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April 4, 2014 Examples of operators and spectra Paul Garrett garrett@math.umn.edu http: /www.math.umn.edu/ garrett/ [This document is http://www.math.umn.edu/˜garrett/m/fun/notes 2012-13/06b examples spectra.pdf] 1. Generalities on spectra 2. Positive examples 3. Simplest Rellich compactness lemma 4. Cautionary examples: non-normal operators The usefulness of the notion of spectrum of an operator on a Hilbert space is the analogy to eigenvalues of operators on ﬁnite-dimensional spaces. Naturally, things become more complicated in inﬁnite-dimensional vector spaces. 1. Generalities on spectra It is convenient to know that spectra of continuous operators are non-empty compact subsets of Knowing this, every non-empty compact subset of is easily made to appear as the spectrum of a continuous operator, even normal ones, as below. [1.0.1] Proposition: The spectrum ) of a continuous linear operator on a Hilbert space is bounded by the operator norm op Proof: For op , an obvious heuristic suggests an expression for the resolvent = ( 1 + ... The inﬁnite series converges in operator norm for T/ op 1, that is, for op . Then 1 + ... = 1 giving a continuous inverse ( , so 6 ). /// [1.0.2] Remark: The same argument applied to shows that ) is inside the closed ball of radius op . By the elementary identity = ( ... T exists for op , that is, for /n op . That is, ) is inside the closed ball of radius inf /n op . The latter expression is the spectral radius of . This notion is relevant to non-normal operators, such as the Volterra operator , whose spectral radius is 0, while its operator norm is much larger. [1.0.3] Proposition: The spectrum ) of a continuous linear operator on a Hilbert space is compact Proof: That 6 ) is that there is a continuous linear operator ( . We claim that for suﬃciently close to , ( exists. Indeed, a heuristic suggests an expression for ( in terms of ( The algebra is helpfully simpliﬁed by replacing by , so that = 0. With near 0 and granting existence of , the heuristic is = (1 µT 1 + µT + ( µT ...

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Paul Garrett: Examples of operators and spectra (April 4, 2014) The geometric series converges in operator norm for µT op 1, that is, for op . Having found the obvious candidate for an inverse, (1 µT 1 + µT + ( µT ... = 1 and 1 + µT + ( µT ... = 1 so there is a continuous linear operator ( , and 6 ). Having already proven that ) is bounded , it is compact /// [1.0.4] Proposition: The spectrum ) of a continuous linear operator on a Hilbert space is non-empty Proof: The argument reduces the issue to Liouville’s theorem from complex analysis, that a bounded entire (holomorphic) function is constant . Further, an entire function that goes to 0 at is identically 0. Suppose the resolvent = ( is a continuous linear operator for all . The operator norm is readily estimated for large op 1 + ... op ≤ | 1 + op op ... op This goes to 0 as | Hilbert’s identity asserts the complex diﬀerentiability as operator-valued function: (as since is continuous for large , by the same identity: op ≤ | |·| op Thus, the scalar-valued functions v,w for v,w are complex-diﬀerentiable, and satisfy | v,w 〉| ≤ | |·| | ≤ | op ·| |·| | op ·| |·| By Liouville, v,w = 0 for all v,w , which is impossible. Thus, the spectrum is not empty. /// [1.0.5] Proposition: The entire spectrum, both point-spectrum and continuous-spectrum, of a self-adjoint operator is a non-empty, compact subset of . The entire spectrum of a unitary operator is a non-empty, compact subset of the unit circle. Proof: For self-adjoint , we claim that the imaginary part of v,v is at least v,v times the imaginary part of . Indeed, Tv,v is real , since Tv,v v,T v,Tv Tv,v so v,v Tv,v · v,v

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Paul Garrett: Examples of operators and spectra (April 4, 2014) and Im v,v 〉| Im |· v,v and by Cauchy-Schwarz-Bunyakowsky |·| | ≥ | v,v 〉| ≥ | Im |· v,v Im |·| Dividing by | ≥ | Im |·| This inequality shows more than the injectivity of . Namely, the inequality gives a bound on the operator norm of the inverse ( deﬁned on the image of . The image is dense since is not an eigenvalue and there is no residual spectrum for normal operators . Thus, the inverse extends by continuity to a continuous linear map deﬁned on the whole Hilbert space. Thus, has a continuous linear inverse, and is not in the spectrum of For unitary, Tv for all implies op = 1. Thus, ) is contained in the unit disk, by the general bound on spectra in terms of operator norms. From ( , the spectrum of is obtained by complex-conjugating the spectrum of . Thus, for unitary , the spectrum of is also contained in the unit disk. At the same time, the natural gives so ) exactly when ). Thus, the spectra of both and are inside the unit circle. /// 2. Positive examples This section gives non-pathological examples. Let be the usual space of square-summable sequences ,a ,... ), with standard orthonormal basis = (0 ,..., ,... {z 1 at jth position [2.1] Multiplication operators with speciﬁed eigenvalues Given a countable, bounded list of complex numbers , the operator by : ( ,a ,... , ,... has -eigenvector the standard basis element . Clearly : ( ,a ,a ,... ,... so is normal , in the sense that TT . To see that there are no other eigenvalues, suppose Tv with not among the . Then · v,e Tv,e v,T v, · v,e Thus, ( · v,e = 0, and v,e = 0 for all . Since form an orthonormal basis, = 0. ///

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Paul Garrett: Examples of operators and spectra (April 4, 2014) [2.2] Every compact subset of is the spectrum of an operator Grant for the moment a countable dense subset of a non-empty compact subset [1] of , and as above let : ( ,a ,a ,... , , ,... We saw that there are no further eigenvalues. Since spectra are closed , the closure of is contained in ). It remains to show that the continuous spectrum is no larger than the closure of the eigenvalues, in this example . That is, for 6 , exhibit a continuous linear ( For 6 , there is a uniform lower bound 0 < ≤| . That is, sup . Thus, the naturally suggested map ,a ,... , ... is a bounded linear map, and gives ( [2.3] Two-sided shift has no eigenvalues Let be the Hilbert space of two-sided sequences ...,a ,a ,a ,... ) with natural inner product ...,a ,a ,a ,... ...,b ,b ,b ,... ... ... The right and left two-sided shift operators are +1 These operators are mutual adjoints, mutual inverses, so are unitary. Being unitary, their operator norms are 1, so their spectra are non-empty compact subsets of the unit circle. They have no eigenvalues: indeed, for Rv , if there is any index with = 0, then the relation Rv gives +1 for = 0 ,... . Since = 1, such a vector is not in Nevertheless, we claim that ) for every with = 1, and similarly for . Indeed, for not in the spectrum, there is a continuous linear operator ( , so | ·| for some δ > 0. It is easy to make approximate eigenvectors for for any = 1: let = ( ..., ,..., ,λ, , ,..., ,... Obviously it doesn’t matter where the non-zero entries begin. From = ( ..., ,..., ,..., , +1 ,... 1 + 1, while + 1. Thus, | 0, and there can be no ( Thus, every on the unit circle is in ). [2.4] Compact multiplication operators on For a sequence of complex numbers 0, we claim that the multiplication operator : ( ,a ,... , ,... [1] To make a countable dense subset of , for = 1 ,... cover by ﬁnitely-many disks of radius 1 /n , each meeting , and in each choose a point of . The union over = 1 ,... of these ﬁnite sets is countable and dense in

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Paul Garrett: Examples of operators and spectra (April 4, 2014) is compact . We already showed that it has eigenvalues exactly , ,... , and spectrum the closure of Thus, the spectrum includes 0, but 0 is an eigenvalue only when it appears among the , which may range from 0 times to inﬁnitely-many times. To prove that the operator is compact, we prove that the image of the unit ball is pre-compact, by showing that it is totally bounded . Given ε > 0, take such that < for i > k . The ball in of radius max {| is precompact, so has a ﬁnite cover by -balls, centered at points ,...,v . For = ( ,v ,... ) with | 1, Tv = ( , ,..., ,... ) + (0 ,..., , +1 +1 , +2 +2 ,... With the closest of the ,...,v to ( , ,..., ,... ), Tv < (0 ,..., , +1 +1 , +2 +2 ,... < ·| (0 ,..., ,v +1 ,v +2 ,... | ·| | Thus, the image of the unit ball under is covered by ﬁnitely-many 2 -balls. /// [2.5] Multiplication operators on a,b For a,b ], we claim that the multiplication operator a,b a,b by ) = is normal , and has spectrum the image a,b ] of . The eigenvalues are such that ) = on a subset of a,b ] of positive measure. The normality is clear, so, beyond eigenvalues, we need only examine continuous spectrum, not residual. On one hand, if ) = on a set of positive measure, there is an inﬁnite-dimensional sub-space of [0 1] of functions supported there, and all these are eigenvectors. On the other hand, if = 0 in [0 1] and ) = ), even if is altered on a set of measure 0, it must be that ) = on a set of positive measure. To understand the continuous spectrum, for ) = make approximate eigenvectors by taking functions supported on [ δ,x ], where δ> 0 is small enough so that < for < . Then ·| dx ·| Thus, inf =0 = 0, so is not invertible. If is not an eigenvalue, it is in the continuous spectrum. On the other hand, if , then there is some δ> 0 such that | for all [0 1], by the compactness of [0 1]. Then ·| dx ·| dx ·| Thus, there is a continuous inverse ( ), and is not in the spectrum. 3. Simplest Rellich compactness lemma One characterization of the th Levi-Sobolev space of functions ) on a product = ( of circles is as the closure of the function space of ﬁnite Fourier series with respect to the Levi-Sobolev norm (squared) i (1 + , on ﬁnite Fourier series)

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Paul Garrett: Examples of operators and spectra (April 4, 2014) The standard orthonormal basis for ) is (2 n/ i (1 + s/ (with By the Plancherel theorem, the map from ) (with counting measure) to ) by } (2 n/ i (1 + s/ is an isometric isomorphism. For s>t , there is a continuous inclusion ). In terms of these orthonormal bases, there is a commutative diagram inc given by (1 + (2 n/ i (1 + s/ inc (2 n/ (1 + i (1 + t/ Since s>t , the number = (1 + are bounded by 1, and have unique limit point 0. In particular, ) is compact Thus, we have the simplest instance of Rellich’s compactness lemma : the inclusion ) is compact for s>t 4. Cautionary examples: non-normal operators [4.1] Shift operators on one-sided We claim the following: The right-shift : ( ,a ,... (0 ,a ,a ,... and the left-shift : ( ,a ,a ,... ,... are mutual adjoints. These operators are not normal, since = 1 but : ( ,a ,... (0 ,a ,... The eigenvalues of the left-shift are all complex numbers in the open unit disk in . In particular, there is a continuum of eigenvalues and eigenvectors, so they cannot be mutually orthogonal . The spectrum is the closed unit disk.

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Paul Garrett: Examples of operators and spectra (April 4, 2014) The right-shift has no eigenvalues, has continuous spectrum the unit circle, and residual spectrum the open unit disk with 0 removed. Indeed, suppose (0 ,a ,a ,... ) = ,a ,... ) = ,a ,... With the lowest index such that = 0, the th component in the eigenvector relation gives 0 = , so = 0. Then, the ( + 1) th component gives +1 = 0, contradiction. This proves that has no eigenvalues. Oppositely, for 1, (1 ,λ, ,... ) = ( λ, ,... ) = (1 ,λ, ,... so every such is an eigenvector for . On the other hand, for = 1, in an eigenvector relation ,... ) = ,a ,... ) = ,a ,... let be the smallest index with = 0. Then +1 +2 +1 ... , so ,a ,... ) = (0 ,..., ,a ,λa , ,... But this is not in for = 1 and = 0, so on the unit circle is not an eigenvalue. For = 1, we can make approximate -eigenvectors for by = (1 ,λ, ,..., ,... since = ( λ, ,..., ,... (1 ,λ, ,..., ,... ) = (0 ,..., , +1 ,... Since +1 (1 + ... + 1 there can be no continuous ( . Thus, on the unit circle is in the spectrum, but not in the point spectrum. That the unit circle is in the spectrum also follows from the observation above that all with 1 are eigenvalues, and the fact that the spectrum is closed The spectrum of is bounded by the operator norm op , and op is visibly 1, so is nothing else in the spectrum. To see that the unit circle is the continuous spectrum of , as opposed to residual , we show that has dense image for = 1. Indeed, for such that, for all 0 = v,w v, v, we would have ( = 0. However, we have seen that has no eigenvalues. Thus, always has dense image, and the unit circle is continuous spectrum for Reversing that discussion, every with 1 is in the residual spectrum of , because such is not an eigenvalue, and does not have dense image: for -eigenvector for v,w v, v, v, = 0 That is, the image ( is in the orthogonal complement to the eigenvector . The same computation shows that the unit circle is continuous spectrum for , because it is not eigenvalues for

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Paul Garrett: Examples of operators and spectra (April 4, 2014) [4.2] Volterra operator We will show that the Volterra operator Vf ) = dt on [0 1] is compact but not self-adjoint, that its spectrum is , and that it has no eigenvalues. A relation Tf for and = 0 implies is continuous |·| Tf Tf | ·| dt ≤ | ·| The fundamental theorem of calculus would imply is continuously diﬀerentiable and = ( Tf Thus, would be a constant multiple of x/ , by the mean value theorem. However, by Cauchy-Schwarz- Bunyakowsky, for a -eigenfunction |·| | ≤ | ·| No non-zero multiple of the exponential satisﬁes this. Thus, there are no eigenvectors for non-zero eigenvalues. For [0 1] and Tf = 0 [0 1], Tf is almost everywhere 0. Since Tg ) is unavoidably continuous Tf ) is 0 for all . Thus, for all x,y in the interval, 0 = 0 0 = Tf Tf ) = dt That is, Tf ) is orthogonal in [0 1] to all characteristic functions of intervals. Finite linear combinations of these are dense in [0 1] in the topology, and [0 1] is dense in [0 1]. Thus = 0, and there are no eigenvectors for the Volterra operator. To see that is compact , rewrite it as being given by an integral kernel x,y ): Tf ) = dy x,y dy (with x,y ) = 0 (for 0 y 1 (for x 1) Thus, is Hilbert-Schmidt, and compact. The adjoint is given by the integral kernel x,y ) = y,x ), visibly diﬀerent from x,y ), so is not self-adjoint. To see that the spectrum is at most , show that the spectral radius is 0: ) = ... dtdx ...dx ... dx ...dx dt 1)! dt From this, op , and log lim (2 )! lim log(2 )! = lim log lim (log + log(2 + 1)) lim (log + log(2 + 1)) lim log 2 lim log 2 since (2 for 1 , noting the sign. That is, lim /n op = 0, so the spectral radius is 0. Since the spectrum is non-empty, it must be exactly

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April 4, 2014 Examples of operators and spectra Paul Garrett garrett@math.umn.edu http: /www.math.umn.edu/ garrett/ [This document is http://www.math.umn.edu/˜garrett/m/fun/notes 2012-13/06b examples spectra.pdf] 1. Generalities on spectra 2. Positive examples 3. Simplest Rellich compactness lemma 4. Cautionary examples: non-normal operators The usefulness of the notion of spectrum of an operator on a Hilbert space is the analogy to eigenvalues of operators on ﬁnite-dimensional spaces. Naturally, things become more complicated in inﬁnite-dimensional vector spaces. 1. Generalities on spectra It is convenient to know that spectra of continuous operators are non-empty compact subsets of Knowing this, every non-empty compact subset of is easily made to appear as the spectrum of a continuous operator, even normal ones, as below. [1.0.1] Proposition: The spectrum ) of a continuous linear operator on a Hilbert space is bounded by the operator norm op Proof: For op , an obvious heuristic suggests an expression for the resolvent = ( 1 + ... The inﬁnite series converges in operator norm for T/ op 1, that is, for op . Then 1 + ... = 1 giving a continuous inverse ( , so 6 ). /// [1.0.2] Remark: The same argument applied to shows that ) is inside the closed ball of radius op . By the elementary identity = ( ... T exists for op , that is, for /n op . That is, ) is inside the closed ball of radius inf /n op . The latter expression is the spectral radius of . This notion is relevant to non-normal operators, such as the Volterra operator , whose spectral radius is 0, while its operator norm is much larger. [1.0.3] Proposition: The spectrum ) of a continuous linear operator on a Hilbert space is compact Proof: That 6 ) is that there is a continuous linear operator ( . We claim that for suﬃciently close to , ( exists. Indeed, a heuristic suggests an expression for ( in terms of ( The algebra is helpfully simpliﬁed by replacing by , so that = 0. With near 0 and granting existence of , the heuristic is = (1 µT 1 + µT + ( µT ...

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Paul Garrett: Examples of operators and spectra (April 4, 2014) The geometric series converges in operator norm for µT op 1, that is, for op . Having found the obvious candidate for an inverse, (1 µT 1 + µT + ( µT ... = 1 and 1 + µT + ( µT ... = 1 so there is a continuous linear operator ( , and 6 ). Having already proven that ) is bounded , it is compact /// [1.0.4] Proposition: The spectrum ) of a continuous linear operator on a Hilbert space is non-empty Proof: The argument reduces the issue to Liouville’s theorem from complex analysis, that a bounded entire (holomorphic) function is constant . Further, an entire function that goes to 0 at is identically 0. Suppose the resolvent = ( is a continuous linear operator for all . The operator norm is readily estimated for large op 1 + ... op ≤ | 1 + op op ... op This goes to 0 as | Hilbert’s identity asserts the complex diﬀerentiability as operator-valued function: (as since is continuous for large , by the same identity: op ≤ | |·| op Thus, the scalar-valued functions v,w for v,w are complex-diﬀerentiable, and satisfy | v,w 〉| ≤ | |·| | ≤ | op ·| |·| | op ·| |·| By Liouville, v,w = 0 for all v,w , which is impossible. Thus, the spectrum is not empty. /// [1.0.5] Proposition: The entire spectrum, both point-spectrum and continuous-spectrum, of a self-adjoint operator is a non-empty, compact subset of . The entire spectrum of a unitary operator is a non-empty, compact subset of the unit circle. Proof: For self-adjoint , we claim that the imaginary part of v,v is at least v,v times the imaginary part of . Indeed, Tv,v is real , since Tv,v v,T v,Tv Tv,v so v,v Tv,v · v,v

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Paul Garrett: Examples of operators and spectra (April 4, 2014) and Im v,v 〉| Im |· v,v and by Cauchy-Schwarz-Bunyakowsky |·| | ≥ | v,v 〉| ≥ | Im |· v,v Im |·| Dividing by | ≥ | Im |·| This inequality shows more than the injectivity of . Namely, the inequality gives a bound on the operator norm of the inverse ( deﬁned on the image of . The image is dense since is not an eigenvalue and there is no residual spectrum for normal operators . Thus, the inverse extends by continuity to a continuous linear map deﬁned on the whole Hilbert space. Thus, has a continuous linear inverse, and is not in the spectrum of For unitary, Tv for all implies op = 1. Thus, ) is contained in the unit disk, by the general bound on spectra in terms of operator norms. From ( , the spectrum of is obtained by complex-conjugating the spectrum of . Thus, for unitary , the spectrum of is also contained in the unit disk. At the same time, the natural gives so ) exactly when ). Thus, the spectra of both and are inside the unit circle. /// 2. Positive examples This section gives non-pathological examples. Let be the usual space of square-summable sequences ,a ,... ), with standard orthonormal basis = (0 ,..., ,... {z 1 at jth position [2.1] Multiplication operators with speciﬁed eigenvalues Given a countable, bounded list of complex numbers , the operator by : ( ,a ,... , ,... has -eigenvector the standard basis element . Clearly : ( ,a ,a ,... ,... so is normal , in the sense that TT . To see that there are no other eigenvalues, suppose Tv with not among the . Then · v,e Tv,e v,T v, · v,e Thus, ( · v,e = 0, and v,e = 0 for all . Since form an orthonormal basis, = 0. ///

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Paul Garrett: Examples of operators and spectra (April 4, 2014) [2.2] Every compact subset of is the spectrum of an operator Grant for the moment a countable dense subset of a non-empty compact subset [1] of , and as above let : ( ,a ,a ,... , , ,... We saw that there are no further eigenvalues. Since spectra are closed , the closure of is contained in ). It remains to show that the continuous spectrum is no larger than the closure of the eigenvalues, in this example . That is, for 6 , exhibit a continuous linear ( For 6 , there is a uniform lower bound 0 < ≤| . That is, sup . Thus, the naturally suggested map ,a ,... , ... is a bounded linear map, and gives ( [2.3] Two-sided shift has no eigenvalues Let be the Hilbert space of two-sided sequences ...,a ,a ,a ,... ) with natural inner product ...,a ,a ,a ,... ...,b ,b ,b ,... ... ... The right and left two-sided shift operators are +1 These operators are mutual adjoints, mutual inverses, so are unitary. Being unitary, their operator norms are 1, so their spectra are non-empty compact subsets of the unit circle. They have no eigenvalues: indeed, for Rv , if there is any index with = 0, then the relation Rv gives +1 for = 0 ,... . Since = 1, such a vector is not in Nevertheless, we claim that ) for every with = 1, and similarly for . Indeed, for not in the spectrum, there is a continuous linear operator ( , so | ·| for some δ > 0. It is easy to make approximate eigenvectors for for any = 1: let = ( ..., ,..., ,λ, , ,..., ,... Obviously it doesn’t matter where the non-zero entries begin. From = ( ..., ,..., ,..., , +1 ,... 1 + 1, while + 1. Thus, | 0, and there can be no ( Thus, every on the unit circle is in ). [2.4] Compact multiplication operators on For a sequence of complex numbers 0, we claim that the multiplication operator : ( ,a ,... , ,... [1] To make a countable dense subset of , for = 1 ,... cover by ﬁnitely-many disks of radius 1 /n , each meeting , and in each choose a point of . The union over = 1 ,... of these ﬁnite sets is countable and dense in

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Paul Garrett: Examples of operators and spectra (April 4, 2014) is compact . We already showed that it has eigenvalues exactly , ,... , and spectrum the closure of Thus, the spectrum includes 0, but 0 is an eigenvalue only when it appears among the , which may range from 0 times to inﬁnitely-many times. To prove that the operator is compact, we prove that the image of the unit ball is pre-compact, by showing that it is totally bounded . Given ε > 0, take such that < for i > k . The ball in of radius max {| is precompact, so has a ﬁnite cover by -balls, centered at points ,...,v . For = ( ,v ,... ) with | 1, Tv = ( , ,..., ,... ) + (0 ,..., , +1 +1 , +2 +2 ,... With the closest of the ,...,v to ( , ,..., ,... ), Tv < (0 ,..., , +1 +1 , +2 +2 ,... < ·| (0 ,..., ,v +1 ,v +2 ,... | ·| | Thus, the image of the unit ball under is covered by ﬁnitely-many 2 -balls. /// [2.5] Multiplication operators on a,b For a,b ], we claim that the multiplication operator a,b a,b by ) = is normal , and has spectrum the image a,b ] of . The eigenvalues are such that ) = on a subset of a,b ] of positive measure. The normality is clear, so, beyond eigenvalues, we need only examine continuous spectrum, not residual. On one hand, if ) = on a set of positive measure, there is an inﬁnite-dimensional sub-space of [0 1] of functions supported there, and all these are eigenvectors. On the other hand, if = 0 in [0 1] and ) = ), even if is altered on a set of measure 0, it must be that ) = on a set of positive measure. To understand the continuous spectrum, for ) = make approximate eigenvectors by taking functions supported on [ δ,x ], where δ> 0 is small enough so that < for < . Then ·| dx ·| Thus, inf =0 = 0, so is not invertible. If is not an eigenvalue, it is in the continuous spectrum. On the other hand, if , then there is some δ> 0 such that | for all [0 1], by the compactness of [0 1]. Then ·| dx ·| dx ·| Thus, there is a continuous inverse ( ), and is not in the spectrum. 3. Simplest Rellich compactness lemma One characterization of the th Levi-Sobolev space of functions ) on a product = ( of circles is as the closure of the function space of ﬁnite Fourier series with respect to the Levi-Sobolev norm (squared) i (1 + , on ﬁnite Fourier series)

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Paul Garrett: Examples of operators and spectra (April 4, 2014) The standard orthonormal basis for ) is (2 n/ i (1 + s/ (with By the Plancherel theorem, the map from ) (with counting measure) to ) by } (2 n/ i (1 + s/ is an isometric isomorphism. For s>t , there is a continuous inclusion ). In terms of these orthonormal bases, there is a commutative diagram inc given by (1 + (2 n/ i (1 + s/ inc (2 n/ (1 + i (1 + t/ Since s>t , the number = (1 + are bounded by 1, and have unique limit point 0. In particular, ) is compact Thus, we have the simplest instance of Rellich’s compactness lemma : the inclusion ) is compact for s>t 4. Cautionary examples: non-normal operators [4.1] Shift operators on one-sided We claim the following: The right-shift : ( ,a ,... (0 ,a ,a ,... and the left-shift : ( ,a ,a ,... ,... are mutual adjoints. These operators are not normal, since = 1 but : ( ,a ,... (0 ,a ,... The eigenvalues of the left-shift are all complex numbers in the open unit disk in . In particular, there is a continuum of eigenvalues and eigenvectors, so they cannot be mutually orthogonal . The spectrum is the closed unit disk.

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Paul Garrett: Examples of operators and spectra (April 4, 2014) The right-shift has no eigenvalues, has continuous spectrum the unit circle, and residual spectrum the open unit disk with 0 removed. Indeed, suppose (0 ,a ,a ,... ) = ,a ,... ) = ,a ,... With the lowest index such that = 0, the th component in the eigenvector relation gives 0 = , so = 0. Then, the ( + 1) th component gives +1 = 0, contradiction. This proves that has no eigenvalues. Oppositely, for 1, (1 ,λ, ,... ) = ( λ, ,... ) = (1 ,λ, ,... so every such is an eigenvector for . On the other hand, for = 1, in an eigenvector relation ,... ) = ,a ,... ) = ,a ,... let be the smallest index with = 0. Then +1 +2 +1 ... , so ,a ,... ) = (0 ,..., ,a ,λa , ,... But this is not in for = 1 and = 0, so on the unit circle is not an eigenvalue. For = 1, we can make approximate -eigenvectors for by = (1 ,λ, ,..., ,... since = ( λ, ,..., ,... (1 ,λ, ,..., ,... ) = (0 ,..., , +1 ,... Since +1 (1 + ... + 1 there can be no continuous ( . Thus, on the unit circle is in the spectrum, but not in the point spectrum. That the unit circle is in the spectrum also follows from the observation above that all with 1 are eigenvalues, and the fact that the spectrum is closed The spectrum of is bounded by the operator norm op , and op is visibly 1, so is nothing else in the spectrum. To see that the unit circle is the continuous spectrum of , as opposed to residual , we show that has dense image for = 1. Indeed, for such that, for all 0 = v,w v, v, we would have ( = 0. However, we have seen that has no eigenvalues. Thus, always has dense image, and the unit circle is continuous spectrum for Reversing that discussion, every with 1 is in the residual spectrum of , because such is not an eigenvalue, and does not have dense image: for -eigenvector for v,w v, v, v, = 0 That is, the image ( is in the orthogonal complement to the eigenvector . The same computation shows that the unit circle is continuous spectrum for , because it is not eigenvalues for

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Paul Garrett: Examples of operators and spectra (April 4, 2014) [4.2] Volterra operator We will show that the Volterra operator Vf ) = dt on [0 1] is compact but not self-adjoint, that its spectrum is , and that it has no eigenvalues. A relation Tf for and = 0 implies is continuous |·| Tf Tf | ·| dt ≤ | ·| The fundamental theorem of calculus would imply is continuously diﬀerentiable and = ( Tf Thus, would be a constant multiple of x/ , by the mean value theorem. However, by Cauchy-Schwarz- Bunyakowsky, for a -eigenfunction |·| | ≤ | ·| No non-zero multiple of the exponential satisﬁes this. Thus, there are no eigenvectors for non-zero eigenvalues. For [0 1] and Tf = 0 [0 1], Tf is almost everywhere 0. Since Tg ) is unavoidably continuous Tf ) is 0 for all . Thus, for all x,y in the interval, 0 = 0 0 = Tf Tf ) = dt That is, Tf ) is orthogonal in [0 1] to all characteristic functions of intervals. Finite linear combinations of these are dense in [0 1] in the topology, and [0 1] is dense in [0 1]. Thus = 0, and there are no eigenvectors for the Volterra operator. To see that is compact , rewrite it as being given by an integral kernel x,y ): Tf ) = dy x,y dy (with x,y ) = 0 (for 0 y 1 (for x 1) Thus, is Hilbert-Schmidt, and compact. The adjoint is given by the integral kernel x,y ) = y,x ), visibly diﬀerent from x,y ), so is not self-adjoint. To see that the spectrum is at most , show that the spectral radius is 0: ) = ... dtdx ...dx ... dx ...dx dt 1)! dt From this, op , and log lim (2 )! lim log(2 )! = lim log lim (log + log(2 + 1)) lim (log + log(2 + 1)) lim log 2 lim log 2 since (2 for 1 , noting the sign. That is, lim /n op = 0, so the spectral radius is 0. Since the spectrum is non-empty, it must be exactly

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