# Nuclear operator

nuclear mapping

A linear operator mapping one locally convex space into another, and having a special form of approximation by operators of finite rank (that is, by continuous linear operators with finite-dimensional ranges). A nuclear operator has certain properties inherent in finite-dimensional operators. In particular, a nuclear operator mapping a space with a basis into itself has a finite trace (see below), which coincides with the sum of the series formed from the diagonal elements of the matrix of this operator relative to an arbitrary basis. Nuclear operators first appeared in mathematical quantum mechanics and were called "operators with a trace" (see , ). On a Hilbert space the operators with a trace are in a one-to-one correspondence with bivalent tensors, and the trace of an operator coincides with contraction of the corresponding tensor. By using this correspondence, A.F. Ruston  carried over the concept of a nuclear operator to Banach spaces. Independently, in connection with the theory of nuclear spaces (cf. Nuclear space), A. Grothendieck carried over the concept to locally convex spaces (see , ). Let $E$ and $F$ be locally convex spaces over the field of real or complex numbers, let $E ^ \prime$ and $F ^ { \prime }$ be their duals endowed with the strong topology, let $L ( E, F )$ be the vector space of all continuous linear mappings from $E$ into $F$, and let $S ( E, F )$ be the space of all weakly continuous mappings from $E$ into $F$. Set $L ( E, E) = L ( E)$ and $S ( E, E) = S ( E)$.

A linear operator $A: E \rightarrow F$ is called nuclear if it can be represented in the form

$$\tag{1 } x \mapsto Ax = \ \sum _ {i = 1 } ^ \infty \lambda _ {i} \langle x,\,x _ {i} ^ \prime \rangle\, y _ {i} ,$$

where $\{ \lambda _ {i} \}$ is a summable numerical sequence, $\{ x _ {i} ^ \prime \}$ is an equicontinuous sequence in $E ^ \prime$, $\{ y _ {i} \}$ is a sequence of elements from a certain complete bounded convex circled set in $F$ (cf. Topological vector space), and $\langle x, x ^ \prime \rangle$ denotes the value of the linear functional $x ^ \prime$ at a vector $x$. The representation (1) can be regarded as an expansion of the operator as a sum of operators of rank 1 (that is, with a one-dimensional range), and the corresponding series is absolutely convergent in $L ( E, F )$ in the topology of uniform convergence on bounded sets. Thus, in this topology, the nuclear operator $A$ is the limit of a sequence of operators of finite rank. If $E$ and $F$ are Banach spaces, then a nuclear operator $A$ can be approximated, in the nuclear norm, by operators of finite rank.

The expansion (1) is called a nuclear representation of $A$. Every nuclear operator has a nuclear representation (1) such that $x _ {i} ^ \prime \rightarrow 0$, $y _ {i} \rightarrow 0$. If $E$ is a barrelled space and is complete, or at least quasi-complete (i.e. closed bounded sets in $E$ are complete), then the expansion (1) is nuclear if and only if $\{ x _ {i} ^ \prime \}$ and $\{ y _ {i} \}$ are bounded.

By changing the conditions on $\{ \lambda _ {i} \}$, $\{ x _ {i} ^ \prime \}$, and $\{ y _ {i} \}$ one can obtain different modifications of the concept of a nuclear operator (see , , ). If instead of the equicontinuity of $\{ x _ {i} ^ \prime \}$ one requires its elements to belong to a complete bounded convex circled set in $E ^ \prime$, then the expansion (1) defines a Fredholm operator; these operators form the natural domain of application of the Fredholm theory (see , ). Every nuclear operator is a Fredholm operator, and when $E$ is endowed with the Mackey topology, any Fredholm operator $A: E \rightarrow F$ is nuclear. A nuclear operator $A$ is called strongly nuclear (or a nuclear operator of order $0$) if it admits a nuclear representation (1) in which $\{ \lambda _ {i} \}$ is a rapidly decreasing sequence, that is, $\sum _ {i = 1 } ^ \infty | \lambda _ {i} | ^ {p} < \infty$ for all $p > 0$.

Integral operators (in particular, Fredholm integral operators) provide many examples of nuclear operators and their modifications (see , , , ).

## Properties of nuclear operators.

Every nuclear operator $A \in L ( E, F )$ is compact, that is, it maps a neighbourhood of zero in $E$ into a set with compact closure in $F$. Thus, every nuclear operator is continuous, and every Fredholm operator is weakly continuous. The product (in any order) of a nuclear operator and a continuous linear operator is a nuclear operator. In particular, the set of all nuclear operators is an ideal in the algebra $L ( E)$; correspondingly, the Fredholm operators form an ideal in $S ( E)$. The strongly nuclear operators also form an ideal in $L ( E)$. Every nuclear operator $A \in L ( E, F )$ has a unique extension $\widehat{A} \in L ( \widehat{E} , F )$, where $\widehat{E}$ is the completion of $E$ and $\widehat{A}$ is nuclear. If $A \in L ( E, F )$ is a Fredholm operator, then the dual mapping $F ^ { \prime } \rightarrow E ^ \prime$ is a nuclear operator. For any nuclear operator $A \in L ( E, F )$ one can find Banach spaces $E _ {1}$ and $F _ {1}$, compact operators $K _ {1} \in L ( E, E _ {1} )$ and $K _ {2} \in L ( F _ {1} , F )$, and a nuclear operator $B \in L ( E _ {1} , F _ {1} )$, such that $A = K _ {2} BK _ {1}$. If $A \in L ( E)$ is a strongly nuclear operator, then the sequence of its eigen values (in general, complex), ordered in decreasing absolute value, is rapidly decreasing.

Let $E$ be a nuclear space and let $F$ be a complete or quasi-complete space. Then for $A \in L ( E, F )$ the following assertions are equivalent: 1) $A$ is a nuclear operator; 2) $A$ is a compact operator; 3) $A$ is a bounded operator; i.e. $A$ maps a neighbourhood of zero in $E$ into a bounded set in $F$; and 4) $A$ is a strongly nuclear operator.

Let $E$, $F$ and $G$ be Hilbert spaces, and let $K _ {1} \in L ( E, F )$ and $K _ {2} \in L ( F, G)$ be Hilbert–Schmidt operators (cf. Hilbert–Schmidt operator). Then $K _ {2} K _ {1} \in L ( E, G)$ is nuclear. Conversely, every nuclear operator is the product of two operators of Hilbert–Schmidt type. An arbitrary completely-continuous (compact) operator $A \in L ( E, F )$ is nuclear if and only if the series of eigen values of the positive-definite operator $T \in L ( E)$ in the polar decomposition $A = UT$ converges, where $U$ is an isometric operator mapping the range of $T$ into $F$ (see ).

## Operators with a trace.

Let $E$ be an arbitrary locally convex space, and let $A$ be a nuclear (respectively, Fredholm) operator mapping $E$ into itself and admitting a representation of the form (1). The series $\sum _ {i = 1 } ^ \infty \lambda _ {i} \langle y _ {i} , x _ {i} ^ \prime \rangle$ converges absolutely; if its sum does not depend on the representation (1), then the sum is called the trace of the nuclear (Fredholm) operator $A$, and is denoted by $\mathop{\rm tr} A$. In this case the trace is well defined (see , ). The operator is said to be an operator of finite trace or a trace class operator. If (1) contains only a finite number of terms, then $A$ is an operator of finite rank, and $\mathop{\rm tr} A$ is the same as the trace of the finite-dimensional operator induced in the range of $A$.

Let $E ^ \prime \overline \otimes \; E$ be the inductive tensor product of $E ^ \prime$ and $E$, that is, the completion of the (algebraic) tensor product $E ^ \prime \otimes E$ in the strongest locally convex topology in which the canonical bilinear mapping $E ^ \prime \times E \rightarrow E ^ \prime \otimes E$ ($( x ^ \prime , x)$ goes into $x ^ \prime \otimes x$) is continuous in each variable separately. The composition of this mapping with any continuous linear form on $E ^ \prime \overline \otimes \; E$ gives a bilinear form on $E ^ \prime \times E$ that is continuous in each variable separately, and the correspondence between forms of this type is one-to-one. In particular, the bilinear form $( x ^ \prime , x) \mapsto \langle x, x ^ \prime \rangle$ corresponds to a continuous linear form on $E ^ \prime \overline \otimes \; E$. The value of this form at a $u \in E ^ \prime \overline \otimes \; E$ is denoted by $\mathop{\rm tr} u$. An element $u \in E ^ \prime \overline \otimes \; E$ is called a Fredholm kernel if it admits an expansion of the form

$$\tag{2 } u = \ \sum _ {i = 1 } ^ \infty \lambda _ {i} \ x _ {i} ^ \prime \otimes y _ {i} ,$$

where $\{ \lambda _ {i} \}$, $\{ x _ {i} ^ \prime \}$ and $\{ y _ {i} \}$ are the same as in the expansion (1) for a Fredholm operator. The Fredholm kernels form a subspace in $E ^ \prime \overline \otimes \; E$, denoted by $E ^ \prime \widetilde \otimes E$.

Suppose that the algebra $S ( E)$ of weakly continuous operators on $E$ is endowed with the weak operator topology defined by semi-norms $A \mapsto | \langle Ay, x ^ \prime \rangle |$, where $A \in S ( E)$, and $x ^ \prime$ and $y$ range over $E ^ \prime$ and $E$, respectively. The mapping $\Gamma : E ^ \prime \widetilde \otimes E \rightarrow S ( E)$ that sends an element $u$ of the form (2) into an operator $A$ of the form (1), is well defined, linear and continuous; also $\mathop{\rm tr} u = \mathop{\rm tr} \Gamma ( u)$ if the trace of the operator $A = \Gamma ( u)$ is well defined. If $E$ and $E ^ \prime$ are complete (for example, if $E$ is a Fréchet space), then $\Gamma$ can be continuously extended to $E ^ \prime \overline \otimes \; E$. The images of elements of $E ^ \prime \overline \otimes \; E$ under this mapping are called operators with a trace (see , ). If $E$ is a Banach space, then every operator with a trace is nuclear, so that in this case the classes of nuclear operators, of Fredholm operators and of operators with a trace coincide. There are operators with a trace that are not Fredholm operators (for example, in nuclear Fréchet spaces). The non-compactness of these operators makes their study difficult.

## The single-valuedness problem (the "problème de biunivocité" ).

If the mapping $\Gamma$ is one-to-one, or at least if $\Gamma ( u) = 0$ implies $\mathop{\rm tr} u = 0$, then the trace of $\Gamma ( u)$ is well defined by $\mathop{\rm tr} \Gamma ( u) = \mathop{\rm tr} u$.

This possibility is closely connected with the approximation property, which is that $L ( E)$ contains a net (cf. Net (directed set)) of operators of finite rank that converges to the identity operator in the topology of uniform convergence on all pre-compact sets. If $E$ is a Banach space, then the trace of any nuclear operator is well defined if and only if the approximation property holds . A reflexive separable space $X$ without the approximation property (and without a Schauder basis, thus solving a well-known problem of S. Banach) has been constructed . This solves the single-valuedness problem: There is an $u \in X ^ \prime \overline \otimes \; X$ such that $\Gamma ( u) = 0$ but $\mathop{\rm tr} u = 1$. If a locally convex space $E$ has the approximation property, then every nuclear operator has a well defined trace; if $\{ B _ \nu \}$ is a net of operators of finite rank that converges to an arbitrary operator $B \in L ( E)$ uniformly on all pre-compact (or, at least, on convex balanced compact) sets, then

$$\tag{3 } \mathop{\rm tr} AB = \lim\limits _ \nu \mathop{\rm tr} AB _ \nu$$

is valid for any nuclear operator $A$ (see ). However, there is a locally convex space with the approximation property in which it is impossible to properly define the trace for all Fredholm operators. Any Fredholm operator on a locally convex space $E$ has a well-defined trace if $E$ has the bounded approximation property, that is, if there is a net of operators of finite rank that converges to the identity operator in the weak operator topology, and that is bounded in this topology; any space with a Schauder basis has this property. If $\{ B _ \nu \}$ is a bounded net that converges in $S ( E)$ to an arbitrary operator $B$ (for example, if $\{ B _ \nu \}$ is an arbitrary countable convergent sequence in $S ( E)$), then (3) holds for any Fredholm operator $A$ provided that the $AB _ \nu$ have a well-defined trace (for example, if the $B _ \nu$ are operators of finite rank, or if $E$ has the bounded approximation property). If $E$ has the approximation (respectively, bounded approximation) property, then for any nuclear (respectively, Fredholm) operator $A$ and any $B \in L ( E)$ (respectively, $B \in S ( E)$) one has $\mathop{\rm tr} AB = \mathop{\rm tr} BA$ (see ).

## Matrix trace.

Suppose that a locally convex space $E$ has a Schauder basis $\{ e _ {i} \} _ {i=1} ^ \infty$ so that any $x \in E$ can be expanded as $x = \sum _ {i = 1 } ^ \infty \langle x, e _ {i} ^ \prime \rangle e _ {i}$, where $e _ {i} ^ \prime \in E ^ \prime$. Then $\sum _ {i = 1 } ^ \infty \langle Ae _ {i} , e _ {i} ^ \prime \rangle$ is called the matrix trace of the operator $A$ if the series is convergent. This series converges absolutely if the basis is unconditional. Any Fredholm operator on a space with a Schauder basis has a well-defined trace that coincides with its matrix trace, which in this case does not depend on the choice of the basis .

An arbitrary continuous operator on a Hilbert space is nuclear if and only if it has a finite matrix trace for any orthonormal basis (see , , ).

## Nuclear trace.

Let $T$ be a compact space with a Borel measure $\mu$, let $C ( T)$ be the Banach space of continuous functions on $T$ equipped with the topology of uniform convergence, and let $K ( t, s)$ be a continuous function on $T \times T$. Then the linear integral operator

$$K: \phi ( t) \mapsto \int\limits _ { T } K ( t, s) \phi ( s) d \mu ( s)$$

on $C ( T)$ (a classical Fredholm integral operator) is nuclear and has a well-defined trace; moreover,

$$\tag{4 } \mathop{\rm tr} K = \int\limits _ { T } K ( t, t) d \mu ( t).$$

If $K$ is the integral operator with kernel $K ( t, s)$, acting on a space of functions on a space $T$ with a measure $\mu$, and if the right-hand side of (4) can be given a reasonable meaning, then this quantity is called the nuclear trace of $K$. For different classes of integral operators, conditions can be obtained that ensure the nuclearity of these operators, and enable one to give a meaning to (4) (see , , , ).

## Spectral trace.

Let $E$ be a locally convex space over the field of complex numbers, and let $A$ be a nuclear operator on $E$. The spectrum of $A$, as of any compact operator (cf. Spectrum of an operator), is either a finite set or is a sequence that converges to zero, and any non-zero value has finite spectral multiplicity. If the series

$$\tag{5 } \sum _ { j } \sigma _ {j} ( A),$$

formed from the non-zero eigen values of $A$ (each eigen value appears in (5) as many times as its spectral multiplicity) converges absolutely, then its sum is called the spectral trace of $A$, and is denoted by $\mathop{\rm tr} _ \sigma A$. Every nuclear operator on a Hilbert space has a spectral trace, which coincides with its matrix trace . Let $E$ be a multi-Hilbert space (a Hilbertiable space), that is, the topology in $E$ can be generated by a family of semi-norms each of which is obtained from a non-negative definite Hermitian form on $E \times E$; any nuclear space is an example of a multi-Hilbert space. Then any nuclear operator $A$ on $E$ has a well-defined trace and a spectral trace, and $\mathop{\rm tr} _ \sigma A = \mathop{\rm tr} A$ (see ). A nuclear operator $A$ need not have a matrix trace. A nuclear operator $A$ on a Banach space need not have a spectral trace even when the space has a basis and $\mathop{\rm tr} A$ is well defined. Also, the equality $\mathop{\rm tr} _ \sigma A = \mathop{\rm tr} A$ can be violated. For example, in the Banach space $c _ {0}$ of sequences that converge to 0 there is a nuclear operator $A$ such that $\mathop{\rm tr} A = 1$ and $A ^ {2} = 0$, so that $A$ does not have non-zero eigen values, and $\mathop{\rm tr} _ \sigma A = 0$. For a nuclear operator $A$ acting on an arbitrary Banach or locally convex space (without, perhaps, any approximation properties), it is possible to give conditions on $A$ under which $\mathop{\rm tr} _ \sigma A$ and $\mathop{\rm tr} A$ exist and are equal (see , , , ).

Example. Let $E$ be a complex Banach space and let $L ( E)$ be the algebra of continuous linear operators on $E$ equipped with the usual operator norm. For any $A \in L ( E)$ let $\alpha _ {r} ( A)$ denote the greatest lower bound of $\| A - F \|$ when $F$ ranges over the set of all operators in $L ( E)$ with rank (that is, dimension of the range) not exceeding $r = 0, 1 ,\dots$. The set of all $A \in L ( E)$ for which $\sum \alpha _ {r} ( A) < \infty$ is denoted by $l _ {1} ( E)$. Every $A \in l _ {1} ( E)$ is nuclear; if $E$ is a Hilbert space, then $l _ {1} ( E)$ coincides with the set of all nuclear operators on $E$. For an arbitrary Banach space $E$, each operator $A \in l _ {1} ( E)$ has a trace, $\mathop{\rm tr} A$, and a spectral trace, and $\mathop{\rm tr} _ \sigma A = \mathop{\rm tr} A$ (see , ).

How to Cite This Entry:
Nuclear operator. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Nuclear_operator&oldid=51492
This article was adapted from an original article by G.L. Litvinov (originator), which appeared in Encyclopedia of Mathematics - ISBN 1402006098. See original article