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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.cite-ref-footnotemittelstaedt2009389-390-1-0[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).

• Definition

• Equivalent formulations

• Examples

• Spectral theorem

• Mercer's theorem

• Finite-rank operators

• Properties

• Lidskii's theorem

• Relationship between common classes of operators

• Trace class as the dual of compact operators

• As the predual of bounded operators

• See also

• References

• Bibliography

Let H {\displaystyle H} be a separable Hilbert space, { e k } k = 1 ∞ ∞ {\displaystyle \left\{e_{k}\right\}_{k=1}^{\infty }} an orthonormal basis and A : H → → H {\displaystyle A:H\to H} a positive bounded linear operator on H {\displaystyle H} . The trace of A {\displaystyle A} is denoted by Tr ⁡ ⁡ ( A ) {\displaystyle \operatorname {Tr} (A)} and defined ascite-ref-footnoteconway200086-2-0[2]cite-ref-footnotereedsimon1980206-3-0[3]

Tr ⁡ ⁡ ( A ) = ∑ ∑ k = 1 ∞ ∞ ⟨ A e k , e k ⟩ , {\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 T : H → → H {\displaystyle T:H\rightarrow H} is called trace class if and only if

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

where | T | := T ∗ ∗ T {\displaystyle |T|:={\sqrt {T^{*}T}}} denotes the positive-semidefinite Hermitian square root.cite-ref-footnotereedsimon1980196-4-0[4]

The trace-norm of a trace class operator T is defined as ‖ ‖ T ‖ ‖ 1 := 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 B 1 ( H ) {\displaystyle B_{1}(H)} and that B 1 ( H ) {\displaystyle B_{1}(H)} , with the trace-norm, becomes a Banach space.

When H {\displaystyle H} is finite-dimensional, every (positive) operator is trace class. For A {\displaystyle A} this definition coincides with that of the trace of a matrix. If H {\displaystyle H} is complex, then A {\displaystyle A} is always self-adjoint (i.e. A = A ∗ ∗ = | A | {\displaystyle A=A^{*}=|A|} ) though the converse is not necessarily true.cite-ref-footnotereedsimon1980195-5-0[5]

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

• Tr ⁡ ⁡ ( | T | ) = ∑ ∑ k ⟨ | T | e k , e k ⟩ {\textstyle \operatorname {Tr} (|T|)=\sum _{k}\left\langle |T|\,e_{k},e_{k}\right\rangle } is finite for every orthonormal basis ( e k ) k {\displaystyle \left(e_{k}\right)_{k}} of H.cite-ref-footnoteconway200086-2-1[2]

• T is a nuclear operator.cite-ref-footnotetr-ves2006494-6-0[6]cite-ref-footnoteconway200089-7-0[7] There exist two orthogonal sequences ( x i ) i = 1 ∞ ∞ {\displaystyle \left(x_{i}\right)_{i=1}^{\infty }} and ( y i ) i = 1 ∞ ∞ {\displaystyle \left(y_{i}\right)_{i=1}^{\infty }} in H {\displaystyle H} and positive real numbers ( λ λ i ) i = 1 ∞ ∞ {\displaystyle \left(\lambda _{i}\right)_{i=1}^{\infty }} in ℓ 1 {\displaystyle \ell ^{1}} such that ∑ ∑ i = 1 ∞ ∞ λ λ i < ∞ ∞ {\textstyle \sum _{i=1}^{\infty }\lambda _{i}<\infty } and x ↦ ↦ T ( x ) = ∑ ∑ i = 1 ∞ ∞ λ λ i ⟨ x , x i ⟩ y i , ∀ ∀ x ∈ ∈ 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 ( λ λ i ) i = 1 ∞ ∞ {\displaystyle \left(\lambda _{i}\right)_{i=1}^{\infty }} are the singular values of T (or, equivalently, the eigenvalues of | T | {\displaystyle |T|} ), with each value repeated as often as its multiplicity.cite-ref-footnotereedsimon1980203-204-209-8-0[8]

• T is a compact operator with Tr ⁡ ⁡ ( | T | ) < ∞ ∞ . {\displaystyle \operatorname {Tr} (|T|)<\infty .} If T is trace class thencite-ref-footnoteconway1990268-9-0[9] ‖ ‖ T ‖ ‖ 1 = sup { | Tr ⁡ ⁡ ( C T ) | : ‖ ‖ C ‖ ‖ ≤ ≤ 1 and C : H → → H is a compact operator } . {\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.cite-ref-footnotetr-ves2006502-508-10-0[10]

• T is equal to the composition of two Hilbert-Schmidt operators.cite-ref-footnoteconway1990267-11-0[11]

• | T | {\textstyle {\sqrt {|T|}}} is a Hilbert-Schmidt operator.cite-ref-footnoteconway1990267-11-1[11]

Let T {\displaystyle T} be a bounded self-adjoint operator on a Hilbert space. Then T 2 {\displaystyle T^{2}} is trace class if and only if T {\displaystyle T} has a pure point spectrum with eigenvalues { λ λ i ( T ) } i = 1 ∞ ∞ {\displaystyle \left\{\lambda _{i}(T)\right\}_{i=1}^{\infty }} such thatcite-ref-footnotesimon201021-12-0[12]

Tr ⁡ ⁡ ( T 2 ) = ∑ ∑ i = 1 ∞ ∞ λ λ i ( T 2 ) < ∞ ∞ . {\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 K {\displaystyle K} is a continuous symmetric positive-definite kernel on L 2 ( [ a , b ] ) {\displaystyle L^{2}([a,b])} , defined as

K ( s , t ) = ∑ ∑ j = 1 ∞ ∞ λ λ j e j ( s ) e j ( t ) {\displaystyle K(s,t)=\sum _{j=1}^{\infty }\lambda _{j}\,e_{j}(s)\,e_{j}(t)}

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

Tr ⁡ ⁡ ( T K ) = ∫ ∫ a b K ( t , t ) d t = ∑ ∑ i λ λ i . {\displaystyle \operatorname {Tr} (T_{K})=\int _{a}^{b}K(t,t)\,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 B 1 ( H ) {\displaystyle B_{1}(H)} (when endowed with the trace norm).cite-ref-footnoteconway1990268-9-1[9]

Given any x , y ∈ ∈ H , {\displaystyle x,y\in H,} define the operator x ⊗ ⊗ y : H → → H {\displaystyle x\otimes y:H\to H} by ( x ⊗ ⊗ y ) ( z ) := ⟨ ⟨ z , y ⟩ ⟩ x . {\displaystyle (x\otimes y)(z):=\langle z,y\rangle x.} Then x ⊗ ⊗ 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), Tr ⁡ ⁡ ( A ( x ⊗ ⊗ y ) ) = ⟨ ⟨ A x , y ⟩ ⟩ . {\displaystyle \operatorname {Tr} (A(x\otimes y))=\langle Ax,y\rangle .} cite-ref-footnoteconway1990268-9-2[9]

1. If A : H → → H {\displaystyle A:H\to H} is a non-negative self-adjoint operator, then A {\displaystyle A} is trace-class if and only if Tr ⁡ ⁡ A < ∞ ∞ . {\displaystyle \operatorname {Tr} A<\infty .} Therefore, a self-adjoint operator A {\displaystyle A} is trace-class if and only if its positive part A + {\displaystyle A^{+}} and negative part 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, Tr ⁡ ⁡ ( a A + b B ) = a Tr ⁡ ⁡ ( A ) + b Tr ⁡ ⁡ ( B ) . {\displaystyle \operatorname {Tr} (aA+bB)=a\operatorname {Tr} (A)+b\operatorname {Tr} (B).} The bilinear map ⟨ ⟨ A , B ⟩ ⟩ = 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. Tr : B 1 ( H ) → → C {\displaystyle \operatorname {Tr} :B_{1}(H)\to \mathbb {C} } is a positive linear functional such that if T {\displaystyle T} is a trace class operator satisfying T ≥ ≥ 0 and Tr ⁡ ⁡ T = 0 , {\displaystyle T\geq 0{\text{ and }}\operatorname {Tr} T=0,} then T = 0. {\displaystyle T=0.} cite-ref-footnoteconway1990267-11-2[11]

4. If T : H → → H {\displaystyle T:H\to H} is trace-class then so is T ∗ ∗ {\displaystyle T^{*}} and ‖ ‖ T ‖ ‖ 1 = ‖ T ∗ ∗ ‖ 1 . {\displaystyle \|T\|_{1}=\left\|T^{*}\right\|_{1}.} cite-ref-footnoteconway1990267-11-3[11]

5. If A : H → → H {\displaystyle A:H\to H} is bounded, and T : H → → H {\displaystyle T:H\to H} is trace-class, then A T {\displaystyle AT} and T A {\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), andcite-ref-footnoteconway1990267-11-4[11]cite-ref-footnotereedsimon1980218-13-0[13] ‖ ‖ A T ‖ ‖ 1 = Tr ⁡ ⁡ ( | A T | ) ≤ ≤ ‖ ‖ A ‖ ‖ ‖ ‖ T ‖ ‖ 1 , ‖ ‖ T A ‖ ‖ 1 = Tr ⁡ ⁡ ( | T A | ) ≤ ≤ ‖ ‖ 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,cite-ref-footnoteconway1990267-11-5[11] Tr ⁡ ⁡ ( A T ) = Tr ⁡ ⁡ ( T A ) {\displaystyle \operatorname {Tr} (AT)=\operatorname {Tr} (TA)} and | Tr ⁡ ⁡ ( A T ) | ≤ ≤ ‖ ‖ 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 ( e k ) k {\displaystyle \left(e_{k}\right)_{k}} and ( f k ) k {\displaystyle \left(f_{k}\right)_{k}} are two orthonormal bases of H and if T is trace class then ∑ ∑ k | ⟨ T e k , f k ⟩ | ≤ ≤ ‖ ‖ T ‖ ‖ 1 . {\textstyle \sum _{k}\left|\left\langle Te_{k},f_{k}\right\rangle \right|\leq \|T\|_{1}.} cite-ref-footnoteconway1990268-9-3[9]

7. If A is trace-class, then one can define the Fredholm determinant of I + A {\displaystyle I+A} : det ( I + A ) := ∏ ∏ n ≥ ≥ 1 [ 1 + λ λ n ( A ) ] , {\displaystyle \det(I+A):=\prod _{n\geq 1}[1+\lambda _{n}(A)],} where { λ λ n ( A ) } n {\displaystyle \{\lambda _{n}(A)\}_{n}} is the spectrum of A . {\displaystyle A.} The trace class condition on A {\displaystyle A} guarantees that the infinite product is finite: indeed, det ( I + A ) ≤ ≤ e ‖ ‖ A ‖ ‖ 1 . {\displaystyle \det(I+A)\leq e^{\|A\|_{1}}.} It also implies that det ( I + A ) ≠ ≠ 0 {\displaystyle \det(I+A)\neq 0} if and only if ( I + A ) {\displaystyle (I+A)} is invertible.

8. If A : H → → H {\displaystyle A:H\to H} is trace class then for any orthonormal basis ( e k ) k {\displaystyle \left(e_{k}\right)_{k}} of H , {\displaystyle H,} the sum of positive terms ∑ ∑ k | ⟨ A e k , e k ⟩ | {\textstyle \sum _{k}\left|\left\langle A\,e_{k},e_{k}\right\rangle \right|} is finite.cite-ref-footnoteconway1990267-11-6[11]

9. If A = B ∗ ∗ C {\displaystyle A=B^{*}C} for some Hilbert-Schmidt operators B {\displaystyle B} and C {\displaystyle C} then for any normal vector e ∈ ∈ H , {\displaystyle e\in H,} | ⟨ ⟨ A e , e ⟩ ⟩ | = 1 2 ( ‖ ‖ B e ‖ ‖ 2 + ‖ ‖ C e ‖ ‖ 2 ) {\textstyle |\langle Ae,e\rangle |={\frac {1}{2}}\left(\|Be\|^{2}+\|Ce\|^{2}\right)} holds.cite-ref-footnoteconway1990267-11-7[11]

Let A {\displaystyle A} be a trace-class operator in a separable Hilbert space H , {\displaystyle H,} and let { λ λ n ( A ) } n = 1 N ≤ ≤ ∞ ∞ {\displaystyle \{\lambda _{n}(A)\}_{n=1}^{N\leq \infty }} be the eigenvalues of A . {\displaystyle A.} Let us assume that λ λ n ( A ) {\displaystyle \lambda _{n}(A)} are enumerated with algebraic multiplicities taken into account (that is, if the algebraic multiplicity of λ λ {\displaystyle \lambda } is k , {\displaystyle k,} then λ λ {\displaystyle \lambda } is repeated k {\displaystyle k} times in the list λ λ 1 ( A ) , λ λ 2 ( A ) , … … {\displaystyle \lambda _{1}(A),\lambda _{2}(A),\dots } ). Lidskii's theorem (named after Victor Borisovich Lidskii) states that Tr ⁡ ⁡ ( A ) = ∑ ∑ n = 1 N λ λ 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 ∑ ∑ n = 1 N | λ λ n ( A ) | ≤ ≤ ∑ ∑ 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 { λ λ n ( A ) } n = 1 N {\displaystyle \{\lambda _{n}(A)\}_{n=1}^{N}} and the singular values { s m ( A ) } m = 1 M {\displaystyle \{s_{m}(A)\}_{m=1}^{M}} of the compact operator A . {\displaystyle A.} cite-ref-14[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 ℓ ℓ 1 ( 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 ℓ ℓ 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 ℓ ℓ ∞ ∞ ( N ) , {\displaystyle \ell ^{\infty }(\mathbb {N} ),} the compact operators that of c 0 {\displaystyle c_{0}} (the sequences convergent to 0), Hilbert–Schmidt operators correspond to ℓ ℓ 2 ( N ) , {\displaystyle \ell ^{2}(\mathbb {N} ),} and finite-rank operators to 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 T {\displaystyle T} on a Hilbert space takes the following canonical form: there exist orthonormal bases ( u i ) i {\displaystyle (u_{i})_{i}} and ( v i ) i {\displaystyle (v_{i})_{i}} and a sequence ( α α i ) i {\displaystyle \left(\alpha _{i}\right)_{i}} of non-negative numbers with α α i → → 0 {\displaystyle \alpha _{i}\to 0} such that T x = ∑ ∑ i α α i ⟨ ⟨ x , v i ⟩ ⟩ u i for all x ∈ ∈ 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 T {\displaystyle T} is trace-class iff the series ∑ ∑ i α α i {\textstyle \sum _{i}\alpha _{i}} is convergent, T {\displaystyle T} is Hilbert–Schmidt iff ∑ ∑ i α α i 2 {\textstyle \sum _{i}\alpha _{i}^{2}} is convergent, and T {\displaystyle T} is finite-rank iff the sequence ( α α i ) 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 H {\displaystyle H} is infinite-dimensional: { finite rank } ⊆ ⊆ { trace class } ⊆ ⊆ { Hilbert--Schmidt } ⊆ ⊆ { 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 ‖ ‖ T ‖ ‖ 1 = Tr ⁡ ⁡ [ ( T ∗ ∗ T ) 1 / 2 ] = ∑ ∑ i α α 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 ‖ ‖ T ‖ ‖ 2 = [ Tr ⁡ ⁡ ( T ∗ ∗ T ) ] 1 / 2 = ( ∑ ∑ i α α i 2 ) 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 ‖ ‖ T ‖ ‖ = sup i ( α α i ) . {\textstyle \|T\|=\sup _{i}\left(\alpha _{i}\right).} By classical inequalities regarding sequences, ‖ ‖ T ‖ ‖ ≤ ≤ ‖ ‖ T ‖ ‖ 2 ≤ ≤ ‖ ‖ T ‖ ‖ 1 {\displaystyle \|T\|\leq \|T\|_{2}\leq \|T\|_{1}} for appropriate 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 c 0 {\displaystyle c_{0}} is ℓ ℓ 1 ( N ) . {\displaystyle \ell ^{1}(\mathbb {N} ).} Similarly, we have that the dual of compact operators, denoted by K ( H ) ∗ ∗ , {\displaystyle K(H)^{*},} is the trace-class operators, denoted by B 1 . {\displaystyle B_{1}.} The argument, which we now sketch, is reminiscent of that for the corresponding sequence spaces. Let f ∈ ∈ K ( H ) ∗ ∗ , {\displaystyle f\in K(H)^{*},} we identify f {\displaystyle f} with the operator T f {\displaystyle T_{f}} defined by ⟨ ⟨ T f x , y ⟩ ⟩ = f ( S x , y ) , {\displaystyle \langle T_{f}x,y\rangle =f\left(S_{x,y}\right),} where S x , y {\displaystyle S_{x,y}} is the rank-one operator given by S x , y ( h ) = ⟨ ⟨ h , y ⟩ ⟩ x . {\displaystyle S_{x,y}(h)=\langle h,y\rangle x.}

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

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

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

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

• Nuclear operators between Banach spaces

cite-note-footnotemittelstaedt2009389-390-11. ↑ Mittelstaedt 2009, pp. 389–390.

cite-note-footnoteconway200086-22. ↑ Conway 2000, p. 86.

cite-note-footnotereedsimon1980206-33. ↑ Reed & Simon 1980, p. 206.

cite-note-footnotereedsimon1980196-44. ↑ Reed & Simon 1980, p. 196.

cite-note-footnotereedsimon1980195-55. ↑ Reed & Simon 1980, p. 195.

cite-note-footnotetr-ves2006494-66. ↑ Trèves 2006, p. 494.

cite-note-footnoteconway200089-77. ↑ Conway 2000, p. 89.

cite-note-footnotereedsimon1980203-204-209-88. ↑ Reed & Simon 1980, pp. 203–204, 209.

cite-note-footnoteconway1990268-99. ↑ Conway 1990, p. 268.

cite-note-footnotetr-ves2006502-508-1010. ↑ Trèves 2006, pp. 502–508.

cite-note-footnoteconway1990267-1111. ↑ Conway 1990, p. 267.

cite-note-footnotesimon201021-1212. ↑ Simon 2010, p. 21.

cite-note-footnotereedsimon1980218-1313. ↑ Reed & Simon 1980, p. 218.

cite-note-1414. ↑ Simon, B. (2005) Trace ideals and their applications, Second Edition, American Mathematical Society.

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