{j}\) and \(w_{j}^{\prime}\) satisfy the requirement. Then \(w_{j}-w_{j}^{\prime}\in N(B+K_{2}- \mu I)\subseteq H_{1}\) (see Lemma 6), hence \(w_{j}-w_{j}^{\prime}\in N_{j}^{\prime}\cap H_{1}= \{ 0 \} \).
Lemma 9
Proof
Let \(\mu \in\rho_{e}(B)\) and \(u\in N(B+K_{2}- \mu I)\). Then \(u\in N(B- \mu I)\cap H_{1}\). But \((B- \mu I)|H_{1}\) is regular because for any \(j\in\mathbb{N}\), \((B- \mu I)|M_{j}\) is also regular. Hence for any \(j\in\mathbb{N}\), \(u\in R[(B-\mu I)^{j}|H_{1}]\subseteq R[(B+K_{2}- \mu I)^{j}]\). As further \(B+K_{2}- \mu I\) is a compact perturbation of the semi-Fredholm operator \(B-\mu I\), it is itself semi-Fredholm and hence \(R(B+K_{2}- \mu I)\) is a closed subspace, which completes the demonstration We now show that \(\operatorname{reg}(B+K_{2})\subseteq M_{e}(B+K_{2})\). Let \(\rho e(B)=\bigcup_{k}C_{k}=\operatorname{reg}(B+K_{2})\) where \(C_{k}\) is the nth k-connected component of \(\rho_{e}(B)\).
For any \(k\in \mathbb{N}^{\ast}\), \(C_{k}\cap \operatorname{reg}(B)=\emptyset\). Let \(\mu \in C_{k}\cap \operatorname{reg}(B)=C_{k}\cap M_{e}(B)\).
Two cases are possible.
First case. \(B- \mu I\) is one-to-one and therefore \(N(B+K_{2}- \mu I)\subseteq N(B- \mu I)=0\Rightarrow B+K_{2}-\mu I\) is one-to-one, hence \(C_{k}\subseteq M_{e}(B+K_{2})\).
Theorem 7
Corollary 1
Let A be a Riesz operator. Then there exists a normal compact operator K such that \((A-K)\in QN(H)\).
Proof
A satisfies the hypothesis of Lemma 9 and also \(\rho_{e}(A)=\mathbb{C}\backslash0\) and \(\operatorname{reg}(A)=\mathbb{C}\backslash\sigma(A)=\rho(A)\).
\(\rho_{e}(A)\cap \operatorname{reg}(A)=\rho(A)=M_{e}(A)\). Hence there is a compact operator K such that \(\rho(A-K)=M_{e}(A-K)=\rho_{e}(A)=\mathbb{C}\backslash0\Rightarrow\sigma(A-K)=0\Rightarrow(A-K)\in QN(H)\). We find the result of [14]. □
Remark 6
The result of Stampfli is a special case of Theorem 7 because if \(\mu\in\Phi_{0}(A)\), then \(\mu \in\rho_{e}(A)\subseteq M_{e}(A+K)\), hence \(A+K- \mu I\) is one-to-one or onto. If \(A+K- \mu I\) is one-to-one (resp. onto), then \(N(A+K- \mu I)=0\) (resp. \(N(A^{\ast}+K^{\ast}- \mu I)=0\)) and \(\operatorname{ind}(A+K- \mu I)=\operatorname{ind}(A- \mu I)=0N(A^{\ast}+K^{\ast}- \mu I)=0\) (resp. \(N(A+K- \mu I)=0\)) and therefore \(A+K- \mu I\) is one-to-one and onto. Hence \(\mu \in\rho(A+K)\Rightarrow\Phi_{0}(A)=\rho(A+K)\).
3 Conclusion
After the famous result of Stampfli, for all bounded operator A the Weyl spectrum is \(\sigma_{W}(A)=\bigcap \sigma(A+K)\) where the intersection is taken over all closed ideal compact operators. In this work, we prove the possibility to extend this result by the use of Kato’s decomposition to the set of semi-Fredholm operators.
Declarations
Acknowledgements
The authors are grateful to the anonymous referees who have contributed to improve the quality of this paper.
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Authors’ Affiliations
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