Trace class

In mathematics, specifically functional analysis, a trace-class operator is a linear operator for which a trace may be defined, such that the trace is a finite number independent of the choice of basis used to compute the trace. This trace of trace-class operators generalizes the trace of matrices studied in linear algebra. All trace-class operators are compact operators.

In quantum mechanics, quantum states are described by density matrices, which are certain trace class operators.^[1]

Trace-class operators are essentially the same as nuclear operators, though many authors reserve the term "trace-class operator" for the special case of nuclear operators on Hilbert spaces and use the term "nuclear operator" in more general topological vector spaces (such as Banach spaces).

Let ${\displaystyle H}${\displaystyle H} be a separable Hilbert space, ${\displaystyle \left\{e_{k}\right\}_{k=1}^{\infty }}${\displaystyle \left\{e_{k}\right\}_{k=1}^{\infty }} an orthonormal basis and ${\displaystyle A:H\to H}${\displaystyle A:H\to H} a positive bounded linear operator on ${\displaystyle H}${\displaystyle H}. The trace of ${\displaystyle A}${\displaystyle A} is denoted by ${\displaystyle \operatorname {Tr} (A)}${\displaystyle \operatorname {Tr} (A)} and defined as^{[2] [3]}

${\displaystyle \operatorname {Tr} (A)=\sum _{k=1}^{\infty }\left\langle Ae_{k},e_{k}\right\rangle ,}${\displaystyle \operatorname {Tr} (A)=\sum _{k=1}^{\infty }\left\langle Ae_{k},e_{k}\right\rangle ,}

independent of the choice of orthonormal basis. A (not necessarily positive) bounded linear operator ${\displaystyle T:H\rightarrow H}${\displaystyle T:H\rightarrow H} is called trace class if and only if

${\displaystyle \operatorname {Tr} (|T|)<\infty ,}${\displaystyle \operatorname {Tr} (|T|)<\infty ,}

where ${\displaystyle |T|:={\sqrt {T^{*}T}}}${\displaystyle |T|:={\sqrt {T^{*}T}}} denotes the positive-semidefinite Hermitian square root.^[4]

The trace-norm of a trace class operator T is defined as $${\displaystyle \|T\|_{1}:=\operatorname {Tr} (|T|).}$${\displaystyle \|T\|_{1}:=\operatorname {Tr} (|T|).} One can show that the trace-norm is a norm on the space of all trace class operators ${\displaystyle B_{1}(H)}${\displaystyle B_{1}(H)} and that ${\displaystyle B_{1}(H)}${\displaystyle B_{1}(H)}, with the trace-norm, becomes a Banach space.

When ${\displaystyle H}${\displaystyle H} is finite-dimensional, every (positive) operator is trace class. For ${\displaystyle A}${\displaystyle A} this definition coincides with that of the trace of a matrix. If ${\displaystyle H}${\displaystyle H} is complex, then ${\displaystyle A}${\displaystyle A} is always self-adjoint (i.e. ${\displaystyle A=A^{*}=|A|}${\displaystyle A=A^{*}=|A|}) though the converse is not necessarily true.^[5]

Given a bounded linear operator ${\displaystyle T:H\to H}${\displaystyle T:H\to H}, each of the following statements is equivalent to ${\displaystyle T}${\displaystyle T} being in the trace class:

• ${\textstyle \operatorname {Tr} (|T|)=\sum _{k}\left\langle |T|,円e_{k},e_{k}\right\rangle }${\textstyle \operatorname {Tr} (|T|)=\sum _{k}\left\langle |T|,円e_{k},e_{k}\right\rangle } is finite for every orthonormal basis ${\displaystyle \left(e_{k}\right)_{k}}${\displaystyle \left(e_{k}\right)_{k}} of H.^[2]
• T is a nuclear operator.^{[6] [7]}

  There exist two orthogonal sequences ${\displaystyle \left(x_{i}\right)_{i=1}^{\infty }}${\displaystyle \left(x_{i}\right)_{i=1}^{\infty }} and ${\displaystyle \left(y_{i}\right)_{i=1}^{\infty }}${\displaystyle \left(y_{i}\right)_{i=1}^{\infty }} in ${\displaystyle H}${\displaystyle H} and positive real numbers ${\displaystyle \left(\lambda _{i}\right)_{i=1}^{\infty }}${\displaystyle \left(\lambda _{i}\right)_{i=1}^{\infty }} in ${\displaystyle \ell ^{1}}${\displaystyle \ell ^{1}} such that ${\textstyle \sum _{i=1}^{\infty }\lambda _{i}<\infty }${\textstyle \sum _{i=1}^{\infty }\lambda _{i}<\infty } and

  ${\displaystyle x\mapsto T(x)=\sum _{i=1}^{\infty }\lambda _{i}\left\langle x,x_{i}\right\rangle y_{i},\quad \forall x\in H,}${\displaystyle x\mapsto T(x)=\sum _{i=1}^{\infty }\lambda _{i}\left\langle x,x_{i}\right\rangle y_{i},\quad \forall x\in H,}

  where ${\displaystyle \left(\lambda _{i}\right)_{i=1}^{\infty }}${\displaystyle \left(\lambda _{i}\right)_{i=1}^{\infty }} are the singular values of T (or, equivalently, the eigenvalues of ${\displaystyle |T|}${\displaystyle |T|}), with each value repeated as often as its multiplicity.^[8]

• T is a compact operator with ${\displaystyle \operatorname {Tr} (|T|)<\infty .}${\displaystyle \operatorname {Tr} (|T|)<\infty .}

  If T is trace class then^[9]

  ${\displaystyle \|T\|_{1}=\sup \left\{|\operatorname {Tr} (CT)|:\|C\|\leq 1{\text{ and }}C:H\to H{\text{ is a compact operator }}\right\}.}${\displaystyle \|T\|_{1}=\sup \left\{|\operatorname {Tr} (CT)|:\|C\|\leq 1{\text{ and }}C:H\to H{\text{ is a compact operator }}\right\}.}

• T is an integral operator.^[10]
• T is equal to the composition of two Hilbert-Schmidt operators.^[11]
• ${\textstyle {\sqrt {|T|}}}${\textstyle {\sqrt {|T|}}} is a Hilbert-Schmidt operator.^[11]

Let ${\displaystyle T}${\displaystyle T} be a bounded self-adjoint operator on a Hilbert space. Then ${\displaystyle T^{2}}${\displaystyle T^{2}} is trace class if and only if ${\displaystyle T}${\displaystyle T} has a pure point spectrum with eigenvalues ${\displaystyle \left\{\lambda _{i}(T)\right\}_{i=1}^{\infty }}${\displaystyle \left\{\lambda _{i}(T)\right\}_{i=1}^{\infty }} such that^[12]

${\displaystyle \operatorname {Tr} (T^{2})=\sum _{i=1}^{\infty }\lambda _{i}(T^{2})<\infty .}${\displaystyle \operatorname {Tr} (T^{2})=\sum _{i=1}^{\infty }\lambda _{i}(T^{2})<\infty .}

Mercer's theorem provides another example of a trace class operator. That is, suppose ${\displaystyle K}${\displaystyle K} is a continuous symmetric positive-definite kernel on ${\displaystyle L^{2}([a,b])}${\displaystyle L^{2}([a,b])}, defined as

${\displaystyle K(s,t)=\sum _{j=1}^{\infty }\lambda _{j},円e_{j}(s),円e_{j}(t)}${\displaystyle K(s,t)=\sum _{j=1}^{\infty }\lambda _{j},円e_{j}(s),円e_{j}(t)}

then the associated Hilbert–Schmidt integral operator ${\displaystyle T_{K}}${\displaystyle T_{K}} is trace class, i.e.,

${\displaystyle \operatorname {Tr} (T_{K})=\int _{a}^{b}K(t,t),円dt=\sum _{i}\lambda _{i}.}${\displaystyle \operatorname {Tr} (T_{K})=\int _{a}^{b}K(t,t),円dt=\sum _{i}\lambda _{i}.}

Every finite-rank operator is a trace-class operator. Furthermore, the space of all finite-rank operators is a dense subspace of ${\displaystyle B_{1}(H)}${\displaystyle B_{1}(H)} (when endowed with the trace norm).^[9]

Given any ${\displaystyle x,y\in H,}${\displaystyle x,y\in H,} define the operator ${\displaystyle x\otimes y:H\to H}${\displaystyle x\otimes y:H\to H} by ${\displaystyle (x\otimes y)(z):=\langle z,y\rangle x.}${\displaystyle (x\otimes y)(z):=\langle z,y\rangle x.} Then ${\displaystyle x\otimes y}${\displaystyle x\otimes y} is a continuous linear operator of rank 1 and is thus trace class; moreover, for any bounded linear operator A on H (and into H), ${\displaystyle \operatorname {Tr} (A(x\otimes y))=\langle Ax,y\rangle .}${\displaystyle \operatorname {Tr} (A(x\otimes y))=\langle Ax,y\rangle .}^[9]

1. If ${\displaystyle A:H\to H}${\displaystyle A:H\to H} is a non-negative self-adjoint operator, then ${\displaystyle A}${\displaystyle A} is trace-class if and only if ${\displaystyle \operatorname {Tr} A<\infty .}${\displaystyle \operatorname {Tr} A<\infty .} Therefore, a self-adjoint operator ${\displaystyle A}${\displaystyle A} is trace-class if and only if its positive part ${\displaystyle A^{+}}${\displaystyle A^{+}} and negative part ${\displaystyle A^{-}}${\displaystyle A^{-}} are both trace-class. (The positive and negative parts of a self-adjoint operator are obtained by the continuous functional calculus.)
2. The trace is a linear functional over the space of trace-class operators, that is, $${\displaystyle \operatorname {Tr} (aA+bB)=a\operatorname {Tr} (A)+b\operatorname {Tr} (B).}$${\displaystyle \operatorname {Tr} (aA+bB)=a\operatorname {Tr} (A)+b\operatorname {Tr} (B).} The bilinear map $${\displaystyle \langle A,B\rangle =\operatorname {Tr} (A^{*}B)}$${\displaystyle \langle A,B\rangle =\operatorname {Tr} (A^{*}B)} is an inner product on the trace class; the corresponding norm is called the Hilbert–Schmidt norm. The completion of the trace-class operators in the Hilbert–Schmidt norm are called the Hilbert–Schmidt operators.
3. ${\displaystyle \operatorname {Tr} :B_{1}(H)\to \mathbb {C} }${\displaystyle \operatorname {Tr} :B_{1}(H)\to \mathbb {C} } is a positive linear functional such that if ${\displaystyle T}${\displaystyle T} is a trace class operator satisfying ${\displaystyle T\geq 0{\text{ and }}\operatorname {Tr} T=0,}${\displaystyle T\geq 0{\text{ and }}\operatorname {Tr} T=0,} then ${\displaystyle T=0.}${\displaystyle T=0.}^[11]
4. If ${\displaystyle T:H\to H}${\displaystyle T:H\to H} is trace-class then so is ${\displaystyle T^{*}}${\displaystyle T^{*}} and ${\displaystyle \|T\|_{1}=\left\|T^{*}\right\|_{1}.}${\displaystyle \|T\|_{1}=\left\|T^{*}\right\|_{1}.}^[11]
5. If ${\displaystyle A:H\to H}${\displaystyle A:H\to H} is bounded, and ${\displaystyle T:H\to H}${\displaystyle T:H\to H} is trace-class, then ${\displaystyle AT}${\displaystyle AT} and ${\displaystyle TA}${\displaystyle TA} are also trace-class (i.e. the space of trace-class operators on H is a two-sided ideal in the algebra of bounded linear operators on H), and^{[11] [13]} $${\displaystyle \|AT\|_{1}=\operatorname {Tr} (|AT|)\leq \|A\|\|T\|_{1},\quad \|TA\|_{1}=\operatorname {Tr} (|TA|)\leq \|A\|\|T\|_{1}.}$${\displaystyle \|AT\|_{1}=\operatorname {Tr} (|AT|)\leq \|A\|\|T\|_{1},\quad \|TA\|_{1}=\operatorname {Tr} (|TA|)\leq \|A\|\|T\|_{1}.} Furthermore, under the same hypothesis,^[11] $${\displaystyle \operatorname {Tr} (AT)=\operatorname {Tr} (TA)}$${\displaystyle \operatorname {Tr} (AT)=\operatorname {Tr} (TA)} and ${\displaystyle |\operatorname {Tr} (AT)|\leq \|A\|\|T\|.}${\displaystyle |\operatorname {Tr} (AT)|\leq \|A\|\|T\|.} The last assertion also holds under the weaker hypothesis that A and T are Hilbert–Schmidt.
6. If ${\displaystyle \left(e_{k}\right)_{k}}${\displaystyle \left(e_{k}\right)_{k}} and ${\displaystyle \left(f_{k}\right)_{k}}${\displaystyle \left(f_{k}\right)_{k}} are two orthonormal bases of H and if T is trace class then ${\textstyle \sum _{k}\left|\left\langle Te_{k},f_{k}\right\rangle \right|\leq \|T\|_{1}.}${\textstyle \sum _{k}\left|\left\langle Te_{k},f_{k}\right\rangle \right|\leq \|T\|_{1}.}^[9]
7. If A is trace-class, then one can define the Fredholm determinant of ${\displaystyle I+A}${\displaystyle I+A}: $${\displaystyle \det(I+A):=\prod _{n\geq 1}[1+\lambda _{n}(A)],}$${\displaystyle \det(I+A):=\prod _{n\geq 1}[1+\lambda _{n}(A)],} where ${\displaystyle \{\lambda _{n}(A)\}_{n}}${\displaystyle \{\lambda _{n}(A)\}_{n}} is the spectrum of ${\displaystyle A.}${\displaystyle A.} The trace class condition on ${\displaystyle A}${\displaystyle A} guarantees that the infinite product is finite: indeed, $${\displaystyle \det(I+A)\leq e^{\|A\|_{1}}.}$${\displaystyle \det(I+A)\leq e^{\|A\|_{1}}.} It also implies that ${\displaystyle \det(I+A)\neq 0}${\displaystyle \det(I+A)\neq 0} if and only if ${\displaystyle (I+A)}${\displaystyle (I+A)} is invertible.
8. If ${\displaystyle A:H\to H}${\displaystyle A:H\to H} is trace class then for any orthonormal basis ${\displaystyle \left(e_{k}\right)_{k}}${\displaystyle \left(e_{k}\right)_{k}} of ${\displaystyle H,}${\displaystyle H,} the sum of positive terms ${\textstyle \sum _{k}\left|\left\langle A,円e_{k},e_{k}\right\rangle \right|}${\textstyle \sum _{k}\left|\left\langle A,円e_{k},e_{k}\right\rangle \right|} is finite.^[11]
9. If ${\displaystyle A=B^{*}C}${\displaystyle A=B^{*}C} for some Hilbert-Schmidt operators ${\displaystyle B}${\displaystyle B} and ${\displaystyle C}${\displaystyle C} then for any normal vector ${\displaystyle e\in H,}${\displaystyle e\in H,} ${\textstyle |\langle Ae,e\rangle |={\frac {1}{2}}\left(\|Be\|^{2}+\|Ce\|^{2}\right)}${\textstyle |\langle Ae,e\rangle |={\frac {1}{2}}\left(\|Be\|^{2}+\|Ce\|^{2}\right)} holds.^[11]

Let ${\displaystyle A}${\displaystyle A} be a trace-class operator in a separable Hilbert space ${\displaystyle H,}${\displaystyle H,} and let ${\displaystyle \{\lambda _{n}(A)\}_{n=1}^{N\leq \infty }}${\displaystyle \{\lambda _{n}(A)\}_{n=1}^{N\leq \infty }} be the eigenvalues of ${\displaystyle A.}${\displaystyle A.} Let us assume that ${\displaystyle \lambda _{n}(A)}${\displaystyle \lambda _{n}(A)} are enumerated with algebraic multiplicities taken into account (that is, if the algebraic multiplicity of ${\displaystyle \lambda }${\displaystyle \lambda } is ${\displaystyle k,}${\displaystyle k,} then ${\displaystyle \lambda }${\displaystyle \lambda } is repeated ${\displaystyle k}${\displaystyle k} times in the list ${\displaystyle \lambda _{1}(A),\lambda _{2}(A),\dots }${\displaystyle \lambda _{1}(A),\lambda _{2}(A),\dots }). Lidskii's theorem (named after Victor Borisovich Lidskii) states that $${\displaystyle \operatorname {Tr} (A)=\sum _{n=1}^{N}\lambda _{n}(A)}$${\displaystyle \operatorname {Tr} (A)=\sum _{n=1}^{N}\lambda _{n}(A)}

Note that the series on the right converges absolutely due to Weyl's inequality $${\displaystyle \sum _{n=1}^{N}\left|\lambda _{n}(A)\right|\leq \sum _{m=1}^{M}s_{m}(A)}$${\displaystyle \sum _{n=1}^{N}\left|\lambda _{n}(A)\right|\leq \sum _{m=1}^{M}s_{m}(A)} between the eigenvalues ${\displaystyle \{\lambda _{n}(A)\}_{n=1}^{N}}${\displaystyle \{\lambda _{n}(A)\}_{n=1}^{N}} and the singular values ${\displaystyle \{s_{m}(A)\}_{m=1}^{M}}${\displaystyle \{s_{m}(A)\}_{m=1}^{M}} of the compact operator ${\displaystyle A.}${\displaystyle A.}^[14]

One can view certain classes of bounded operators as noncommutative analogue of classical sequence spaces, with trace-class operators as the noncommutative analogue of the sequence space ${\displaystyle \ell ^{1}(\mathbb {N} ).}${\displaystyle \ell ^{1}(\mathbb {N} ).}

Indeed, it is possible to apply the spectral theorem to show that every normal trace-class operator on a separable Hilbert space can be realized in a certain way as an ${\displaystyle \ell ^{1}}${\displaystyle \ell ^{1}} sequence with respect to some choice of a pair of Hilbert bases. In the same vein, the bounded operators are noncommutative versions of ${\displaystyle \ell ^{\infty }(\mathbb {N} ),}${\displaystyle \ell ^{\infty }(\mathbb {N} ),} the compact operators that of ${\displaystyle c_{0}}${\displaystyle c_{0}} (the sequences convergent to 0), Hilbert–Schmidt operators correspond to ${\displaystyle \ell ^{2}(\mathbb {N} ),}${\displaystyle \ell ^{2}(\mathbb {N} ),} and finite-rank operators to ${\displaystyle c_{00}}${\displaystyle c_{00}} (the sequences that have only finitely many non-zero terms). To some extent, the relationships between these classes of operators are similar to the relationships between their commutative counterparts.

Recall that every compact operator ${\displaystyle T}${\displaystyle T} on a Hilbert space takes the following canonical form: there exist orthonormal bases ${\displaystyle (u_{i})_{i}}${\displaystyle (u_{i})_{i}} and ${\displaystyle (v_{i})_{i}}${\displaystyle (v_{i})_{i}} and a sequence ${\displaystyle \left(\alpha _{i}\right)_{i}}${\displaystyle \left(\alpha _{i}\right)_{i}} of non-negative numbers with ${\displaystyle \alpha _{i}\to 0}${\displaystyle \alpha _{i}\to 0} such that $${\displaystyle Tx=\sum _{i}\alpha _{i}\langle x,v_{i}\rangle u_{i}\quad {\text{ for all }}x\in H.}$${\displaystyle Tx=\sum _{i}\alpha _{i}\langle x,v_{i}\rangle u_{i}\quad {\text{ for all }}x\in H.} Making the above heuristic comments more precise, we have that ${\displaystyle T}${\displaystyle T} is trace-class iff the series ${\textstyle \sum _{i}\alpha _{i}}${\textstyle \sum _{i}\alpha _{i}} is convergent, ${\displaystyle T}${\displaystyle T} is Hilbert–Schmidt iff ${\textstyle \sum _{i}\alpha _{i}^{2}}${\textstyle \sum _{i}\alpha _{i}^{2}} is convergent, and ${\displaystyle T}${\displaystyle T} is finite-rank iff the sequence ${\displaystyle \left(\alpha _{i}\right)_{i}}${\displaystyle \left(\alpha _{i}\right)_{i}} has only finitely many nonzero terms. This allows to relate these classes of operators. The following inclusions hold and are all proper when ${\displaystyle H}${\displaystyle H} is infinite-dimensional:$${\displaystyle \{{\text{ finite rank }}\}\subseteq \{{\text{ trace class }}\}\subseteq \{{\text{ Hilbert--Schmidt }}\}\subseteq \{{\text{ compact }}\}.}$${\displaystyle \{{\text{ finite rank }}\}\subseteq \{{\text{ trace class }}\}\subseteq \{{\text{ Hilbert--Schmidt }}\}\subseteq \{{\text{ compact }}\}.}

The trace-class operators are given the trace norm ${\textstyle \|T\|_{1}=\operatorname {Tr} \left[\left(T^{*}T\right)^{1/2}\right]=\sum _{i}\alpha _{i}.}${\textstyle \|T\|_{1}=\operatorname {Tr} \left[\left(T^{*}T\right)^{1/2}\right]=\sum _{i}\alpha _{i}.} The norm corresponding to the Hilbert–Schmidt inner product is $${\displaystyle \|T\|_{2}=\left[\operatorname {Tr} \left(T^{*}T\right)\right]^{1/2}=\left(\sum _{i}\alpha _{i}^{2}\right)^{1/2}.}$${\displaystyle \|T\|_{2}=\left[\operatorname {Tr} \left(T^{*}T\right)\right]^{1/2}=\left(\sum _{i}\alpha _{i}^{2}\right)^{1/2}.} Also, the usual operator norm is ${\textstyle \|T\|=\sup _{i}\left(\alpha _{i}\right).}${\textstyle \|T\|=\sup _{i}\left(\alpha _{i}\right).} By classical inequalities regarding sequences, $${\displaystyle \|T\|\leq \|T\|_{2}\leq \|T\|_{1}}$${\displaystyle \|T\|\leq \|T\|_{2}\leq \|T\|_{1}} for appropriate ${\displaystyle T.}${\displaystyle T.}

It is also clear that finite-rank operators are dense in both trace-class and Hilbert–Schmidt in their respective norms.

The dual space of ${\displaystyle c_{0}}${\displaystyle c_{0}} is ${\displaystyle \ell ^{1}(\mathbb {N} ).}${\displaystyle \ell ^{1}(\mathbb {N} ).} Similarly, we have that the dual of compact operators, denoted by ${\displaystyle K(H)^{*},}${\displaystyle K(H)^{*},} is the trace-class operators, denoted by ${\displaystyle B_{1}.}${\displaystyle B_{1}.} The argument, which we now sketch, is reminiscent of that for the corresponding sequence spaces. Let ${\displaystyle f\in K(H)^{*},}${\displaystyle f\in K(H)^{*},} we identify ${\displaystyle f}${\displaystyle f} with the operator ${\displaystyle T_{f}}${\displaystyle T_{f}} defined by $${\displaystyle \langle T_{f}x,y\rangle =f\left(S_{x,y}\right),}$${\displaystyle \langle T_{f}x,y\rangle =f\left(S_{x,y}\right),} where ${\displaystyle S_{x,y}}${\displaystyle S_{x,y}} is the rank-one operator given by $${\displaystyle S_{x,y}(h)=\langle h,y\rangle x.}$${\displaystyle S_{x,y}(h)=\langle h,y\rangle x.}

This identification works because the finite-rank operators are norm-dense in ${\displaystyle K(H).}${\displaystyle K(H).} In the event that ${\displaystyle T_{f}}${\displaystyle T_{f}} is a positive operator, for any orthonormal basis ${\displaystyle u_{i},}${\displaystyle u_{i},} one has $${\displaystyle \sum _{i}\langle T_{f}u_{i},u_{i}\rangle =f(I)\leq \|f\|,}$${\displaystyle \sum _{i}\langle T_{f}u_{i},u_{i}\rangle =f(I)\leq \|f\|,} where ${\displaystyle I}${\displaystyle I} is the identity operator: $${\displaystyle I=\sum _{i}\langle \cdot ,u_{i}\rangle u_{i}.}$${\displaystyle I=\sum _{i}\langle \cdot ,u_{i}\rangle u_{i}.}

But this means that ${\displaystyle T_{f}}${\displaystyle T_{f}} is trace-class. An appeal to polar decomposition extend this to the general case, where ${\displaystyle T_{f}}${\displaystyle T_{f}} need not be positive.

A limiting argument using finite-rank operators shows that ${\displaystyle \|T_{f}\|_{1}=\|f\|.}${\displaystyle \|T_{f}\|_{1}=\|f\|.} Thus ${\displaystyle K(H)^{*}}${\displaystyle K(H)^{*}} is isometrically isomorphic to ${\displaystyle B_{1}.}${\displaystyle B_{1}.}

Recall that the dual of ${\displaystyle \ell ^{1}(\mathbb {N} )}${\displaystyle \ell ^{1}(\mathbb {N} )} is ${\displaystyle \ell ^{\infty }(\mathbb {N} ).}${\displaystyle \ell ^{\infty }(\mathbb {N} ).} In the present context, the dual of trace-class operators ${\displaystyle B_{1}}${\displaystyle B_{1}} is the bounded operators ${\displaystyle B(H).}${\displaystyle B(H).} More precisely, the set ${\displaystyle B_{1}}${\displaystyle B_{1}} is a two-sided ideal in ${\displaystyle B(H).}${\displaystyle B(H).} So given any operator ${\displaystyle T\in B(H),}${\displaystyle T\in B(H),} we may define a continuous linear functional ${\displaystyle \varphi _{T}}${\displaystyle \varphi _{T}} on ${\displaystyle B_{1}}${\displaystyle B_{1}} by ${\displaystyle \varphi _{T}(A)=\operatorname {Tr} (AT).}${\displaystyle \varphi _{T}(A)=\operatorname {Tr} (AT).} This correspondence between bounded linear operators and elements ${\displaystyle \varphi _{T}}${\displaystyle \varphi _{T}} of the dual space of ${\displaystyle B_{1}}${\displaystyle B_{1}} is an isometric isomorphism. It follows that ${\displaystyle B(H)}${\displaystyle B(H)} is the dual space of ${\displaystyle B_{1}.}${\displaystyle B_{1}.} This can be used to define the weak-* topology on ${\displaystyle B(H).}${\displaystyle B(H).}

• Nuclear operators between Banach spaces

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