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# Open Mathematics

### formerly Central European Journal of Mathematics

Editor-in-Chief: Gianazza, Ugo / Vespri, Vincenzo

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Volume 16, Issue 1

# Some remarks on spectra of nuclear operators

Oleg I. Reinov
Published Online: 2018-04-26 | DOI: https://doi.org/10.1515/math-2018-0043

## Abstract

We give criteria for the spectra of some nuclear operators in subspaces of quotients of Lp-spaces to be central-symmetric, as well as for the spectra of linear operators in Banach spaces to be Zd-symmetric in the sense of B. Mityagin. Also, we present a short proof of a corresponding Mityagin’s theorem.

MSC 2010: 47B06; 47B10

## 1 Introduction

It was shown by M. I. Zelikin in [1] that the spectrum of a nuclear operator in a separable Hilbert space is central-symmetric if and only if the spectral traces of all odd powers of the operator equal zero. Recall that the spectrum of every nuclear operator in a Hilbert space consists of non-zero eigenvalues of finite algebraic multiplicity, which have no limit point except possibly zero, and maybe zero. This system of all eigenvalues (written according to their multiplicities) is absolutely summable, and the spectral trace of any nuclear operator is, by definition, the sum of all its eigenvalues (taken according to their multiplicities).

The space of nuclear operators in a Hilbert space may be defined as the space of all trace-class operators (see [2, p. 77]); in this case we speak about the “nuclear trace” of an operator). Trace-class operators in a Hilbert space can be considered also as the elements of the completion of the tensor product of the Hilbert space and its Banach dual with respect to the greatest crossnorm on this tensor product [2, p. 119]. The well-known Lidskiǐ theorem [3] says that the nuclear trace of any nuclear operator in a Hilbert space (or, what is the same, of the corresponding tensor element) coincides with its spectral trace. Thus, Zelikin’s theorem [1] can be reformulated in the following way: the spectrum of a nuclear operator in a separable Hilbert space is central-symmetric if and only if the nuclear traces of all odd powers of the corresponding tensor element are zero.

One of the aim of our notes is to give an exact generalization of this result to the case of tensor elements of so-called s-projective tensor products of subspaces of quotients of Lp(μ)-spaces. In particular, we get as a consequence Zelikin’s theorem (taking p = 2).

Another problem which is under consideration in our notes is concentrated around the so-called ℤd-symmetry of the spectra of the linear operators. The notion of ℤd-symmetry of the spectra was introduced by B. S. Mityagin in a preprint [4] and in his paper [5]. He is interested there in a generalization of the result from [1] in two directions: to extend Zelikin’s theorem to the case of general Banach spaces and to change the property of a compact operator to have central-symmetric spectrum to have ℤd-symmetric spectrum. Roughly speaking, ℤd-symmetry of a spectrum of a compact operator T means that for any non-zero eigenvalue λ of T the spectrum contains also as eigenvalues of the same algebraic multiplicities all “d-shifted”, numbers tλ for $\begin{array}{}t\in \sqrt[d]{1}.\end{array}$ B. S. Mityagin has obtained a very nice result, showing that the spectrum of a compact operator T in an arbitrary Banach space, some power Tm of which is nuclear, is ℤd-symmetric if and only if for all large enough integers of type kd + r (0 < r < d) the nuclear traces of Tkd+r are zero. We present some thoughts around this theorem, giving, in particular, a short (but not so elementary as in [4, 5]) proof for the case where the operator is not necessarily compact. Let us mention, however, that the proof from [4, 5] can be adapted for this situation too.

Some words about the content of the paper.

In Section 2, we introduce some notation, definitions and terminology in connection with so-called s-projective tensor products, s-nuclear operators and the approximation properties of order s, s ∈ (0, 1]. We formulate here two auxiliary assertions from the paper [6]; they give us generalized Grothendieck-Lidskiǐ trace formulas which will be useful in the next section.

Section 3 contains an exact generalization of Zelikin’s theorem. In this section we present a criterion for the spectra of s-nuclear operators in subspaces of quotients of Lp-spaces to be central-symmetric.

Results of Section 4 show that the criterion of the central symmetry, obtained in the previous section, is optimal. In particular, we present here (Theorem 4.1) sharp examples of s-nuclear operators T in the spaces lp, 1 ≤ p ≤ + ∞, p ≠ 2, for which trace T = 1 and T2 = 0.

Finally, Section 5 is devoted to the study of Mityagin’s ℤd-symmetry of the spectra of linear operators. Our aim here is to give a short (but using the Fredholm Theory) proof of Mityagin’s theorem [4, 5] for arbitrary linear continuous (Riesz) operators. Firstly, we consider a ℤ2 situation (central symmetry) to clarify an idea which is to be used then in the general case. We finish the paper with a short proof of the theorem from [4, 5] for continuous (not necessarily compact) operators and with some simple examples of applications.

## 2 Preliminaries

By X, Y, … we denote the Banach spaces, L(X, Y) is a Banach space of all linear continuous operators from X to Y; L(X): = L(X, X). For a Banach dual to a space X we use the notation X*. If xX and x′ ∈ X, then 〈x′, x〉 denotes the value x′(x).

By X*⊗̂X we denote the projective tensor product of the spaces X* and X [7] (see also [8, 9]). It is a completion of the algebraic tensor product X*X (considered as a linear space of all finite rank continuous operators w in X) with respect to the norm

$||w||∧:=inf{∑k=1N||xk′||||xk||:w=∑k=1Nxk′⊗xk}.$

Every element u of the projective tensor product X*⊗̂X can be represented in the form

$u=∑iλixi′⊗xi,$

where (λi) ∈ l1 and $\begin{array}{}||{x}_{i}^{\prime }||\le 1,\end{array}$xi∥ ≤ 1 [7].

More generally, if 0 < s ≤ 1, then X*⊗̂s X is a subspace of the projective tensor product, consisting of the tensor elements u, uX*⊗̂X, which admit representations of the form $\begin{array}{}u=\sum _{k=1}^{\mathrm{\infty }}{x}_{k}^{\prime }\otimes {x}_{k},\end{array}$ where $\begin{array}{}\left({x}_{k}^{\prime }\right)\end{array}$X*, (xk) ⊂ X and $\begin{array}{}\sum _{k=1}^{\mathrm{\infty }}||{x}_{k}^{\prime }|{|}^{s}\phantom{\rule{thinmathspace}{0ex}}||{x}_{k}|{|}^{s}<\mathrm{\infty }\end{array}$ [7, 8, 9]. Thus, X*⊗̂X = X*⊗̂1X.

On the linear space X*X, a linear functional “trace” is defined in a natural way. It is continuous on the normed space (X*X, ∥·∥) and has the unique continuous extension to the space X*⊗̂X, which we denote by trace.

Every tensor element u, uX*⊗̂X, of the form $\begin{array}{}u=\sum _{k=1}^{\mathrm{\infty }}{x}_{k}^{\prime }\otimes {x}_{k}\end{array}$ generates naturally an operator : XX, (x) := $\begin{array}{}\sum _{k=1}^{\mathrm{\infty }}〈{x}_{k}^{\prime },x〉\phantom{\rule{thinmathspace}{0ex}}{x}_{k}\end{array}$ for xX. This defines a natural mapping j1 : X*⊗̂XL(x). The operators, lying in the image of this map are called nuclear [7, 10]. More generally, if 0 < s ≤ 1, u = $\begin{array}{}\sum _{k=1}^{\mathrm{\infty }}{x}_{k}^{\prime }\otimes {x}_{k}\end{array}$ and $\begin{array}{}\sum _{k=1}^{\mathrm{\infty }}||{x}_{k}^{\prime }|{|}^{s}\phantom{\rule{thinmathspace}{0ex}}||{x}_{k}|{|}^{s}<\mathrm{\infty },\end{array}$ then the corresponding operator is called s-nuclear [11, 12]. By js we denote a natural map from X*⊗̂sX to L(x). We say that a space X has the approximation property of order s, 0 < s ≤ 1 (the APs), if the canonical mapping js is one-to-one [11, 12]. Note that the AP1 is exactly the approximation property AP of A. Grothendieck [2, 8]. Classical spaces, such as Lp(μ) and C(K), have the approximation property. If a space X has the APs, then we can identify the tensor product X*⊗̂sX with the space Ns(x) of all s-nuclear operators in X (i.e. with the image of this tensor product under the map js). In this case for every operator TNs(x) = X*⊗̂sX the functional trace T is well defined and called the nuclear trace of the operator T.

It is clear that if a Banach space has the approximation property, then it has all the properties APs, s ∈ (0, 1]. Every Banach space has the property AP2/3 (A. Grothendiek [7], see also [11]). Since each Banach space is a subspace of an L(μ)-space, the following fact (to be used below) is a generalization of the mentioned result of A. Grothendieck:

#### Lemma 2.1

([6, Corollary 10]). Let s ∈ (0, 1], p ∈ [1, ∞] and 1/s = 1+|1/p – 1/2|. If a Banach space Y is isomorphic to a subspace of a quotient (or to a quotient of a subspace) of some Lp(μ)-space, then it has the APs.

Thus, for such spaces we have an equality Y*⊗̂sY = Ns(Y) and the nuclear trace of any operator TNs(Y) is well defined.

We will need also the following auxiliary assertion (the first part of which is a consequence of the previous lemma).

#### Lemma 2.2

([6, Theorem 1]). Let Y be a subspace of a quotient (or a quotient of a subspace) of some Lp(μ)-space, 1 ≤ p ≤ ∞. If TNs(Y), where 1/s = 1 + |1/2 – 1/p|, then

1. the nuclear trace of the operator T is well defined,

2. $\begin{array}{}\sum _{n=1}^{\mathrm{\infty }}|{\lambda }_{n}\left(T\right)|<\mathrm{\infty },\end{array}$ wheren(T)} is the system of all eigenvalues of the operator T (written according to their algebraic multiplicities), and

$traceT=∑n=1∞λn(T).$

Following [1], we say that a spectrum of a compact operator in a Banach space is central-symmetric, if for each of its eigenvalue λ the number –λ is also its eigenvalue and of the same algebraic multiplicity. We shall use the same terminology in the case of operators, all non-zero spectral values which are eigenvalues of finite multiplicity and have no limit point except possibly zero; the corresponding eigenvalue sequence for such an operator T will be denoted by sp (T); thus it is an unordered sequence of all eigenvalues of T taken according to their multiplicities.

## 3 On central symmetry

Let us note firstly that the theorem of Zelikin (in the form as it was formulated in [1]) can not be extended to the case of general Banach spaces, even if the spaces have the Grothendieck approximation property.

#### Example 3.1

Let U be a nuclear operator in the space l1, constructed in [10, Proposition 10.4.8]. This operator has the property that trace U = 1 and U2 = 0. Evidently, the spectrum of this operator is {0}. Let us note that the operator is not only nuclear, but also belongs to the space Ns(l1) for all s ∈ (2/3, 1]. It is not possible to present such an example in the case of 2/3-nuclear operators (see Corollary 3.6 below). Note also that, however, the traces of all operators Um, m = 2, 3, …, (in particular, U2n–1) are equal to zero.

#### Remark 3.2

For every nuclear operator T : XX and for any natural number n > 1, the nuclear trace of Tn is well defined (see [7, Chap. II, Cor. 2, p. 16]) and equals the sum of all its eigenvalues (according to their multiplicity) [7, Chap. II, Cor. 1, p. 15]. Therefore, if the spectrum of a nuclear operator T : XX is central-symmetric, then for each odd m = 3, 5, 7, … the nuclear trace of the operator Tm is equal to zero. This follows from the fact that the eigenvalue sequences of T and Tm can be arranged in such a way thatn(T)m} = {λn(Tm)} (see, e.g., [12, 3.2.24, p. 147]).

Let us formulate and prove the central result of this section.

#### Theorem 3.3

Let Y be a subspace of a quotient (or a quotient of a subspace) of an Lp-space, 1 ≤ p ≤ ∞, and uY*⊗̂sY, where 1/s = 1 + |1/2 – 1/p|, The spectrum of the operator u͠ is central-symmetric if and only if trace u2n–1 = 0, n = 1,2,….

#### Proof

If the spectrum of is central-symmetric, then, by Lemma 2.2, trace u = trace $\begin{array}{}\stackrel{~}{u}=\sum _{n=1}^{\mathrm{\infty }}{\lambda }_{n}\left(T\right)=0;\end{array}$ also, by Remark 3.2, trace $\begin{array}{}{u}^{m}=\sum _{n=1}^{\mathrm{\infty }}{\lambda }_{n}\left({T}^{m}\right)=0\end{array}$ for m = 3, 5, ….

To prove the converse, we need some information from the Fredholm Theory. Let u be an element of the projective tensor product X*⊗̂X, where X is an arbitrary Banach space. Recall that the Fredholm determinant det (1 – zu) of u (see [7, Chap. II, p. 13] or [10, 12, 13]) is an entire function

$det(1−zu)=1−ztraceu+⋯+(−1)nznαn(u)+…,$

all zeros of which are exactly (according to their multiplicities) the inverses of nonzero eigenvalues of the operator , associated with the tensor element u. By [7, Chap. II, Cor. 2, pp. 17-18], this entire function is of the form

$det(1−zu)=e−ztraceu∏i=1∞(1−zzi)ezzi,$

where {zi = λi()} is a system of all eigenvalues of the operator (written according to their algebraic multiplicities). Also, there exists a δ > 0 such that for all z, |z| ≤ δ, we have

$det(1−zu)=exp(−∑n=1∞1nzntraceun)$(1)

(see [13, p. 350]; cf. [14, Theor. I.3.3, p. 10]).

Now, let uY*⊗̂sY be as in the formulation of our theorem and suppose that trace u2n–1 = 0, n = 1, 2, …. By (1), we get: for a neighborhood U = U(0) of zero in ℂ, det (1 – zu) = det (1 + zu) for zU. Therefore, the entire function det (1 – zu) is even. By definition of det (1 – zu), the sequence of zeros of this function is exactly the sequence of inverses of nonzero eigenvalues of . Hence, the spectrum of is central-symmetric. □

Since under the conditions of Theorem 3.3 the space Y has the APs, the tensor product Y*⊗̂sY can be identified naturally with the space of all s-nuclear operators in Y. Hence, the statement of Theorem 3.3 may be reformulated in the following way:

#### Corollary 3.4

Let s ∈ [2/3, 1], p ∈ [2, ∞], 1/s = 1 + |1/2 – 1/p|, Y be a subspace of a quotient (or a quotient of a subspace) of an Lp-space, T be an s-nuclear operator in Y. The spectrum of T is central-symmetric if and only if trace T2n–1 = 0, nN.

#### Corollary 3.5

([1]). The spectrum of a nuclear operator T, acting on a Hilbert space, is central-symmetric if and only if trace T2n–1 = 0, nN.

For a proof, it is enough to apply Theorem 3.3 for the case p = 2.

#### Corollary 3.6

The spectrum of a 2/3-nuclear operator T, acting on an arbitrary Banach space, is central-symmetric if and only if trace T2n–1 = 0, nN.

For a proof, it is enough to apply Theorem 3.3 for the case p = ∞, taking into account the fact that every Banach space is isometric to a subspace of an L(μ)-space.

In connection with Corollary 3.6, let us pay attention again to the nuclear operator from Example 3.1.

## 4 Sharpness of results of Section 3

Now we will show that the statement of Theorem 3.3 is sharp and that the exponent s is optimal if p is fixed (if of course p ≠ 2, i.e. s ≠ 1).

Consider the case 2 < p ≤ ∞. In this case 1/s = 1 + |1/2 – 1/p| = 3/2 – 1/p. In a paper of the author [9, Example 2] the following result was obtained (see a proof in [9]):

(⋆) Let r ∈ [2/3, 1), p ∈ (2, ∞], 1/r = 3/2 – 1/p. There exist a subspace Yp of the space lp (c0 if p = ∞) and a tensor element $\begin{array}{}{w}_{p}\in {Y}_{p}^{\ast }{\stackrel{^}{\otimes }}_{1}{Y}_{p}\end{array}$ such that $\begin{array}{}{w}_{p}\in {Y}_{p}^{\ast }{\stackrel{^}{\otimes }}_{s}{Y}_{p}\end{array}$ for every s > r, trace wp = 1, p = 0 and the space Yp (as well as $\begin{array}{}{Y}_{p}^{\ast }\end{array}$) has the APr (but evidently does not have the APs if 1 ≥ s > r). Moreover, this element admits a nuclear representation of the form

$wp=∑k=1∞μkxk′⊗xk,where||xk′||=||xk||=1,∑k=1∞|μk|s<∞∀s>r.$

Evidently, we have for a tensor element u := wp from the assertion (⋆): trace u = 1 and the spectrum of the operator equals {0}.

The case where 2 < p ≤ ∞ can be considered analogously (with an application of the assertion (⋆) to a “transposed” tensor element $\begin{array}{}{w}_{p}^{t}\in {Y}_{p}{\stackrel{^}{\otimes }}_{1}{Y}_{p}^{\ast }.\right)\end{array}$

As was noted above (Example 3.1), there exists a nuclear operator U in l1 such that U2 = 0 and trace U = 1. The following theorem is an essential generalization of this result and gives us the sharpness of the statement of Corollary 3.4 (even in the case where Y = lp).

#### Theorem 4.1

Let p ∈ [1, ∞], p ≠ 2, 1/r = 1 + |1/2 – 1/p|. There exists a nuclear operator V in lp (in c0 if p = ∞) such that

1. VNs(lp) for each s ∈ (r, 1];

2. VNr(lp);

3. trace V = 1 and V2 = 0.

#### Proof

Suppose that p > 2. Consider the tensor element w := wp from the assertion (⋆) and its representation w = $\begin{array}{}w=\sum _{k=1}^{\mathrm{\infty }}{\mu }_{k}\phantom{\rule{thinmathspace}{0ex}}{x}_{k}^{\prime }\otimes {x}_{k},\end{array}$ where $\begin{array}{}||{x}_{k}^{\prime }||\end{array}$ = ∥xk∥ = 1 and $\begin{array}{}\sum _{k=1}^{\mathrm{\infty }}|{\mu }_{k}{|}^{s}<\mathrm{\infty }\end{array}$ for each s, s > r. Let l : Y : = Yplp be the identity inclusion. Let $\begin{array}{}{y}_{k}^{\prime }\end{array}$ be an extension of the functional $\begin{array}{}{x}_{k}^{\prime }\end{array}$ (k = 1, 2, …) from the subspace Y to the whole space lp with the same norm and set $\begin{array}{}v:=\sum _{k=1}^{\mathrm{\infty }}{\mu }_{k}\phantom{\rule{thinmathspace}{0ex}}{y}_{k}^{\prime }\otimes l\left({x}_{k}\right).\end{array}$ Then vlp′⊗̂slp (1/p + 1/p′ = 1) for each s ∈ (r, 1], trace v = $\begin{array}{}\sum {\mu }_{k}\phantom{\rule{thinmathspace}{0ex}}〈{y}_{k}^{\prime },l\left({x}_{k}\right)〉=1\end{array}$ and (lp) ⊂ l(Y) ⊂ lp. On the other hand, we have a diagram:

$Y→llp→v~0Y→llp→v~0Y→v~0lp,$

where 0 is an operator generated by , = lv͠0 and 0l = = 0. Put V := . Clearly, trace V = 1 and the spectrum sp V2 = {0}. Let us note that VNr(lp) (by Lemma 2.2). If p ∈ [1, 2), then it is enough to consider the adjoint operator. □

It follows from Theorem 4.1 that the assertion of Corollary 3.4 is optimal already in the case of the space Y = lp (which, by the way, has the Grothendieck approximation property).

## 5 Generalizations: around Mityagin’s theorem

Recall that if TL(x) and, for some m ∈ ℕ, Tm is a Riesz operator (see, e.g., [15, p. 943] for a definition), then T is a Riesz operator too (see, e.g., [12, 3.2.24, p. 147]). In particular, if Tm is compact, then all non-zero spectral values λ(T) ∈ sp (T) are eigenvalues of finite (algebraic) multiplicity and have no limit point except possibly zero. Also, in this case the eigenvalue sequences of T and Tm can be arranged in such a way that {λn(T)m} = {λn(Tm)} (see [12, 3.2.24, p. 147]). Recall that in this case we denote by sp (T) (resp., by sp (Tm) the sequence {λn(T)m} (resp., {λn(Tm)}).

We are going to present a short proof of the theorem of B. Mityagin from [4, 5]. To clarify our idea of the proof, let us consider firstly the simplest case where d = 2.

#### Theorem 5.1

Let X be a Banach space and TL(x). Suppose that some power of T is nuclear. The spectrum of T is central-symmetric if and only if there is an integer K ≥ 0 such that for every l > K the value trace Tl is well defined and trace T2l+1 = 0 for all l > K.

#### Proof

Suppose that TL(x) and there is an M ∈ ℕ so that TMN(x). Fix an odd N0, N0 > M, with the property that TN0N2/3(x) (it is possible since a product of three nuclear operators is 2/3-nuclear) and trace TN0+2k = 0 for all k = 0, 1, 2, …. By Corollary 3.6, the spectra of all TN0+2k are central-symmetric (since, e.g., trace TN0 = trace (TN0)3 = trace (TN0)5 = … = 0 by assumption). Assume that the spectrum of T is not central-symmetric. Then there exists an eigenvalue λ0 ∈ sp (T) such that –λ0 ∉ sp (T).

Now, $\begin{array}{}{\lambda }_{0}^{{N}_{0}}\end{array}$ ∈ sp (TN0), so $\begin{array}{}-{\lambda }_{0}^{{N}_{0}}\end{array}$ ∈ sp (TN0). Hence, there exist μN0 ∈ sp (T) and θN0 so that |θN0| = 1, $\begin{array}{}{\mu }_{{N}_{0}}^{{N}_{0}}=-{\lambda }_{0}^{{N}_{0}}\end{array}$ and μN0 = θN0 λ0, θN0 ≠ –1. Analogously, $\begin{array}{}{\lambda }_{0}^{{N}_{0}+2}\in \mathrm{sp}\phantom{\rule{thinmathspace}{0ex}}\left({T}^{{N}_{0}+2}\right),\text{so}-{\lambda }_{0}^{{N}_{0}+2}\in \mathrm{sp}\phantom{\rule{thinmathspace}{0ex}}\left({T}^{{N}_{0}+2}\right).\end{array}$ Hence, there exist μN0+2 ∈ sp (T) and θN0+2 so that |θN0+2| = 1, $\begin{array}{}{\mu }_{{N}_{0}+2}^{{N}_{0}+2}=-{\lambda }_{0}^{{N}_{0}+2}\end{array}$ and μN0+2 = θN0+2λ0, θN0+2 ≠ –1 etc. By induction we get the sequences $\begin{array}{}\left\{{\mu }_{{N}_{0}+2k}{\right\}}_{k=0}^{\mathrm{\infty }}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\text{and}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\left\{{\theta }_{{N}_{0}+2k}{\right\}}_{k=0}^{\mathrm{\infty }}\end{array}$ with the properties that μN0+2k ∈ sp (T), |θN0+2k| = 1, $\begin{array}{}{\mu }_{{N}_{0}+2k}^{{N}_{0}+2k}=-{\lambda }_{0}^{{N}_{0}+2k}\end{array}$ and μN0+2k = θN0+2k λ0, θN0+2k ≠ –1. Since μN0+2k ∈ sp (T) and |μN0+2k| = |λ0| > 0, the sequence {μN0+2k is finite as a set, i.e., we have that μN0+2K = μN0+2K+2 = … for some K > 1. It follows that θN0+2K = θN0+2K+2 = …. But $\begin{array}{}{\theta }_{{N}_{0}+2k}^{{N}_{0}+2k}=-1\end{array}$ for all k. Thus $\begin{array}{}{\theta }_{{N}_{0}+2K}^{l}=-1\end{array}$ for every odd lN0 + 2K. Therefore, θN0+2K = –1. A contradiction. □

Now we are going to consider a general case of a notion of ℤd-symmetry of a spectra, introduced and investigated by B. Mityagin in [4, 5]. Let T be an operator in X, all non-zero spectral values which are eigenvalues of finite multiplicity and have no limit point except possibly zero. Recall that we denote by sp (T) the corresponding unordered eigenvalue sequence for T (possibly, including zero). For a fixed d = 2, 3, … and for the operator T, the spectrum of T is called ℤd-symmetric, if λ ∈ sp (T) implies tλ ∈ sp (T) for every $\begin{array}{}t\in \sqrt[d]{1}.\end{array}$

Let r ∈ (0, ∞], 𝔻 := {z ∈ ℂ: |z| < r}, f : 𝔻 → ℂ. and d ∈ ℕ ∖ {1}. We say that f is d-even if f(tz) = f(z) for every $\begin{array}{}t\in \sqrt[d]{1}.\end{array}$

#### Lemma 5.2

Let Φ(X) be a linear subspace of X*⊗̂X of spectral type l1, i.e., for every vΦ(X) the seriesλ∈sp() |λ| is convergent. Let d ∈ ℕ, d > 1. If uΦ(X), then the Fredholm determinant det (1 – zu) is d-even if and only if the eigenvalue sequence of isd-symmetric if and only if trace ukd+r = 0 for all k = 0, 1, 2, … and r = 1, 2, …, d – 1.

#### Proof

If the function det (1 – zu) is d-even, then the eigenvalue sequence of is ℤd-symmetric, since this sequence coincides with the sequence of inverses of zeros of det (1 – zu) (according to their multiplicities).

If the eigenvalue sequence of is ℤd-symmetric, then trace u = ∑λ∈sp (u͠) λ = 0 (since Φ(X) is of spectral type l1 and $\begin{array}{}\sum _{t\in \sqrt[d]{1}}t=0\right).\end{array}$ Also, by the same reason trace ukd+r = 0 for all k = 0, 1, 2, … and r = 1, 2, …, d – 1, since the spectrum of l is absolutely summable for every l ≥ 2 and we may assume that {λm(l)} = {λm()l} (for every fixed l) [12, 3.2.24, p. 147].

Now, let trace ukd+r = 0 for all k = 0, 1, 2, … and r = 1, 2, …, d – 1. By (1), det (1 – zu) = $\begin{array}{}\mathrm{exp}\left(-\sum _{m=1}^{\mathrm{\infty }}\frac{1}{md}\phantom{\rule{thinmathspace}{0ex}}{z}^{md}\mathrm{trace}\phantom{\rule{thinmathspace}{0ex}}{u}^{md}\right)\end{array}$ in a neighborhood U of zero. Therefore, this function is d-even in the neighborhood U. By the uniqueness theorem det (1 – zu) is d-even in ℂ. □

#### Corollary 5.3

For any Banach space X and for every uX*⊗̂2/3X the conclusion of Lemma 5.2 holds.

#### Corollary 5.4

Let Y be a subspace of a quotient of an Lp-space, 1 ≤ p ≤ ∞. For any uY*⊗̂sY, where 1/s = 1 + |1/2 – 1/p|, the conclusion of Lemma 5.2 holds.

Now we are ready to present a short proof of the theorem of B. Mityagin [4, 5]. Note that the theorem in [4, 5] is formulated and proved for compact operators, but the proof from [4, 5] can be easily adapted for the general case of linear operators.

#### Theorem 5.5

Let X be a Banach space and TL(X). Suppose that some power of T is nuclear. The spectrum of T isd-symmetric if and only if there is an integer K ≥ 0 such that for every l > Kd the value trace Tl is well defined and trace Tkd+r = 0 for all k = K, K + 1, K + 2, … and r = 1, 2, …, d – 1.

#### Proof

Fix N0 ∈ ℕ such that TN0 is 2/3-nuclear (it is possible by a composition theorem from [7, Chap, II, Theor. 3, p. 10]). Note that, by A. Grothendieck, the trace of Tl is well defined for all lN0.

Suppose that the spectrum of T is ℤd-symmetric. Take an integer l : = kd + rN0 with 0 < r < d. Since the spectrum of Tl is absolutely summable, trace Tl = ∑λ∈sp (Tl) λ and we may assume that {λm(Tl) = {λm(T)l}, we get that trace Tkd+r = 0.

In proving the converse, we may (and do) assume that Kd > N0. Consider an infinite increasing sequence {pm} of prime numbers with p1 > (K + 1)d. Assuming that trace Tkd+r = 0 if k = K, K + 1, K + 2, … and r = 1, 2, …, d – 1, for a fixed pm we get from Lemma 5.2 (more precisely, from Corollary 5.3) that the function det (1 – zTpm) is d-even. Suppose that the spectrum of T is not ℤd-symmetric. Then there exist an eigenvalue λ0 ∈ sp (T) and a root $\begin{array}{}\theta \in \sqrt[d]{1}\end{array}$ so that θλ0 ∉ sp (T). On the other hand, again by Lemma 5.2, the spectrum of Tpm is ℤd-symmetric. Since $\begin{array}{}{\lambda }_{0}^{{p}_{m}}\in \mathrm{sp}\phantom{\rule{thinmathspace}{0ex}}\left({T}^{{p}_{m}}\right),\end{array}$ there exists μm ∈ sp (T) such that $\begin{array}{}{\mu }_{m}^{{p}_{m}}={\theta }^{{p}_{m}}{\lambda }_{0}^{{p}_{m}}\in \mathrm{sp}\phantom{\rule{thinmathspace}{0ex}}\left({T}^{{p}_{m}}\right);\end{array}$ hence, μm = θm λ0 for some θm with |θm| = 1. But |λ0| > 0. Therefore, the set {μm} is finite and it follows that there is an integer M > 1 such that θM = θM+1 = θM+2 = …. Hence, $\begin{array}{}{\theta }^{{p}_{m}}={\theta }_{M}^{{p}_{m}}\end{array}$ for all mM. Thus, θM = θ. A contradiction. □

Let us give some examples in which we can apply Theorem 5.5, but the main result of [4, 5] does not work.

#### Example 5.6

Let Πp be the ideal of absolutely p-summing operators (p ∈ [1, ∞); see [10] for a definition and related facts). Then for some n one has $\begin{array}{}{\mathit{\Pi }}_{p}^{n}\subset N.\end{array}$ In particular, $\begin{array}{}{\mathit{\Pi }}_{2}^{2}\left(C\left[0,1\right]\right)\subset N\left(C\left[0,1\right]\right),\end{array}$ but not every absolutely 2-summing operator in C[0, 1] is compact. Another interesting example: $\begin{array}{}{\mathit{\Pi }}_{3}^{3}\end{array}$ is of spectral type l1 [16]. We do not know (maybe it is unknown to everybody), whether the finite rank operators are dense in this ideal. However, Theorem 5.5 may be applied. Moreover, it can be seen that, for example, the spectrum of an operator T from $\begin{array}{}{\mathit{\Pi }}_{3}^{3}\end{array}$ is central-symmetric if and only if the spectral traces of the operators T2k–1 are zero for all k > 0.

## Acknowledgement

I would like to express my gratitude to B. S. Mityagin for a two-days helpful discussion on the topic of this paper and on the problems related to the notion of nuclear operators at Aleksander Pełczyński Memorial Conference (13 July – 19 July, 2014, Będlewo, Poland) as well as for very useful information which B. S. Mityagin gave me in letters. Also, I am very grateful to him for drawing my attention to the paper of M. I. Zelikin [1].

The author would like to thank the referee for helpful remarks.

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Accepted: 2018-03-27

Published Online: 2018-04-26

Citation Information: Open Mathematics, Volume 16, Issue 1, Pages 453–460, ISSN (Online) 2391-5455,

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