Abstract
We study variants of s-numbers in the context of multilinear operators. The notion of an \(s^{(k)}\)-scale of k-linear operators is defined. In particular, we shall deal with multilinear variants of the \(s^{(k)}\)-scales of the approximation, Gelfand, Hilbert, Kolmogorov and Weyl numbers. We investigate whether the fundamental properties of important s-numbers of linear operators are inherited to the multilinear case. We prove relationships among some \(s^{(k)}\)-numbers of k-linear operators with their corresponding classical Pietsch’s s-numbers of a generalized Banach dual operator, from the Banach dual of the range space to the space of k-linear forms, on the product of the domain spaces of a given k-linear operator.
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1 Introduction
A natural question that appears in Functional Analysis and theory of operators is whether there are variants of some classical properties of linear operators in the setting of multilinear operators. We note that in many cases dealing with multilinear, instead of linear mappings, has proved to be a subtle subject and require different methods, and the proofs are more complicated. Moreover, many linear results are no longer true in the setting of multilinear operators. For instance, the linear Marcinkiewicz multiplier theorem, whose natural bilinear version fails, as shown by Grafakos and Kalton in [6]. Furthermore, in some questions, completely new definitions and new techniques need to be developed.
An axiomatic approach to s-numbers and Banach operator ideals of linear operators was developed by Pietsch [17]. There are many of these s-numbers sequences. The approximation numbers are the largest (under coordinate-wise ordering as sequences) s-numbers on Banach spaces. These numbers have proved to give a very useful information about the degree of compactness of operators acting between Banach spaces. The Gelfand and Kolmogorov numbers play an important role in the study of the local theory of Banach spaces. Pietsch [17, p. 220] shows that an operator T acting between Banach spaces is compact if and only if \(s_n(T)\rightarrow 0\) as \(n\rightarrow \infty \) for \((s_n)\) either the sequence of Gelfand numbers or the sequence of Kolmogorov numbers. The Weyl numbers, defined by Pietsch, are truly important in the study of Riesz operators in Banach spaces. We refer to [4, 7, 8, 10, 21, 23] for an in-depth study of s-numbers and their applications.
We point out that multilinear operators appear naturally in harmonic analysis and functional analysis, including the theory of ideals of operators in Banach spaces. In recent times singular multilinear operators have been intensively studied, including the famous bilinear Hilbert transform (see [6, 9]).
In 1983 Pietsch [20] proposed and sketched a theory of ideals and s-numbers of multilinear functionals. In his work, he wrote "Therefore this paper should be mainly considered as a research program for the future". Pietsch’s motivation was the extension of the theory of s-numbers and operator ideals to the nonlinear case. Pietsch’s paper is the origin of studies of several authors on some classes of ideals of multilinear operators. We refer to the study of such ideals to [2, 11, 12] and also to the survey paper [16] and references therein.
While the properties of s-numbers of linear operators are well-known, Pietsch’s research problem on a theory of s-numbers of multilinear operators had not been studied extensively. This is the main motivation to continue the study of variants of s-numbers in the context of multilinear operators initiated in the paper [5] and continued in [1] in the setting of polynomials acting between Banach spaces.
We introduce a notion of an \(s^{(k)}\)-scale of k-linear operators. Exploring ideas of Pietsch developed in the linear setting we investigate whether the fundamental properties of important s-numbers ideals of linear operators are inherited to the multilinear case. It should be noted that whereas the work is based on some ideas from the theory of s-numbers ideals of bounded linear operators, some proofs may be extended from the linear case to multilinear operators, and others require some new ideas and methods. The difficulty comes from the fact that, even in the bilinear case, the range or the kernel of a multilinear operator is not necessarily a linear subspace. In particular, as a consequence, the well-known relations between the dimensions of the kernel and the range in the linear case are not true in general, in the multilinear case.
The organization of the paper is as follows: in Sect. 2, we introduce a modified variant of the notion of s-numbers in the setting of k-linear operators. This notion is more general that the one given in [5]. We also introduce the notion of symmetric, injective, surjective \(s^{(k)}\)-scale, and also the mixed multiplicative property. In Sect. 3, we consider the \(s^{(k)}\)-scales of approximation numbers and as application, we study the approximation numbers of bilinear diagonal operators. In other sections, we study in detail multilinear variants of important classical Pietsch’s s-numbers which are shown in the of Contents of the paper.
2 s-Numbers of multilinear operators
We shall use the notation and terminology commonly used in Banach space theory. If X is a Banach space we denote by \(X^{*}\) its dual Banach space, and by \(U_X\), \({{\mathop {U}\limits ^{\circ }}}_X\), \(S_X\) the closed, the open unit ball and the unit sphere of X, respectively. As usual, \(J_X :X \rightarrow \ell _{\infty }(U_{X^{*}})\) denotes the metric injection defined by \(J_X(x) := (x^{*}(x))_{x^* \in U_{X^{*}}}\) with values in the space \(\ell _\infty (U_{X^{*}})\) of bounded sequences. The canonical embedding of X to the bidual \(X^{**}\) of X is denoted by \(\kappa _X\). \({\mathbb {K}}\) denotes the field of all scalars (complex or real) and \({\mathbb {N}}\) the set of all positive integers and for each \(n\in {\mathbb {N}}\), \([n]:= \{1, \ldots , n\}\). For \(1\le p\le \infty \) and a nonempty set \(\Gamma \), we let \(\ell _p(\Gamma )\) be the space of all p-summable sequences \((x_\gamma )_{\gamma \in \Gamma }\). If \(\Gamma = [n]\) (resp., \(\Gamma = {\mathbb {N}}\)) we use the standard notion \(\ell _p^n\) (resp., \(\ell _p\)) instead of \(\ell _p([n])\) (resp., \(\ell _p({\mathbb {N}})\)). The standard unit vector basis in \(c_0\) is denoted by \((e_i)\). If no confusion arises we also use \((e_i)_{i=1}^n\) to denote the unit basis vectors in \({\mathbb {R}}^n\). The product \(X_1 \times \cdots \times X_k\) of Banach spaces is equipped with the standard norm \(\Vert (x_j)\Vert := \max _{1\le j\le k}\Vert x_j\Vert _{X_j}\), for all \((x_j) \in X_1 \times \cdots \times X_k\). The Banach space of all continuous k-linear mappings T from \(X_1 \times \cdots \times X_k\) to a Banach space Y is endowed with the norm
and it is denoted by \({\mathcal {L}}(X_1,\ldots , X_k; Y)\). As usual, in the case \(k=1\), the space of all bounded linear operators from \(X_1\) to Y is denoted by \({\mathcal {L}}(X_1; Y)\). If \(k\ge 2\) and \(Y = {\mathbb {K}}\), then we shorten by \({\mathcal {L}}(X_1, \ldots , X_k)\), which is the space of all k-linear forms. If in addition \(X_1 = \cdots = X_k =X\) (resp., \(Y={\mathbb {K}}\)), we write \({\mathcal {L}}(^kX; Y)\) (resp., \({\mathcal {L}}(^kX)\)).
From now on we assume that all spaces which will appear are Banach spaces. Let \(k \ge 2\) be an integer and let \(T\in {\mathcal {L}}(X_1, \ldots , X_k; Y)\). For each \(i\in [k]\), \(1<i \le k\), and every \(x^{i}:= (x_1,\ldots , x_{i-1}, x_{i+1},\ldots , x_k) \in \prod _{j=1, j\ne i}^{k} X_j\), the associated operator \(T_{x^i}\in {\mathcal {L}}(X_i; Y)\) is defined by
with an obvious modification for \(i=1\) and \(i=k\).
For every k-linear operator \(T \in {\mathcal {L}}(X_1, \ldots , X_{k}; Y)\), its rank is given by
where [A] denotes the linear subspace generated by a nonempty subset \(A \subset Y\).
Throughout the paper, for each \(n\in {\mathbb {N}}\), \(I_n:\ell _2^n \rightarrow \ell _2^n\) denotes the identity map. For each \(k, n \in {\mathbb {N}}\) we let \(\otimes _k I_n\) to denote the k-linear operator, from \(\prod _{i=1}^k \ell _2^n\) to \(\ell _2 ([n]^k)\), defined for all \(x_1=(x_1(j))_{j=1}^n, \ldots , x_k = (x_k(j))_{j=1}^n \in \ell _2^n\) by
where \(x_1\otimes \cdots \otimes x_k := ((x_1\otimes \cdots \otimes x_k)(j))_{j\in [n]^k}\) and
Now, we are ready to introduce a modified variant of the notion of s-numbers in the setting of k-linear operators, which appeared in [5].
A rule \(s^{(k)} = (s_n^{(k)}) :{\mathcal {L}}(X_1, \ldots , X_k; Y) \rightarrow [0, \infty )^{{\mathbb {N}}}\), assigning to every operator \(T\in {\mathcal {L}}(X_1, \ldots , X_k; Y)\) a non-negative scalar sequence \((s_n^{(k)}(T))\), is said to be an \(s^{(k)}\)-scale if the following conditions are satisfied:
(S1) Monotonicity: For every \(T\in {\mathcal {L}}(X_1, \ldots , X_k; Y)\),
(S2) Additivity: For every \(S, T\in {\mathcal {L}}(X_1, \ldots , X_k; Y)\),
(S3) Ideal-property: For every \(T\in {\mathcal {L}}(X_1, \ldots , X_k; Y),\, S\in {\mathcal {L}}(Y; Z)\), \(R_1 \in {\mathcal {L}}(W_1; X_1),\) \(\ldots , R_k \in {\mathcal {L}}(W_k; X_k)\),
(S4) Rank-property: For every \(T\in {\mathcal {L}}(X_1, \ldots , X_k; Y)\) with \(\text {rank}(T) < n\), \(s_n^{(k)}(T) =0\).
(S5) Norming property: For each \(n\in {\mathbb {N}}\) one has \(s_n^{(k)} (\otimes _k I_n) = 1\).
We note that in the setting of linear operators the property (S5) is equivalent to the original property defined by Pietsch [19],
where \(I_n:\ell _2^n \rightarrow \ell _2^n\) denotes the identity map.
Following the classical case in the linear setting, we call \(s^{(k)}_n(T)\) the n-th \(s^{(k)}\)-number of the k-linear operator T. To show the domain \(X_1 \times \cdots \times X_k\) and the range space Y, we write \(s_n^{(k)}(T:X_1 \times \cdots \times X_k \rightarrow Y)\).
We will use an obvious estimate, which follows from condition (S2):
Given an \(s^{(k)}\)-scale \((s_n^{(k)})\), we also introduce the following definitions:
-
(J1)
\((s_n^{(k)})\) is called injective if, given any metric injection \(J\in {\mathcal {L}}(Y; Z)\), that is, \(\Vert Jy\Vert = \Vert y\Vert \) for all \(y\in Y\), \(s_n^{(k)}(T) = s_n^{(k)}(JT)\) for all \(T\in {\mathcal {L}}(X_1, \ldots , X_k; Y)\) and all Banach spaces \(X_1,\ldots , X_k\).
-
(J2)
\((s_n^{(k)})\) is called injective in the strict sense if \(s_n^{(k)}(T) = s_n^{(k)}(J_Y T)\) for all \(T\in {\mathcal {L}}(X_1, \ldots , X_k; Y)\).
-
(S)
\((s_n^{(k)})\) is called surjective if, given any metric surjections \(Q_j \in {\mathcal {L}}(Y_j; X_j)\), (i.e., \(Q_j({\mathop {U}\limits ^{\circ }}_{Y_j}) = \, {\mathop {U}\limits ^{\circ }}_{X_j}\) for each \(j\in [k]\)), \(s_n^{(k)}(T) = s_n^{(k)}(T(Q_1, \ldots , Q_k))\) for all \(T\in {\mathcal {L}}(X_1, \ldots , X_k; Y)\) and any Banach space Y.
Now we are ready to give the following definition:
-
(M)
If \((s_n)\) is an s-scale and \((s^{(k)}_n)\) is an \(s^{(k)}\)-scale, then the pair \(((s_n), (s^{(k)}_n))\) has the mixed multiplicative property, if for any \(T\in {\mathcal {L}} (X_1, \ldots , X_k; Y)\) and \(S\in {\mathcal {L}}(Y; Z)\),
$$\begin{aligned} s_{m+n-1}^{(k)}(ST)\le s_m(S)\,s_n^{(k)}(T), \quad \, m, n \in {\mathbf {N}}. \end{aligned}$$A sequence \((s^{(k)})\) of \(s^{(k)}\)-scales is said to be multiplicative if, for each \(k\in {\mathbb {N}}\) the pair \(((s^{(1)}_n), (s^{(k)}_n))\) has the mixed multiplicative property.
Given \(T\in {\mathcal {L}}(X_1, \ldots , X_k; Y)\), we define the generalized adjoint (adjoint for short) operator \(T^{\times }:Y^{*} \rightarrow {\mathcal {L}}(X_1, \ldots , X_k)\) by
This operator was introduced in [24], where a variant of Schauder Theorem is proved, which states that T is compact if and only if \(T^{\times }\) is compact.
Note that, for any \(T\in {\mathcal {L}}(X_1, \ldots , X_k; Y)\) and \(S\in {\mathcal {L}}(Y; Z)\), we have
where \(S^{*}\) is the classical linear adjoint of S.
In what follows for a k-tuple \((X_1, \ldots , X_k)\) of Banach spaces, we define a k-linear mapping \({\widehat{\kappa }}_{X_1 \times \cdots \times X_k} :X_1\times \cdots \times X_k \rightarrow {\mathcal {L}}(X_1, \ldots , X_k)^{*}\) given by
Following Pietsch [19], we recall that in the setting of linear operators an s-scale \((s_n)\) is said to be symmetric (resp., fully symmetric) if, for every operator S, \((s_n(S))\ge (s_n(S^{*}))\) (resp., \((s_n(S)) = (s_n(S^{*}))\)).
Let \((s_n)\) be an s-scale and \((s^{(k)}_n)\) be an \(s^{(k)}\)-scale. The following notion is motivated by the above definition of Pietsch.
-
(S6)
A pair \((s, s^{(k)})\) is said to be symmetric (resp., fully symmetric) if, for every operator \(T\in {\mathcal {L}}(X_1, \ldots , X_k; Y)\) and each \(n\in {\mathbb {N}}\),
$$\begin{aligned} s_n(T^\times ) \le s_n^{(k)}(T) \quad \, (\text {resp.,}\, s_n(T^\times ) = s_n^{(k)}(T)). \end{aligned}$$
If \((s^{(k)})_{k=1}^{\infty }\) is a sequence of \(s^{(k)}\)-scales then, it is said to be symmetric (resp., fully symmetric) if for each \(k\in {\mathbb {N}}\) the pair \((s^{(1)}, s^{(k)})\) is symmetric (resp., fully symmetric).
Proposition 2.1
Suppose that an s-scale and \(s^{(k)}\)-scale form a symmetric pair \((s, s^{(k)})\). Then, for any k-linear operator with \(s^{(k)}_n(T) = 0\), we have \(\text {rank}\,(T) <n\).
Proof
By our hypothesis \(s_n(T^\times ) \le s_n^{(k)}(T)\) and so \(s_n(T^\times ) = 0\). Since \(T^\times \) is a linear operator, it follows that \(\text {rank}\,(T^\times ) < n\). Applying [5, Lemma 3.1], we conclude that \(\text {rank}(T^\times ) = \text {rank}(T)\) and the result follows. \(\square \)
Since the adjoint operator \(T^\times \) of a k-linear operator reflects some properties of T, as quoted before, and it is a linear operator, if \((s_n)\) is an s-scale defined for linear operators, then following [17], we define for every \(T \in {\mathcal {L}}(X_1, \ldots , X_k; Y)\) the numbers
In what follows \(s^{(\times k)}:=(s^{(\times k)}_n)\) is said to be an adjoint of s-scale, whenever the conditions (S1), (S2), (S4) and (S5) are satisfied, and in addition
\((S3^{\prime })\) \(s^{(\times k)}_n(ST) \le \Vert S\Vert s^{(\times k)}_n(T)\).
3 Approximation numbers
The n-th approximation number of a k-linear operator \(T\in {\mathcal {L}}(X_1, \ldots , X_k; Y)\) is defined by
In the study of these numbers and others \(s^{(k)}\)-scales, we will use without any references an obvious fact that, for an arbitrary operator \(S:X \rightarrow Y\) between Banach spaces such that \(\text {rank}\,(S)< \text {dim}\,(X)\), there exists \(x\in S_X\) with \(Sx=0\).
We prove that \((a^{(k)}_n)\) forms an \(s^{(k)}\)-scale which is called the \(a^{(k)}\)-scale of approximation numbers. To show this, we prove a preliminary result which we will also need later.
Proposition 3.1
Let \(T\in {\mathcal {L}}(X_1, \ldots , X_k; Y)\) with \(\text {rank}\,(T) \ge n\). Suppose that \(\Vert T\Vert =1\) and there exists \({x}^{i} \in \prod _{j=1,\,j\ne i}^{k} U_{X_j}\) such that the associate operator \(T_{{x}^i}\in {\mathcal {L}}(X_i; Y)\) is a metric injection. Then \(a_n^{{k}}(T) = 1\).
Proof
Clearly, \(a^{(k)}_n(T) \le \Vert T\Vert = 1\). We claim that \(a^{(k)}_n(T)=1\). Suppose, on the contrary that
Then, there exists a k-linear operator \(A:X_1 \times \cdots \times X_k \rightarrow Y\) with \(\text {rank}\,(A) < n\) such that \(\Vert T - A\Vert < 1\). In particular, we have
Since \(A_{{x}^i} :X_i \rightarrow Y\) is a bounded operator with
we can find \(x \in S_{X_i}\) such that \(A_{{x}^i}x=0\). Combining with the hypothesis that \(T_{{x}^i}:X_i \rightarrow Y\) is a metric injection yields
This is a contradiction which proves the claim. \(\square \)
As an application we obtain the following corollary.
Corollary 3.2
For each \(k, n\in {\mathbb {N}}\) and every \(1\le p\le \infty \), one has
Proof
For each \(k\in {\mathbb {N}}\) and for all \((x_1, \ldots , x_k) \in \ell _p^n \times \cdots \times \ell _p^n\) one has
This implies that, if we let \(e^i:= (e_1, \ldots , e_1) \in \prod _{j=1, j\ne i}^{k} U_{\ell ^n_p}\), then the associate operator
is a metric injection. Since \(\text {rank}\,(\otimes _k I_n)\ge n\), the required statement follows by Proposition 3.1. \(\square \)
In consequence we get the following statement. We omit the simple proof.
Lemma 3.3
For each k, the sequence \((a_n^{(k)})\) of approximation numbers is a fully symmetric \(s^{(k)}\)-scale.
We have an elementary multilinear variant of a very useful multiplicativity property of an arbitrary s-number sequence called mixing multiplicativity due to Carl [3], which states: If \((s^{(k)}_n)\) is an arbitrary \(s^{(k)}\)-scale, then for every \(T\in {\mathcal {L}}(X_1, \ldots , X_k; Y)\) and \(S\in {\mathcal {L}}(Y; Z)\), we have
that is, the pair \(((a_n), (s^{(k)}_n))\) has the mixed multiplicative property.
We note the following straightforward statement, which is a multilinear variant of a well known Pietsch’s result in the linear setting.
Lemma 3.4
For each \(k\in {\mathbb {N}}\), the sequence \((a^{(k)}_n)_{n=1}^{\infty }\) is the largest sequence of \(s^{(k)}\)-scales and satisfy, for all \(T\in {\mathcal {L}}(X_1, \ldots , X_k; Y)\) and \(S\in {\mathcal {L}}(Y; Z)\), the multiplicativity property:
We need to calculate the n-th approximation numbers of special k-linear operators on the product of finite dimensional \(\ell _p^n\)-spaces with \(1\le p<\infty \). Let \(n, r \in {\mathbb {N}}\) with \(r \le n^k\) and let \(\sigma _k :[n]^k \rightarrow [n^k]\) be a bijection, then \(I^{(r)}_{\sigma _k} :\prod _{i=1}^{k} \ell _p^n \rightarrow \ell ^r_p\) denotes the k-linear mapping defined by the formula:
where for each \(j\in [r]\) we let \(z_j:= x_1(i_1) \cdots x_k(i_k)\) for the unique \((i_1, \ldots , i_k)\in [n]^k\) such that \(\sigma _k(i_1, \ldots , i_k)=j\). Note that if \(k=1\) and \(\sigma _1 :[n] \rightarrow [n]\) is given by \(\sigma _1(j) := j\) for each \(j\in [n]\), then \(I_{\sigma _1}^{(n)}:\ell _p^n \rightarrow \ell _p^n\) is the identity map \(I_n\) on \(\ell _p^n\).
We claim that \(\Vert I^{(r)}_{\sigma _k}\Vert \le 1\). Clearly, the case \(k=1\) is trivial, so let \(k \ge 2\). A standard calculation shows that, for \(r=n^k\), one has
so \(\Vert I^{(n^k)}_{\sigma _k}\Vert = 1\). If \(r < n^k\), then for all \(x_1, \ldots , x_k \in \ell _p^n\), we have
and hence \(\Vert I^{(r)}_{\sigma _k}\Vert \le \Vert I^{(n^k)}_{\sigma _k}\Vert = 1\).
Since we are interested in the multilinear case, in what follows for each positive integer \(k\ge 2\) we will consider the standard bijection \(\sigma _k :[n]^k \rightarrow [n^k]\) given by the formula:
for each \((i_1, \ldots , i_k) \in [n]^k\). We will need the following lemma.
Lemma 3.5
Let \(k, n, r \in {\mathbb {N}}\) with \(k \ge 2\), \(n \le r\le n^k\) and let \(\sigma _k :[n]^k \rightarrow [n^k]\) be the standard bijection. Then, for every \(1\le p < \infty \) one has
Proof
Fix \(k \ge 2\). We claim that, for each \(y\in \ell _p^n\) one has
Indeed, we have that
where \(u_j = e_1 = (x_1(j),\ldots , x_n(j)) = (1,0,\ldots ,0) \in \ell _p^n\) for \(j \in [k-1]\), \(y = (y_1,\ldots , y_n)\) and for each \(\nu \in [r]\) with \(\nu = \sigma _k(i_1,\ldots ,i_k)\), \(z_{\nu }:= x_1(i_1) \ldots x_k(i_k)\). From definition of \(\sigma _k\), it follows that \(\sigma _k(1, \ldots , 1, i_k) = i_k\).
Since \(n\le r \le n^k\), the first n elements of \((z_1,\ldots , z_r)\) are \(y_1,\ldots , y_n\). Indeed, by the formula for the standard bijection \(\sigma _k\) it follows by \(1 = \sigma _k(1,1,\ldots ,1)\) that \(z_1 = 1 \cdot 1 \cdot 1 \cdots 1 \cdot y_1 = y_1\). We conclude in a similar fashion that by \(n = \sigma _k(1,1,\ldots , n)\), \(z_n = 1 \cdot 1 \cdot 1 \cdots 1 \cdot y_n = y_n\). For all other \(z_j\) in \((z_1,\ldots , z_r)\) with \(n < j \le r\), it follows by \(z_j:= x_1(i_1) \ldots x_k(i_k)\) that at least one factor in the product is zero. Hence
and this proves the claim. Thus, the statement follows by Proposition 3.1. \(\square \)
We calculate now the r-th approximation number of the bilinear diagonal operators on the products of \(\ell _p\)-spaces. Let \(r \in {\mathbb {N}}\) be fixed and let \(n \in {\mathbb {N}}\) be the least number such that \(r \le n^2\). Given the standard bijection \(\sigma _2 :[n] \times [n] \rightarrow [n^2]\), let \({\widetilde{\sigma }}_2:{\mathbb {N}}\times {\mathbb {N}}\longrightarrow {\mathbb {N}}\) be a bijection which is an extension of \(\sigma _2\) such that \({\widetilde{\sigma }}_2|_{[n] \times [n]} = \sigma _2\). We point out that the extension \({\widetilde{\sigma }}_2\) can be obtained using an inductive procedure, considering an arbitrary fixed bijection \(\beta :{\mathbb {N}}\times {\mathbb {N}}\rightarrow {\mathbb {N}}\). In what follows, we will use the bijection \({\widetilde{\sigma }}_2\) without any references.
Given a bounded sequence \(\lambda :=(\lambda _k)_{k=1}^{\infty }\), for each \(k\in {\mathbb {N}}\), we find a unique \((i_k, j_k)\in {\mathbb {N}}\times {\mathbb {N}}\) such that \(k={\widetilde{\sigma }}_2(i_k, j_k)\).
As usual we let \(\omega ({\mathbb {N}})\) to denote the space of all sequences modeled on \({\mathbb {N}}\). In what follows for a fixed bijection \({\widetilde{\sigma }}_2\) defined above, we define a mapping
where, for each \(k \in {\mathbb {N}}\), \((x {\tilde{*}} y)_k := x_{i_k} y_{j_k}\) with \({\widetilde{\sigma }}_2 (i_k, j_k) = k\). Given \(1\le p<\infty \) and a bounded sequence \(\lambda = (\lambda _{k})_{k=1}^\infty \), we define a mapping \(D_\lambda :\ell _p \times \ell _p \rightarrow \ell _p\) by
Let \(n\in {\mathbb {N}}\). Similarly as above, for each positive integer \(r\le n^2\), we define a bilinear operator \(D_{\lambda }^{(r)} :\ell _p^n \times \ell _p^n \rightarrow \ell ^r_p\) by
where, for each \(k\in [r]\) we let \((x {\tilde{*}} y)_k := x_{i_k} y_{j_k}\) with \({\widetilde{\sigma }}_2(i_k, j_k) = k\).
Given \(\alpha :=(\alpha _1, \ldots ,\alpha _r)\in {\mathbb {R}}^r\), with \(\alpha _j> 0\) for each \(j\in [r]\), we also define a linear operator \(R_\alpha ^{(r)} :\ell _p^r \rightarrow \ell _p^r\) by
Observe that for every \(x, y \in \ell _p^n\), one has
In what follows, we will use the following standard finite dimensional operators \(J_n:\ell _p^n \rightarrow \ell _p\) and \(Q_n :\ell _p \rightarrow \ell _p^n\) defined for each \(n\in {\mathbb {N}}\) by
Theorem 3.6
Let \(\lambda := (\lambda _j)\) be a sequence such that \(\lambda _1 \ge \lambda _2 \ge \cdots > 0\), and let \(r, n \in {\mathbb {N}}\) with \(1\le r \le n^2\). Then, for the bilinear operator \(D_\lambda :\ell _p \times \ell _p \rightarrow \ell _p\) with \(1\le p \le \infty \) one has
Proof
The case \(r=1\) is obvious. Thus, we may assume that \(1<r\le n^2\). We claim that \(\lambda _r \le a_r^{(2)}(D_\lambda )\). Since \((Q_r D_\lambda (J_n, J_n)) = D_\lambda ^{(r)}\), Lemma 3.5 yields
which implies the required inequality.
We now show that \(a_r^{(2)}(D_\lambda ) \le \lambda _r\). Let us consider the composition
Then, for all \(x,y \in \ell _p\) one has
This implies that \(\Vert D_\lambda - J_{r-1}D_\lambda ^{(r-1)}(Q_n,Q_n)\Vert \le \lambda _r\). Since
and \([(D_\lambda ^{(r-1)}(Q_n,Q_n))(\ell ^p \times \ell ^p)] \subset \ell _p^{r-1}\),
In consequence \(a_r^{(2)}(D_\lambda ) \le \lambda _r\) and so this completes the proof. \(\square \)
4 Gelfand numbers
Given an operator \(T\in {\mathcal {L}}(X_1, \ldots , X_k; Y)\), following [5], the n-th Gelfand number \((c_n^{(k)}(T))\) of \(T\in {\mathcal {L}}(X_1, \ldots , X_k; Y)\) is given by
Recall that \(J_Y:Y \rightarrow \ell _{\infty }(U_{Y^{*}})\) is the metric injection given by \(J_{Y}y := (y^{*}(y))_{y^{*} \in U_{Y^{*}}}\) for all \(y\in Y\).
Lemma 4.1
The sequence \((c^{(k)}_n)\) of Gelfand numbers is an \(s^{(k)}\)-scale.
Proof
Properties (S1) and (S2) follow from the definition. For (S3), since \(\Vert J_Y\Vert = 1\), it follows by (S3) of \(a^{(k)}_n\). The property (S4) also follows from (S4) of \(a^{(k)}_n\). We prove the norming property (S5). Clearly, for each \(n \in {\mathbb {N}}\), we have \(\Vert \otimes _k I_n\Vert = 1\), which implies \(c^{(k)}_n(\otimes _k I_n) \le 1\). We need to show that \(c^{(k)}_n(\otimes _k I_n) = 1\).
Given \(\varepsilon >0\), we can find \(A \in {\mathcal {L}}(\ell _2^n, \ldots , \ell _2^n; \ell _\infty (U_{\ell _2([n]^k)^{*}})\), with \(\text {rank}(A) < n\), such that,
We define \(B:\ell _2^n \rightarrow \ell _\infty (U_{\ell _2([n]^k)^{*}})\) by \(By := A(e_1, \ldots , e_1 ,y)\) for all \(y\in \ell ^n_2\). Since
there exists \(\xi \in S_{\ell _2^n}\), such that, \(B\xi = 0\). Thus, letting \(\Gamma := U_{\ell _2([n]^k)^{*}}\), we get
Since \(\varepsilon > 0\) is arbitrary, \(c^{(k)}(\otimes _k I_n)= a^{(k)}_n(J_{\ell _2^n} \circ (\otimes _k I_n)) \ge 1\). This completes the proof. \(\square \)
Proposition 4.2
Let \(T\in {\mathcal {L}}(X_1, \ldots , X_k; Y)\) such that \(\text {rank}\,(T) \ge n\), and suppose that \(T^{\times }\) is a metric injection. Then, we have \(c^{(k)}_n(T) =1\).
Proof
We first observe that, for every \(T\in {\mathcal {L}}(X_1, \ldots , X_k; Y)\),
Suppose \(T\in {\mathcal {L}}(X_1, \ldots , X_k; Y)\) satisfies the hypotheses. Then, one has
where the infimum is taken over all \(A\in {\mathcal {L}}(X_1, \ldots , X_k; \ell _{\infty }(U_{Y^{*}}))\). Since \(\text {rank}\,(A^{\times }) = \text {rank}\,(A) < n\), we conclude that
Clearly, \(J_Y^{*}:\ell _{\infty }(U_{Y^{*}})^{*} \rightarrow Y^{*}\) is a metric surjection since \(J_Y:Y \rightarrow \ell _{\infty }(U_{Y^{*}})\) is a metric injection. As \((J_YT)^{\times } = T^{\times }J_Y^{*}\), it follows that \((J_YT)^{\times }\) is a metric injection. Now, observe that \(\text {rank}\,((J_YT)^{\times }) = \text {rank}\,(J_YT) = \text {rank}\,(T)\ge n\). Combining with Proposition 3.1, we get
Since \(c_n^{(k)}(T) \le \Vert T\Vert =1\), the statement follows. \(\square \)
Lemma 4.3
For every \(T\in {\mathcal {L}}(X_1, \ldots , X_k; Y)\) one has
Proof
We observe that for any operator \(A \in {\mathcal {L}}(X_1, \ldots , X_k; Y)\), we have \(J_Z A^\times :Y^{*}\) \( \rightarrow \ell _\infty (U_{Z^{*}})\) with \(\text {rank}\,(J_Z A^\times ) = \text {rank}\,(A^\times )\), where \(Z:= {\mathcal {L}}(X_1, \ldots , X_k)\). This implies that, for every \(T\in {\mathcal {L}}(X_1, \ldots , X_k; Y)\), we get
as required. \(\square \)
Similarly as in the linear setting (see [21, Proposition 11.5.3]) one can show that the following statement is true for multilinear operators: if a Banach space F has the metric extension property, then for any \(T\in {\mathcal {L}}(X_1, \ldots , X_k; F)\) we have
As an application, we have the following multilinear variant of Pietsch’s result [19, Theorem 11.5.5], that the Gelfand numbers \((c_n)\) forms the largest injective s-scale in the setting of linear operators.
Proposition 4.4
\((c^{(k)}_n)\) is the largest injective \(s^{(k)}\)-scale.
Proof
Let \(T\in {\mathcal {L}}(X_1, \ldots , X_k; Y)\) and \(J \in {\mathcal {L}}(Y; Z)\) be an injection. Since \(\ell _\infty (U_{Y^*})\) satisfies the metric extension property, there is a linear operator \(L:Z \rightarrow \ell _\infty (U_{Y^*})\), with \(\Vert L\Vert = \Vert J_Y\Vert =1\) such that \(J_Y= L J\). Then using the fact mentioned above yields
Since \(c^{(k)}_n(J T) \le c^{(k)}_n(T)\), \((c^{(k)}_n)\) is an injective \(s^{(k)}\)-scale.
Next, we observe that the above gives that, if \((s^{(k)}_n)\) is an arbitrary injective \(s^{(k)}\)-scale, then for any \(T\in {\mathcal {L}}(X_1, \ldots , X_k; Y)\),
and the result follows. \(\square \)
As an application we obtain the following corollary.
Corollary 4.5
For an arbitrary injective \(s^{(k)}\)-scale, the pair \(((c_n), (s^{(k)}_n))\) has the mixed multiplicative property.
Proof
As we have noticed, the pair \(((a_n), (s^{(k)}_n))\) has the mixed multiplicative property for an arbitrary \(s^{(k)}\)-scale. This fact combined with Proposition 4.4 yields that if \((s^{(k)}_n)\) is an injective \(s^{(k)}\)-scale, then for any \(T\in {\mathcal {L}}(X_1, \ldots , X_k; Y)\) and \(S\in {\mathcal {L}}(Y; Z)\), we have
\(\square \)
5 Kolmogorov numbers
Let N be a closed subspace of a Banach space Y, then \(Q_{N}^Y\) (\(Q_N\) for short) denotes the canonical quotient map from Y onto Y/N. The n-th Kolmogorov number of \(T\in {\mathcal {L}}(X_1, \ldots , X_k; Y)\) is defined by
We shall use the following formula (see [5, Theorem 4.1]):
where \(Q_j\) denotes the canonical metric surjection from \(\ell _1(U_{X_j})\) onto \(X_j\), \(j\in [k]\).
Lemma 5.1
Let \(T\in {\mathcal {L}}(X_1, \ldots , X_k; Y)\) be a k-linear operator. Then, \(d_n^{(k)}(T)=0\) if and only if \(\text {rank}\,(T)<n\).
Proof
Clearly, \(\text {rank}(T)<n\) implies that, for \(N:= [T(X_1 \times \cdots \times X_k)]\), we have \(\text {dim} (N)<n\) and whence \(\Vert Q_{N}T\Vert =0\). In consequence \(d_n^{(k)}(T)=0\).
Now assume that \(d^{(k)}_n(T) = 0\). We apply the above formula \((*)\) to get
where \(Q_j:\ell _1(U_{X_j}) \rightarrow X_j\) are canonical metric surjections, \(j\in [k]\). By Theorem 3.3 and Proposition 2.1, we have \(\text {rank}(T(Q_1, \ldots , Q_k)) < n\). Combining,
yields \(\text {rank}(T) < n\), and this completes the proof. \(\square \)
Lemma 5.2
Let H be a Hilbert space and \(T\in {\mathcal {L}}(X_1, \ldots , X_k; H)\) be a k-linear operator. Then, for each \(n\in {\mathbb {N}}\), we have
where the infimum is taken over all orthogonal projections in H.
Proof
By the above formula \((*)\), it follows that for a given \(\varepsilon >0\), there exists \(A \in {\mathcal {L}}(\ell _1(U_{X_1}), \ldots , \ell _1(U_{X_k}); H)\), such that \(\text {rank}(A) < n\) and
Define \(V:= [A(\ell _1(U_{X_1}) \times \cdots \times \ell _1(U_{X_k}))] \subset H\) and let P be the orthogonal projection onto V. Then, we have \(\text {rank}(P) < n\) and
This implies that
Since \(\varepsilon > 0\) is arbitrary the proof is completed. \(\square \)
Proposition 5.3
For each k, the sequence \((d_n^{(k)})\) of the Kolmogorov numbers forms an \(s^{(k)}\)-scale.
Proof
Again, for simplicity of presentation we only consider the case \(k = 2\). The properties \((S1)-(S2)\) are obvious. Let \(T\in {\mathcal {L}}(X_1, X_2; Y)\), \(S\in {\mathcal {L}}(Y; Z)\) and \(R_1 \in {\mathcal {L}}(W_1; X_1)\), \(R_2 \in {\mathcal {L}}(W_2; X_2)\). Then, for any subspace \(N \subset Y\) with \(\text {dim} (N)< n\) one has
Since \(\text {dim} (S(N)) \le \text {dim}(N) <n \) and subspace N is arbitrary, this yields
so the property (S3) is satisfied. The property (S4) follows from Lemma 5.1.
To finish we observe that from Lemma 5.2, we have
and so, we conclude by Corollary 3.2 that
Thus the property (S5) is also satisfied and this completes the proof. \(\square \)
Since \((a^{(k)}_n)\) is the largest \(s^{(k)}\)-scale, we obtain from Lemma 5.2 and Proposition 5.3 a multilinear variant of the linear result by Pietsch (see [19, Proposition 11.6.2]).
Corollary 5.4
Let H be a Hilbert space and \(T\in {\mathcal {L}}(X_1, \ldots , X_k; H)\) be a k-linear operator. Then, for each \(n\in {\mathbb {N}}\), we have
where the infimum is taken over all orthogonal projections in H.
We conclude this section with a corollary which follows from the formula \((*).\)
Corollary 5.5
\((d^{(k)}_n)\) is the largest surjective \(s^{(k)}\)-scale.
6 Symmetrized approximation numbers
Symmetrized approximation numbers \((t_n)\) were introduced by Pietsch in [19, Proposition 11.7.9]. For any operator \(T \in {\mathcal {L}}(X; Y)\), these are defined by
where \(Q_X\) denotes the canonical metric surjection from \(\ell _1(U_{X})\) onto X.
Note that \((t_n)\) are the largest injective and surjective s-numbers with the property
It is worth noting that we have here a refined version of Schauder’s theorem (see [3, p. 84]), which states that an operator T between arbitrary Banach spaces X and Y is compact if and only if \(\lim _{n\rightarrow \infty }t_n(T) = 0\). Then, by the above formula, it follows that the degree of compactness of T and \(T^{*}\) are the same when they are measured by the symmetrized approximation numbers \(t_n\).
Following the linear case, we introduce a variant of symmetrized approximation numbers in the multilinear setting. For each \(k\in {\mathbb {N}}\), we define a rule \((t^{(k)}) :{\mathcal {L}}^k \rightarrow [0,\infty )^{{\mathbb {N}}}\) assigning to every k-linear operator \(T:X_1 \times \cdots \times X_k \rightarrow Y\) a non-negative scalar sequence \((t^{(k)}_n(T ))\) given by
Clearly, this definition is equivalent to
as well as to
Our aim is to prove the following main result of this section.
Theorem 6.1
For each \(k\in {\mathbb {N}}\), and any operator \(T\in {\mathcal {L}}(X_1, \ldots , X_k; Y)\), one has
To prove Theorem 6.1 we need two preliminary results. Before we state these results, we define a special operator between spaces of k-linear forms. Given a positive integer \(k\ge 2\) and operators \(A_1\in {\mathcal {L}}(Y_1; X_1), \ldots , A_k\in {\mathcal {L}}(Y_k; X_k)\), we define the mapping \(\Phi _{A_1,\ldots , A_k}\) from \({\mathcal {L}}(X_1, \ldots , X_k)\) to \({\mathcal {L}}(Y_1, \ldots , Y_k)\) by
for all \(S\in {\mathcal {L}}(X_1, \ldots , X_k)\) and \((y_1, \ldots , y_k) \in Y_1 \times \cdots \times Y_k\).
Under the above notation, we have the following lemma.
Lemma 6.2
We have that \(\Phi _{A_1,\ldots , A_k} :{\mathcal {L}}(X_1, \ldots , X_k)\rightarrow {\mathcal {L}}(Y_1, \ldots , Y_k)\) is a bounded linear operator, and for any operator \(T \in {\mathcal {L}}(X_1, \ldots , X_k; Y)\), it holds
Proof
The first statement is obvious. For any \(T \in {\mathcal {L}}(X_1, \ldots , X_k; Y)\),
is a bounded operator. Thus, for all \(y^{*}\in Y^{*}\) and all \((y_1, \ldots , y_k) \in Y_1 \times \cdots \times Y_k\),
and so the required formula follows. \(\square \)
To state the next result, recall that given Banach spaces \(X_1,\ldots , X_k\), we let \({\widehat{\kappa }}_{X_1 \times \cdots \times X_k}\) to denote the k-linear operator from \(X_1 \times \cdots \times X_k\) to \({\mathcal {L}}(X_1, \ldots , X_k)^{*}\) given, for all \((x_1, \ldots , x_k)\in X_1 \times \cdots \times X_k\), by
The following factorization result is a preliminary to Theorem 6.1, but is of interest in itself.
Proposition 6.3
For any Banach spaces \(X_1, \ldots , X_k\), let \(X:= X_1 \times \cdots \times X_k\) and, for each \(j\in [k]\), let \(Q_{X_j} :\ell _1(U_{X_j}) \rightarrow X_j\) be the canonical surjection. Then, the operator \(\Phi := \Phi _{Q_{X_1}, \ldots , Q_{X_k}}\) admits the following factorization with \(E:= {\mathcal {L}}(X_1, \ldots , X_k)\):
where the norms of operators P and R are less than or equal 1.
Proof
We first observe that for any \(S\in {\mathcal {L}}(X_1, \ldots , X_k)\) and \((\lambda _{x_j}) \in \ell _1(U_{X_j})\), \(j\in [k]\), we have
We define on \(\ell _{\infty }(U_{{\mathcal {L}}(X_1, \ldots , X_k)^{*}})\) a mapping P by
Clearly, \(P:\ell _{\infty }(U_{{\mathcal {L}}(X_1, \ldots , X_k)^{*}}) \rightarrow \ell _{\infty }(U_X)\) is a bounded operator with \(\Vert P\Vert \le 1\).
We also define an operator \(R:\ell _{\infty }(U_{X_1 \times \cdots \times X_k}) \rightarrow {\mathcal {L}}(\ell _1(U_{X_1}), \ldots , \ell _1(U_{X_k}))\) with norm \(\Vert R\Vert \le 1\) given by
for all \(\xi \in \ell _{\infty }(U_{X_1 \times \cdots \times X_k})\) and \(((\lambda _{x_1}), \ldots , (\lambda _{x_k})) \in \ell _1(U_{X_1}) \times \cdots \times \ell _1(U_{X_k})\).
Finally, we see that, for all \(S\in {\mathcal {L}}(X_1, \ldots , X_k)\) and all \(x \in U_X\), we have
Combining formulas \((*)\) and R yields the required factorization
and this completes the proof. \(\square \)
Now, we are ready to prove the main result of this section.
Proof of Theorem 6.1
We first recall that if Y is a Banach space and \(J_Y :Y \rightarrow \ell _\infty (U_{Y^*})\) is the canonical injection, then for the dual operator \(J_Y^{*}:\ell _\infty (U_{Y^*})^{*} \rightarrow Y^{*}\), we have
Given \(T\in {\mathcal {L}}(X_1, \ldots , X_k; Y)\), we have \(T^{\times }:Y^{*} \rightarrow {\mathcal {L}}(X_1, \ldots , X_k)\). Combining with the above formula and [5, Theorem 5.1] yield
To finish, we apply Proposition 6.3 and [5, Theorem 5.1] to get (by \(\Vert P\Vert , \Vert R\Vert \le 1\))
and this completes the proof. \(\square \)
The following theorem is a consequence of Borsuk antipodal theorem (see, e.g., [22, Theorem 1.4]).
Theorem 6.4
Let Y and Z be closed subspaces of a Banach space X, where Z is finite dimensional, and \(\mathrm{{dim}}\,Y > \mathrm{{dim}}\,Z\). Then, there exists \(y\in Y\) such that \(\Vert y\Vert _Y=1=\mathrm{{dist}}(y, Z)\).
As an application of Theorem 6.4 we get the following result.
Proposition 6.5
Let \(T\in {\mathcal {L}}(X_1, \ldots , X_k; Y)\) be a surjective operator with \(\mathrm{{rank}}(T)\) \(\ge n\). If \(T^{\times }:Y^{*} \rightarrow {\mathcal {L}}(X_1, \ldots , X_k)\) is a metric injection, then \(t^{(k)}_n(T) = 1\).
Proof
For simplicity of notation, we let \(F:= {\mathcal {L}}(X_1, \ldots , X_k)\). By Proposition 6.1, we have
Clearly, \(d_n(J_F T^{\times })\le 1\). We claim that \(d_n(J_F T^{\times }) =1\). To prove this, we apply Proposition 2.2.2 from [4], which says that the nth Kolgomorov number \(d_n(S)\) of an operator \(S\in {\mathcal {L}}(E; F)\) can be expressed as
Suppose \(d_n(J_F T^{\times }) <1\). Then, by the above formula, we can find \(\gamma \in (0, 1)\) and a subspace \(N_{\gamma } \subset \ell _{\infty }(U_{F^{*}})\) with \(\text {dim}\,N_{\gamma } <n\), such that
Since \(\text {rank}\,(T) \ge n\) and \(T^{\times }\) is a metric injection, \(\text {rank}\,(J_{F} T^{\times } (Y^{*})) \ge n\). Combining with Theorem 6.4, we deduce that for \(\varepsilon = \gamma ^{-1} - 1\), there exists \(y^{*} \in Y^{*}\) such that
and, for all \(v\in N_{\gamma }\), we have
This a contradiction with the above inclusion. \(\square \)
We conclude this section with a remark that it is an immediate consequence of the properties of approximation numbers \(a_n^{(k)}\) that the sequence of symmetrized approximation numbers \((t^{(k)}_n)\) satisfy the corresponding properties (S1), (S2), (S3) and (S4).
7 Hilbert numbers
The n-th Hilbert number of a k-linear operator \(T\in {\mathcal {L}}(X_1, \ldots , X_k; Y)\) is defined by
where the supremum is taken over all linear operators \(B \in U_{{\mathcal {L}}(Y; \ell _2)}\) and \(A_1\in U_{{\mathcal {L}}(\ell _2; X_1)}, \ldots , A_k\in U_{{\mathcal {L}}(\ell _2; X_k)}\).
We have the following theorem.
Theorem 7.1
Let \(T\in {\mathcal {L}}(X_1, \ldots , X_k; Y)\) be a k-linear operator. Then, \(h_n^{(k)}(T)=0\) implies \(\text {rank}\,(T)<n\).
Proof
Fix \(n\in {\mathbb {N}}\) and assume \(h^{(k)}_n(T) = 0\). Then, for all operators \(B \in {\mathcal {L}}(Y; \ell _2)\) and \(A_i \in {\mathcal {L}}(\ell _2; X_i)\) with \(i\in [k]\) one has
Since \((a^{(k)}_n)_{n=1}^\infty \) is a fully symmetric \(s^{(k)}\)-scale, it follows from Proposition 2.1 that for all operators \(A_1, \ldots , A_k\) and B as above,
We claim that \(\text {rank}\,(T) <n\). Suppose this is false. Then, there are \((x^j_1, \ldots , x^j_k) \in X_1 \times \cdots \times X_k\), \(j\in [n]\), such that \((y_j):= (T(x^j_1, \ldots , x^j_k))_{j=1}^n\) forms a basis in \([T(X_1 \times \cdots \times X_k)]\). Let \((y_j^{*})\) be a set of biorthogonal functionals to the basis \((y_j)\), that is,
For each \(i\in [k]\), we define the operator \(A_i \in {\mathcal {L}}(\ell _2; X_i)\) by
We also define \(B \in {\mathcal {L}}(Y; \ell _2)\) by
Then, for each \(j\in [n]\), we have
Hence, \(\text {rank}\,(BT(A_1, \ldots , A_k)) \ge n\), and so we arrive a contradiction which completes the proof. \(\square \)
It turns out that the Hilbert numbers for k-linear operators are \(s^{(k)}\)-numbers in the sense given in Sect. 2.
Theorem 7.2
For each \(k\in {\mathbb {N}}\), the sequence \((h^{(k)}_n)\) of Hilbert numbers is an \(s^{(k)}\)-scale.
Proof
Let \(T\in {\mathcal {L}}(X_1, \dots , X_k; Y)\). We claim that the property (S1) holds. Clearly, \((h^{(k)}_n(T))\) is a non-increasing sequence. We show that \(h^{(k)}_1(T) = \Vert T\Vert \).
Let \((x_1, \ldots , x_k) \in U_{X_1} \times \cdots \times U_{X_k}\). By the Hahn–Banach theorem, we can find a norm one functional \(y^{*} \in Y^{*}\) such that
We consider the operators \(B\in U_{{\mathcal {L}}(Y; \ell _2)}\) and \(A_1 \in U_{{\mathcal {L}}(\ell _2; X_1)}, \ldots , A_k\in U_{{\mathcal {L}}(\ell _2; X_k)}\), given, for all \(y\in Y\) and for all \((\xi _j)\in \ell _2\), by
Since \(BT(A_1, \ldots , A_k)(e_1, \ldots , e_k) = \Vert T(x_1, \ldots , x_k)\Vert _Y \, e_1\), we get
This proves that \(\Vert T\Vert \le h^{(k)}_1(T)\). Since the opposite inequality is obvious, the claim is proved.
Using properties of \((a^{(k)}_n)\), we deduce that the properties (S2) and (S3) hold. To prove (S4), we observe that \(\text {rank}\,(T) < n\) implies that, for all \(B \in U_{{\mathcal {L}}(Y; \ell _2)}\) and \(A_1\in \, U_{{\mathcal {L}}(\ell _2; X_1)}, \ldots , A_k\in \, U_{{\mathcal {L}}(\ell _2; X_k)}\), we have \(\text {rank}\,(BT(A_1, \ldots , A_k)) < n\). We have seen that \((a^{(k)}_n)\) is an \(s^{(k)}\)-scale for each k. Hence, \(a^{(k)}_n(T(A_1, \ldots , A_k)) =0\) and so \(h^{(k)}_n(T)=0\) as required.
To finish we need to prove the property (S5). Fix \(n\in {\mathbb {N}}\) and define operators \(A_1= \cdots = A_k:=P_n\in U_{{\mathcal {L}}(\ell _2; \ell ^n_2)}\) and \(B\in U_{{\mathcal {L}}(\ell _2([n]^k); \ell _2)}\) by
where, for each \(i \in [n^k]\) with \(i=\sigma _k(j)\) for the unique \(j:=(j_1, \ldots , j_k)\in [n]^k\), we take \(z_i := x_j\). Now observe that, for \(r:=n^k\), we have
where \(J_r:\ell _2^r \rightarrow \ell _2\) is a metric injection given by
Since \(\text {rank}\, (J_{r}\circ I^{(r)}_{\sigma _k}) \ge n\) and for all \(x\in \ell _2^n\),
it follows from Proposition 3.1 that \(a^{(k)}_n(J_{r}\circ I^{(r)}_{\sigma _k})=1\). In consequence, for \(r=n^k\) we conclude that,
This completes the proof. \(\square \)
8 Weyl and Chang numbers
An important role is played in the theory of eigenvalues of operators in Banach spaces by the famous Weyl numbers defined by Pietsch [18]. We introduce the Weyl numbers in the setting of multilinear operators. The n-th Weyl number of a k-linear operator \(T:X_1 \times \cdots \times X_k\rightarrow Y\) is defined by
We have the following observation.
Proposition 8.1
For each \(k\in {\mathbb {N}}\), the sequence \((x^{(k)}_n)\) of Weyl numbers is an \(s^{(k)}\)-scale.
Proof
Since \((a^{(k)}_n)\) is an \(s^{(k)}\)-scale, the properties (S1)–(S3) easily follows. If \(T\in {\mathcal {L}}(X_1, \ldots , X_k; Y)\) is such that \(\text {rank}\,(T) < n\), then \(\text {rank}\, (T(A_1, \ldots , A_k)) < n\) and \(a^{(k)}_n(T(A_1, \ldots , A_k)) =0\), for all \(A_j\in \, U_{{\mathcal {L}}(\ell _2; X_j)}\), \(j\in [k]\). Consequently \(x^{(k)}_n(T)=0\). This shows that the property (S4) is satisfied.
Clearly, for each \(n\in N\), we have \(h^{(k)}_n(T) \le x^{(k)}_n(T)\). Since \((h^{(k)}_n)\) is an \(s^{(k)}\)-scale,
and so the property (S5) is also satisfied. \(\square \)
Following the proof of Theorem 7.1, we get the following.
Theorem 8.2
Let \(T\in {\mathcal {L}}(X_1, \ldots , X_k; Y)\) be a k-linear operator. Then, \(x_n^{(k)}(T)=0\) implies \(\text {rank}\,(T)<n\).
The n-th Chang number of an operator \(T\in {\mathcal {L}}(X_1, \ldots , X_k; Y)\) is given by
Proposition 8.3
For each \(k\in {\mathbb {N}}\), the sequence \((y^{(k)}_n)\) of Chang numbers is an \(s^{(k)}\)-scale, which has the property: \(y^{(k)}_n(T) = 0\) implies \(\text {rank}\,(T) <n\).
Proof
The properties (S1)–(S3) are easily verified. To show the property (S4), we fix \(T\in {\mathcal {L}}(X_1, \ldots , X_k; Y)\) with \(\text {rank}\,(T) < n\). Then, \(\text {rank}(ST) < n\) for all \(S\in U_{{\mathcal {L}}(Y; \ell _2)}\). Thus \(a^{(k)}_n(ST) =0\) for all \(S\in U_{{\mathcal {L}}(Y; \ell _2)}\) yields \(y_n(T)=0\).
To finish observe that, for every \(T\in {\mathcal {L}}(X_1, \ldots , X_k; Y)\) one has
This clearly combined with the fact that \((h^{(k)}_n)\) is a \(s^{(k)}\)-scale implies that the property (S5) is satisfied for \((y^{(k)}_n)\). \(\square \)
It is well known that Hilbert numbers fail to be multiplicative (see [17, Remark 2.9.19]). This is a consequence of [17, Proposition 2.9.19], which states that
However, the following inequality is true (see [17, Lemma 2.6.6]) for any operators \(T\in {\mathcal {L}}(X; Y)\) and \(S\in {\mathcal {L}}(Y; Z)\)
We have a multilinear variant of this inequality. The proof is similar to the linear case, but we include a proof for the sake of completeness.
Lemma 8.4
For each \(k\in {\mathbb {N}}\) the pair \(((y_n), (h^{(k)}_n))\) satisfies property (M), that is, for every \(T\in {\mathcal {L}}(X_1, \ldots , X_k; Y)\) and \(S\in {\mathcal {L}}(Y; Z)\), we have
Proof
Fix positive integers \(k\ge 2\), m and n. Let \(A_j\in {\mathcal {L}}(\ell _2; X_j)\), \(j\in [k]\) and \(Q\in {\mathcal {L}}(Z; \ell _2)\) with norm less or equal 1. Given \(\varepsilon > 0\), let \(B\in {\mathcal {L}}(Y; \ell _2)\) be such that \(\text {rank}\,(B) < m\) and
Now, let \(A\in {\mathcal {L}}(^k\ell _2; \ell _2)\) with \(\text {rank}\,(A) < n\) be such that
Clearly, \(\text {rank}\,(BT(A_1, \ldots , A_k) + A) \le \text {rank}\,(B) + \text {rank}\,(A) < m + n - 1\). Thus letting \(R:= (QS - B)/\Vert QS-B\Vert \in U_{{\mathcal {L}}(Y; \ell _2)}\), we obtain
Since \(\varepsilon > 0\), Q and \(A_1, \ldots , A_k\) are arbitrary, the desired estimate follows. \(\square \)
For next theorem we need the following known result.
Lemma 8.5
If an operator \(T\in {\mathcal {L}}(Y^{*}; X^{*})\), then \(T= S^{*}\) for some \(S\in {\mathcal {L}}(X; Y)\) whenever T is \(weak^{*}\)-weakly continuous. In particular the statement is true if Y is a reflexive space.
Theorem 8.6
If \(T\in {\mathcal {L}}(X_1, \ldots , X_k; Y)\), then for each \(n\in {\mathbb {N}}\) we have
-
(i)
\(x_n(T^\times ) \le y_n^{(k)}(T)\) ;
-
(ii)
\(y_n^{(k)}(T) \le x_n(T^\times )\).
Proof
(i). We have that \(T^\times :Y^{*} \rightarrow {\mathcal {L}}(X_1, \ldots , X_k)\). Let \(V:\ell _2 \rightarrow Y^{*}\) with \(\Vert V\Vert \le 1\). By Lemma 8.5, there is \(R:Y \rightarrow \ell _2\) such that \(R^{*} = V\). Thus, by [5, Proposition 3.2], we obtain
(ii). Let \(S:Y \rightarrow \ell _2\) with \(\Vert S\Vert \le 1\). Since \(\ell _2\) is reflexive, it follows from [5, Proposition 3. 3] that
Since S is arbitrary, \(y_n^{(k)}(T) \le x_n(T^\times )\) as required. \(\square \)
9 Bernstein numbers
In the theory of Banach operator ideals a closed ideal of super strictly singular (or finitely strictly singular) operator is an interesting class of operators, which contains compact operators and is contained in the class of strictly singular operators. Recall that an operator is strictly singular if its restriction to any infinite-dimensional subspace is not an isomorphism.
Super strictly operators were introduced implicitly by Mityagin and Pełczyński in [15], and explicitly by Milman in [13, 14]. Recall that an operator \(T\in {\mathcal {L}}(X; Y)\) is said to be super strictly singular, if there does not exist a number \(\gamma >0\) and a sequence \((E_n)\) of subspaces, with \(\dim \,(E_n) = n\), such that
Thus, T is super strictly singular if and only if the Bernstein numbers \(b_n(T) \rightarrow 0\), as \(n\rightarrow \infty \), where
where the supremum is taken over all n-dimensional subspaces of X.
We define a variant of Bernstein’s numbers for bilinear operators. We start with some notations from the theory of linear operators. We recall that if \(T :E\rightarrow F\) is an operator between Banach spaces, then the injection modulus of T is given by
An operator T is called an injection if \(j(T)>0\). Clearly, an injection can be characterized as a one-to-one operator from E into F with closed range.
Recall that the surjection modulus of T is given by
An operator T is called a surjection if \(q_1(T)>0\), which is equivalent to \(T(E)=F\).
The above modules are important characteristics in the theory of linear operators, and they are used in the study of isomorphic embeddings, quotients of Banach spaces and, in particular, in the study of isomorphic classification of Banach spaces by the fact that both \(j_1(T)>0\) and \(q_1(T)>0\) if and only if T is an isomorphism.
In what follows, if X is a non-trivial Banach space, then we let \({\mathcal {F}}\!in(X)\) to denote the set of all non-trivial finite-dimensional subspaces of X. If \(E\in {\mathcal {F}}\!in(X)\) with \(\text {dim}\,(E)=n\), then we write \(E\in {\mathcal {F}}\!in_{n}(X)\).
Let \(T\in {\mathcal {L}}(X_1, \ldots , X_k; Y)\). Following the linear case, for every closed subspaces \(N_1 \subset X_1, \ldots , N_k \subset X_k\), we let
We call \(j_k^{X_1\times \cdots \times X_k}(T)\) the modulus of injection of T and denote it by \(j_k(T)\).
In a similar fashion, we define the surjection modulus of T by
(we put \(q_k(O):=0\), where O is the null operator). T is said to be a surjection if \(q_k(T)>0\), that is, T maps \(X_1\times \cdots \times X_k\) onto Y. If \(\Vert T\Vert = q_k(T)=1\), then T is said to be a metric surjection. This means that T maps \(U_{X_1} \times \cdots \times U_{X_k}\) onto \(U_Y\).
In what follows we restrict our discussion to the bilinear operators. At first we note that the modules satisfy the following properties:
Lemma 9.1
The following statements are true:
-
(ii)
If \(S, T \in {\mathcal {L}}(X, Y; Z)\), then \(j_2(S+T) \le j_2(S) + j_2(T)\) and \(q_2(S+T) \le q_2(S) + \Vert T\Vert \).
-
(ii)
If \(T \in {\mathcal {L}}(X, Y; Z)\) and \(S \in {\mathcal {L}}(Z; W)\), then \(j_2(ST) \le \Vert S\Vert j_2(T)\) and \(q_2(ST) \le q_1(S)\Vert T\Vert \). Moreover if T is surjective, then \(j_2(ST) \le j_1(S)\Vert T\Vert \), while if S is surjective, then \(q_2(ST) \le \Vert S\Vert q_2(T)\).
-
(iii)
If \(T \in {\mathcal {L}}(X, Y; Z)\), \(R_1 \in {\mathcal {L}}(X_0; X)\) and \(R_2 \in {\mathcal {L}}(Y_0; Y)\), then \(j_2(T(R_1, R_2)) \le \Vert T\Vert j_1(R_1) j_1(R_2)\). Moreover, if \(R_1\) and \(R_2\) are surjective, then \(j_2(T(R_1,R_2)) \le j_2(T)\Vert R_1\Vert \,\Vert R_2\Vert \).
These properties can be verified easily following the definitions.
Suppose that X, Y and Z are Banach spaces with \(\text {dim}(X) \ge N\) and \(\text {dim}(Y) \ge N\). Then, for each \(n\in [N]\), the n-th Bernstein number of every bilinear operator \(T:X \times Y \rightarrow Z\) is given by
Thus, if both X and Y are infinite-dimensional Banach spaces, then \((b^{(2)}_n(T))\) is well defined for each \(n \in {\mathbb {N}}\).
Proposition 9.2
The sequence \((b^{(2)}_n)\) of Bernstein’s numbers satisfies the following properties: (S1), \(\mathrm{{(S2^\prime )}}\) and (S3), where for \(\mathrm{{(S2^\prime )}}\) we mean \(b^{(2)}_{n}(S + T) \le b^{(2)}_{n}(S) + \Vert T\Vert \) for all \(S, T \in {\mathcal {L}}(X, Y; Z)\). In addition, for all \(T\in {\mathcal {L}}(X, Y; Z)\) with \(\mathrm{{rank}}\,(T) <n\) one has \(b^{(2)}_{n}(T)=0\).
Proof
Let \(T\in {\mathcal {L}}(X, Y; Z)\). Clearly, the sequence \((b^{(2)}_n(T))\) is non-increasing with \(b^{(2)}_n(T) \le \Vert T\Vert \) for each \(n\in {\mathbb {N}}\). We claim that \(b^{(2)}_1(T) = \Vert T\Vert \). In fact, for each \(n\in {\mathbb {N}}\), we can find \((x_n, y_n) \in X \times Y\) such that \(\Vert x_n\Vert _X = \Vert y_n\Vert _Y =1\) and \(\Vert T(x_n, y_n)\Vert _Z \rightarrow \Vert T\Vert \) as \(n\rightarrow \infty \). For each \(n\in {\mathbb {N}}\), let \(V_n = \{\alpha x_n; \, \alpha \in {\mathbb {K}}\}\) and \(W_n = \{\beta y_n; \,\beta \in {\mathbb {K}}\}\). Clearly, \(\text {dim}\,(V_n) = \text {dim}\,(W_n) = 1\) and so
and this proves the claim. Since for every \(M \times N \in {\mathcal {F}}\!in(X) \times {\mathcal {F}}\!in(Y)\),
then the property (S2’) follows.
In what follows, for simplicity, we write \(J_{M \times N}\) instead of the inclusion \(J_{M \times N}^{X \times Y}\) whenever the spaces X and Y are clear. To prove the property (S3), we take any operators \(R_1 \in {\mathcal {L}}(X_0; X)\), \(R_2 \in {\mathcal {L}}(Y_0; Y)\), \(S\in {\mathcal {L}}(Z; W)\) and fix \(0< \varepsilon < b_n^{(2)}(ST(R_1,R_2))\). Then, there is a subspace \(M_0 \times N_0 \in {\mathcal {F}}\!in_n(X_0) \times {\mathcal {F}}\!in_n(Y_0)\), for which
Let \(A_1:= R_1|_{M_0}\) and \(A_2:= R_2|_{N_0}\), and let \(M := R_1(M_0)\) and \(N := R_2(N_0)\). Then
and \(\Vert A_1\Vert \le \Vert R_1\Vert \), \(\Vert A_2\Vert \le \Vert R_2\Vert \). By Lemma 9.1 (iii), it follows that
which implies that \(j_1(A_1) >0\) and \(j_1(A_2) > 0\). In consequence \(A_1\) and \(A_2\) are injective operators, and so \(\text {dim}\,(M) \ge \text {dim}\,(M_0)\) and \(\text {dim}\,(N) \ge \text {dim}\,(N_0)\). Thus, \(\text {dim}(M \times N) \ge n\). Since \(A_1\) and \(A_2\) are surjective, Lemma 9.1 (ii) and (iii), gives
which completes the proof that the property (S3) holds.
To show the last property, fix \(T\in {\mathcal {L}}(X, Y; Z)\) with \(\text {rank}\,(T) <n\). Let \(M \times N \in {\mathcal {F}}\!in_{n}(X) \times {\mathcal {F}}\!in_{n}(Y)\). Then, \(\text {rank}\,(T\!J_{M \times N}) \le \text {rank}\,(T) <n\).
Now observe that, for a given \(v\in N{\setminus } \{0\}\), there exists \(u\in M {\setminus }\{0\}\) such that \(T(u, v)=0\). Otherwise, we would have \(\text {rank}\,T \ge \text {rank}\,T_v \ge n\), where the mapping \(T_v:M \rightarrow Z\) is defined by \(T_v(x) = T(x, v)\) for all \(x\in M\). But this is a contradiction. In consequence, we conclude that
Since M and N are arbitrary, the required statement follows. \(\square \)
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M. Mastyło was supported by the National Science Centre, Poland, Grant no. 2019/33/B/ST1/00165.
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Fernandez, D.L., Mastyło, M. & Silva, E.B. Pietsch’s variants of s-numbers for multilinear operators. Rev. Real Acad. Cienc. Exactas Fis. Nat. Ser. A-Mat. 115, 189 (2021). https://doi.org/10.1007/s13398-021-01123-2
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DOI: https://doi.org/10.1007/s13398-021-01123-2