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Kyungpook Mathematical Journal 2023; 63(1): 1-10

Published online March 31, 2023 https://doi.org/10.5666/KMJ.2022.63.1.1

Copyright © Kyungpook Mathematical Journal.

Criteria for Algebraic Operators to be Unitary

Zenon Jan JabŁoński and Jan Stochel, Il Bong Jung*

Instytut Matematyki, Uniwersytet Jagielloński, ul. Łojasiewicza 6, PL-30348 Kraków, Poland
e-mail : Zenon.Jablonski@im.uj.edu.pl and Jan.Stochel@im.uj.edu.pl

Department of Mathematics, Kyungpook National University, Daegu 41566, Korea
e-mail : ibjung@knu.ac.kr

Received: July 5, 2022; Accepted: September 19, 2022

Criteria for an algebraic operator T on a complex Hilbert space H to be unitary are established. The main one is written in terms of the convergence of sequences of the form {Tnh}n=0 with hH. Related questions are also discussed.

Keywords: Algebraic operator, unitary operator, normaloid operator, strong stability

By the spectral theorem, a unitary operator with a finite spectrum is algebraic and its spectrum is contained in T, the unit circle centered at 0. The most fundamental example of a unitary algebraic operator is the Fourier transform. According to the famous theorem of Plancherel, the Fourier transform extends uniquely to a unitary operator on L2() (see e.g., [18, Theorem IX.6]). Denote it by F. The Fourier transform F has the following properties:

F0=I,F1=F,F2=P,F3=F-1andF4=I,

where I is the identity operator on L2() and P(f)(x)=f(x) for fL2(). This implies that p(x)=x4 is the minimal polynomial of F. As a consequence, the Fourier transform is a unitary algebraic operator with (purely point) spectrum σ(F)={1,1,i,i} (see [18, Theorems IX.1 and IX.6]).

A natural question arises under what additional assumptions an algebraic (bounded linear) operator T on a complex Hilbert space H with spectrum in T is unitary. To answer this question let us look at some broader classes A of operators that can be characterized as follows: an operator T belongs to A if and only if the sequences of the form {Tnh2}n=0 (hH) belong to the corresponding class S of scalar sequences; in most cases, the class S appears naturally in harmonic analysis on *-semigroups. In particular, the celebrated theorem of Lambert states that the class of subnormal operators corresponds to Stieltjes moment sequences (see [17]). In this line of correspondence, we can list the classes of m-isometric operators [2, 3, 4, 13], completely hypercontractive operators [1], completely hyperexpansive operators [5], alternatingly hyperexpansive operators [20], conditionally positive definite operators [14], and so on. The answer to our question (see Theorem 1.1 below) is written in terms of the convergence of the sequences of the form {Tnh}n=0 (hH). The condition of their convergence seems to be optimal, since in the light of Remark 3.3 the assumption of their boundedness ceases to be sufficient. It is also worth mentioning that there are contractions (for which the sequences {Tnh}n=0, hH, automatically converge) with spectrum in T, called unimodular contractions, which are not unitary (see [19]). Clearly, unimodular contractions are normaloid. Let us further note that if we replace the class of normaloid operators by a class of more regular operators, it may turn out that members of the latter class with spectrum in T are unitary. In particular, by Stampfli's theorem (see [21, Corollary, p. 473]), every hyponormal operator with spectrum in T is unitary.

Before formulating the main result, we establish some notation and terminology. Denote by the field of complex numbers. Set T={z:|z|=1}. Write , + and + for the sets of positive integers, nonnegative integers and nonnegative real numbers, respectively. Let B(H) stand for the C*-algebra of all bounded linear operators on a complex Hilbert space H. For TB(H), denote by N(T), σ(T) and r(T) the kernel, the spectrum and the spectral radius of T, respectively. An operator TB(H) is said to be normaloid if r(T)=T, or equivalently, by Gelfand's formula for spectral radius, if and only if Tn=Tn for all n. Call TB(H) algebraic if there exists a nonzero polynomial p (in one indeterminate with complex coefficients) such that p(T)=0; such a p is said to be minimal if p is the (unique) monic polynomial of least degree among all nonzero polynomials q such that q(T)=0.

The following theorem, which is the main result of the paper, characterizes unitary algebraic operators in terms of the convergence of the sequences of the form {Tnh}n=0. Its proof is given in Section 3.

Theorem 1.1. Suppose that TB(H) is algebraic and σ(T)T. Then the following statements are equivalent:

  • (i) T is unitary,

  • (ii) T is normaloid,

  • (iii) T1 (or equivalently, T=1),

  • (iv) the sequence {Tnh}n=0 is convergent in + for every hH.

In this section we give some basic facts about algebraic operators needed in this paper. We begin with a purely linear algebra result, the proof of which is left to the reader. If M is a complex vector space, then the identity transformation on M is denoted by IM (or simply by I if no ambiguity arises). We write

M=M1Mm

in the case when M is a direct sum of (finitely many) vector subspaces M1,,Mm.

Lemma 2.1. Suppose that (2.1) holds. Let A:MM be a linear transformation such that A(Mj)Mj for all j=1,,m and let z. Then AzIM is a bijection if and only if A|MjzIMj is a bijection for all j=1,,m. Moreover, if AzIM is a bijection, then (AzIM)1(Mj)=Mj for all j=1,,m and

(A|MjzIMj)1=(AzIM)1|Mj,j=1,,m.

Corollary 2.2. Suppose that H is a complex Hilbert space which is a direct sum of finitely many nonzero closed vector subspaces H1,,Hm. Let TB(H) be such that T(Hj)Hj for all j=1,,m. Then σ(T|Hj)σ(T) for all j=1,,m.

For the sake of self-containedness, we sketch the proof of the following lemma that collects indispensable facts about algebraic operators.

Lemma 2.3. Let TB(H). Then the following conditions are equivalent:

  • (i) T is algebraic,

  • (ii) there exist an integer m1, integers i1,,im1, distinct complex numbers z1,,zm and closed nonzero vector subspaces H1,,Hm of H such that

    • (ii-a)
      H=H1Hm,

    • (ii-b) T(Hj)Hj for all j=1,,m,

    • (ii-c) (TjzjIj)ij=0 for all j=1,,m, where Tj:=T|Hj and Ij:=IHj,

    • (ii-d) σ(T)={z1,,zm} and σ(Tj)={zj} for all j=1,,m,

    • (ii-e) there exists a constant c(0,) such that

hj c k=1mhk,j=1,,m,h1H1,,hmHm.

Proof. (i)(ii) Let T be an algebraic operator and p be its minimal polynomial. Clearly, degp1. It follows from the fundamental theorem of algebra that

p(x)=(xz1)i1(xzm)im

with unique integers i1,,im1 and distinct complex numbers z1,,zm. In view of [6, Lemma 6.1], the condition (ii-a) holds with Hj:=N((TzjI)ij){0} for j=1,,m. This implies that H1,,Hm are closed vector subspaces of H which are invariant for T. As a consequence, (ii-b) and (ii-c) hold. By the spectral mapping theorem, σ(T){z1,,zm}. Since (TjzjIj)ij=0 and Hj{0}, we infer from the spectral mapping theorem and Corollary 2.2 that

{zj}=σ(Tj)σ(T),j=1,,m,

which implies (ii-d).

Now, we proceed to the proof of (ii-e). Define for j=1,,m the linear projection Pj:HH by

Pj(h1++hm)=hj,h1H1,,hmHm.

By (ii-a), this definition is correct. Using [6, Lemma 6.1(iii)] we see that

K2:=H2Hm=N( j=2 m(Tzj I)ij ),

and so K2 is a closed vector subspace of H. Since, by (ii-a), H=H1K2, we infer from [7, Theorem III.13.2] that P1B(H). A similar argument shows that PjB(H) for all j{1,,m}. This implies (2.2).

(ii)(i) It is enough to note that p(T)=0 with p(x)=(xz1)i1(xzm)im. This completes the proof.

We begin this section by stating an auxiliary lemma.

Lemma 3.1 [11, Lemma 2.1] Let b,w be such that |w|=1 and w±1. Assume that the sequence {Re(wnb)}n=0 is convergent. Then b=0.

Before proving the main result of this paper, we characterize power bounded algebraic operators with spectrum in the unit circle. Recall that TB(H) is said to be power bounded if supn+Tn<.

Lemma 3.2. Let TB(H). Then the following conditions are equivalent:

  • (i) T is a power bounded algebraic operator such that σ(T)T,

  • (ii) there exist an integer m1, closed nonzero vector subspaces H1,,Hm of H and distinct complex numbers z1,,zm such that

    • (ii-a) H=H1Hm,

    • (ii-b) T(h1++hm)=z1h1++zmhm for all h1H1,,hmHm,

    • (ii-c) {z1,,zm}T.

Proof. (i)(ii) Assume that T is a power bounded algebraic operator such that σ(T)T. By Lemma 2.3, there exist an integer m1, integers i1,,im1, distinct complex numbers z1,,zm and closed nonzero vector subspaces H1,,Hm of H that satisfy the conditions (ii-a)-(ii-d) of this lemma. Since σ(T)T, we have

|zl|=1,l=1,,m.

Fix k{1,,m}. Take hkHk{0}. Using (3.1), Lemma 2.3(ii-c) and [6, Sublemma 6.3], we deduce that

αn(hk):=1nN(hk )Tknhkconvergestoapositiverealnumberasn,

where N(hk) is the unique nonnegative integer such that

(Tk-zkIk)N(hk)hk0and(Tk-zkIk)N(hk)+1hk=0.

In particular, we have

Tknhk=nN(hk)αn(hk),n1.

Since Tk is power bounded, the sequence {Tknhk}n=0 is bounded. Hence, one can infer from (3.2) and (3.4) that N(hk)=0. This, together with (3.3), implies that Tkhk=zkhk. As a consequence, the system T,z1,,zm, H1,,Hm satisfies the conditions (ii-a), (ii-b) and (ii-c).

(ii)(i) By (ii-a) and (ii-b), p(T)=0 with p(x)=(xz1)(xzm), which means that T is an algebraic operator. According to the spectral mapping theorem, σ(T){z1,,zm}. In turn, by (ii-b) and the assumption that each Hi is nonzero, we deduce that z1,,zm are eigenvalues of T. Therefore, by (ii-c), we have

σ(T)={z1,,zm}T.

Now, using Lemma 2.3(ii-e) (or the uniform boundedness principle), we deduce the power boundedness of T from (ii-a), (ii-b) and (ii-c). This completes the proof.

Remark 3.3. Regarding Theorem 1.1, we note that in view of Lemma 3.2 there exist algebraic operators T with spectrum in T that are not unitary but have the property that each sequence {Tnh}n=0 is bounded (or equivalently, T is power bounded).

Now, we are ready to prove the main result of this paper.

Proof of Theorem 1.1. The implications (i)(ii)(iii)(iv) are obvious.

(iv)(i) Assume that T is algebraic, σ(T)T and the sequence {Tnh}n=0 is convergent in + for every hH. By the uniform boundedness principle, T is power bounded. In view of the implication (i)(ii) of Lemma 3.2, there exist an integer m1, distinct complex numbers z1,,zm and closed nonzero vector subspaces H1,,Hm of H that satisfy the conditions (ii-a), (ii-b) and (ii-c) of Lemma 3.2. Fix distinct k,l{1,,m}. Take hkHk and hlHl. Then, by the conditions (ii-a), (ii-b) and (ii-c) of Lemma 3.2, we have

Tn(hk+hl)2=zknhk+zlnhl2=hk2+2Re((zk z¯l)nhkhl)+hl2,n0.

Combined with (iv), this implies that the sequence {Re((zk z¯l)n hkhl)}n=0 is convergent. If zk z¯l=1, then we see that Rehkhl=0. Substituting ihk in place of hk, we deduce that hkhl=0. The only possibility left is that zk z¯l±1. Since, by the condition (ii-c) of Lemma 3.2, |zk z¯l|=1, we infer from Lemma 3.1 that hkhl=0. This shows that H=H1Hm. Hence, by the conditions (ii-b) and (ii-c) of Lemma 3.2, we conclude that T is a unitary operator. This completes the proof.

For the reader's convenience, we record here some useful facts related to the main topic of this paper concerning certain classes of operators. We begin by discussing the question of the existence of the limits limnTnh in the case of normaloid operators and the issue of strong stability1) in the context of subnormal operators. Both are intimately related to Theorem 1.1.

Proposition 4.1. Let TB(H) be a normaloid operator. Then the following conditions are equivalent:

  • (i) the sequence {Tnh}n=0 is convergent in + for every hH,

  • (ii) T is power bounded,

  • (iii) T is a contraction.

Proof. By the uniform boundedness principle and Gelfand's formula for r(T), (i) implies (ii) and (ii) implies (iii). That (iii) implies (i) is obvious.

Before formulating the next result, we give the necessary definitions and facts related to the concept of subnormality. Recall that an operator TB(H) is said to be subnormal if there exist a complex Hilbert space K and a normal operator NB(K) such that HK (isometric embedding) and Th=Nh for all hH. Such an N is called a normal extension of T; if K has no proper closed vector subspace containing H and reducing N, then N is called minimal. By a semispectral measure of a subnormal operator TB(H) we mean the Borel B(H)-valued measure F on defined by

F(Δ)=PE(Δ)|H,Δ-Borelsubsetof,

where E is the spectral measure of a minimal normal extension NB(K) of T and PB(K) is the orthogonal projection of K onto H. In view of [12, Proposition 5], a subnormal operator has exactly one semispectral measure. We also need the following fact (see [12, Proposition 4]).

If N is minimal, then for every Borel subset Δof ,F(Δ) = 0 if and only if E(Δ) = 0.

According to [8, Proposition II.4.6]), the following holds.

Any subnormal operator is normaloid.

We refer the reader to [8] for the foundations of the theory of subnormal operators.

Proposition 4.2. Let SB(H) be a subnormal operator, NB(K) be a minimal normal extension of S and F be the semispectral measure of S. Then the following assertions hold:

  • (i) if S is a contraction, then limnSnh2=F(T)hh for every hH,

  • (ii) the following conditions are equivalent:

    • (ii-a) S is strongly stable, i.e., limnSnh=0 for every hH,

    • (ii-b) S is power bounded and F(T)=0,

    • (ii-c) S is a contraction and F(T)=0,

  • (iii) S is strongly stable if and only if N is strongly stable.

Proof. (i) Suppose S1. Then, by [8, Corollary II.2.17], N=S1. It follows from the spectral theorem that

Snh2=Nnh2=(4.1)D¯|z|2nF(dz)h,h,n+,hH,

where D¯={z:|z|1}. Since for zD¯, the sequence {|z|2n}n=0 converges to χT(z)asn, where χT is the characteristic function of T, we deduce from Lebesgue's dominated convergence theorem that (i) holds.

(ii) Since subnormal operators are normaloid (see 4.3), the assertion (ii) follows from (i) and Proposition 4.1.

(iii) Recall that S=N, and also that F(T)=0 if and only if E(T)=0, where E is the spectral measure of N (see 4.2). Combined with (ii) applied to both S and N, this yields (iii).

We now estimate the growth of norms of powers of an algebraic operator whose spectral radius is less than or equal to 1.

Proposition 4.3. Suppose that TB(H) is an algebraic operator such that 0<r(T)1. Then there exists α(0,) such that

Tnαnκr(T)n,n1,

where κ=(degp)1 and p is the minimal polynomial of T.

Proof. Since r(T)>0, in light of Lemma 2.3 and its proof it suffices to consider the case when (TzI)i=0 for some 1idegp and for some z such that 0<|z|1. Setting N=TzI, it is easily seen that

Tn=(zI+N)n= j=0 i1 njznjNj,ni1,

which together with r(T)=|z| implies that

Tn j=0 i1 njNj|z|nj( j=0 i1Nj j!|z|j )ni1r(T)n,ni1.

This completes the proof.

Remark 4.4. a) It is worth mentioning that if TB(H) is an algebraic operator such that r(T)=0, then, in view of Lemma 2.3, the estimate (4.4) still holds, but only for ndegp.

b) It follows from Gelfand's formula for spectral radius that if TB(H) is such that r(T)<1, then T is uniformly stable2), i.e., limnTn=0, and hence the sequence {Tnh}n=0 is convergent for every hH. The latter statement ceases to be true if the spectrum of T has a nonempty intersection with T{1}, even if T is algebraic. z=e2πiθnamely, if T=zI, where z=e2πiθ and 𝜃 is an irrational number, then by Jacobi's theorem (see [10, Theorem I.3.13]) the closure of the set {Tnh:n+} is equal to {zh:zT} for every hH.

c) Note that if TB(H) is algebraic, then the sequence {Tnh1/n}n=1 is convergent in + for all hH (see [6, Proposition 6.2]). Without the assumption that T is algebraic, the sequence {Tnh1/n}n=1 may not converge. For more details on this issue, we refer the reader to [9].

We conclude this section by providing an example illustrating Theorem 1.1 and Proposition 4.1.

Example 4.5. Let NB(H) be a nonzero operator such that N2=0 and let αT. Then the operator Tα:=αI+N is algebraic because p(Tα)=0 with p(x)=(xα)2. Hence, by the spectral mapping theorem σ(Tα)={α} and so Tα is invertible in B(H) and r(Tα)=1. However Tα>1 whenever N>2 (this can be achieved simply by rescaling N). As a consequence, Tα is not normaloid, hence not subnormal (see (4.3)). Observe that HN(N) and

limnTαnh=ifandonlyifhHN(N).

Indeed, by Newton's binomial formula (or simply by induction), we have

(I+α¯N)n=I+nα¯N,n+,

which implies that

(αI+N)nh2=h2+2nRe(αh,Nh)+n2Nh2,n+,hH.

This yields (4.5). Since HN(N), we infer from (4.5) that Tα is not power bounded. A simple example of this kind is H=2, N=[0100] and α=1. In this particular case T1>1 without rescaling.

The research of the first and second authors was supported by the National Science Center (NCN) Grant OPUS No. DEC-2021/43/B/ST1/01651. The research of the third author was supported by Basic Science Research Program through the National Research Foundation of Korea (NRF) funded by the Ministry of Education (NRF-2021R111A1A01043569).

1) We refer the reader to [15, 16] for a discussion of the different types of stability of operators.

2) A more detailed discussion of this issue can be found in [16, Proposition 6.22].

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