Nuclear operator

• σ(X, X′) denotes the coarsest topology on X making every map in X′ continuous and ${\displaystyle X_{\sigma \left(X,X^{\prime }\right)}}$ or ${\displaystyle X_{\sigma }}$ denotes X endowed with this topology.
• σ(X′, X) denotes weak-* topology on X* and ${\displaystyle X_{\sigma \left(X^{\prime },X\right)}}$ or ${\displaystyle X_{\sigma }^{\prime }}$ denotes X′ endowed with this topology.
  • Note that every ${\displaystyle x_{0}\in X}$ induces a map ${\displaystyle X^{\prime }\to \mathbb {R} }$ defined by ${\displaystyle \lambda \mapsto \lambda \left(x_{0}\right)}$. σ(X′, X) is the coarsest topology on X′ making all such maps continuous.
• b(X, X′) denotes the topology of bounded convergence on X and ${\displaystyle X_{b\left(X,X^{\prime }\right)}}$ or ${\displaystyle X_{b}}$ denotes X endowed with this topology.
• b(X′, X) denotes the topology of bounded convergence on X′ or the strong dual topology on X′ and ${\displaystyle X_{b\left(X^{\prime },X\right)}}$ or ${\displaystyle X_{b}^{\prime }}$ denotes X′ endowed with this topology.
  • As usual, if X* is considered as a topological vector space but it has not been made clear what topology it is endowed with, then the topology will be assumed to be b(X′, X).

Let X and Y be vector spaces (no topology is needed yet) and let Bi(X, Y) be the space of all bilinear maps defined on ${\displaystyle X\times Y}$ and going into the underlying scalar field.

For every ${\displaystyle (x,y)\in X\times Y}$, let ${\displaystyle \chi _{(x,y)}}$ be the canonical bilinear form on Bi(X, Y) defined by ${\displaystyle \chi _{(x,y)}(u):=u(x,y)}$ for every u ∈ Bi(X, Y). This induces a canonical map ${\displaystyle \chi :X\times Y\to Bi(X,Y)^{\#}}$ defined by ${\displaystyle \chi (x,y):=\chi _{(x,y)}}$, where ${\displaystyle Bi(X,Y)^{\#}}$ denotes the algebraic dual of Bi(X, Y). If we denote the span of the range of 𝜒 by X ⊗ Y then it can be shown that X ⊗ Y together with 𝜒 forms a tensor product of X and Y (where x ⊗ y := 𝜒(x, y)). This gives us a canonical tensor product of X and Y.

If Z is any other vector space then the mapping Li(X ⊗ Y; Z) → Bi(X, Y; Z) given by u ↦ u ∘ 𝜒 is an isomorphism of vector spaces. In particular, this allows us to identify the algebraic dual of X ⊗ Y with the space of bilinear forms on X × Y.[4] Moreover, if X and Y are locally convex topological vector spaces (TVSs) and if X ⊗ Y is given the 𝜋-topology then for every locally convex TVS Z, this map restricts to a vector space isomorphism ${\displaystyle L(X\otimes _{\pi }Y;Z)\to B(X,Y;Z)}$ from the space of continuous linear mappings onto the space of continuous bilinear mappings.[5] In particular, the continuous dual of X ⊗ Y can be canonically identified with the space B(X, Y) of continuous bilinear forms on X × Y; furthermore, under this identification the equicontinuous subsets of B(X, Y) are the same as the equicontinuous subsets of ${\displaystyle (X\otimes _{\pi }Y)^{\prime }}$.[5]

Suppose that X and Y are Banach spaces and that ${\displaystyle N:X\to Y}$ is a continuous linear operator.

• The following are equivalent:

1. ${\displaystyle N:X\to Y}$ is nuclear.
2. There exists an sequence ${\displaystyle \left(x_{i}^{\prime }\right)_{i=1}^{\infty }}$ in the closed unit ball of ${\displaystyle X^{\prime }}$, a sequence ${\displaystyle \left(y_{i}\right)_{i=1}^{\infty }}$ in the closed unit ball of ${\displaystyle Y}$, and a complex sequence ${\displaystyle \left(c_{i}\right)_{i=1}^{\infty }}$ such that ${\displaystyle \sum _{i=1}^{\infty }|c_{i}|<\infty }$ and ${\displaystyle N}$ is equal to the mapping:[8] ${\displaystyle N(x)=\sum _{i=1}^{\infty }c_{i}x_{i}^{\prime }(x)y_{i}}$ for all ${\displaystyle x\in X}$. Furthermore, the trace-norm ${\displaystyle \|N\|_{\operatorname {Tr} }}$ is equal to the infimum of the numbers ${\displaystyle \sum _{i=1}^{\infty }|c_{i}|}$ over the set of all representations of ${\displaystyle N}$ as such a series.[8]

• If Y is reflexive then ${\displaystyle N:X\to Y}$ is a nuclear if and only if ${\displaystyle {}^{t}N:Y_{b}^{\prime }\to X_{b}^{\prime }}$ is nuclear, in which case ${\displaystyle \|{}^{t}N\|_{\operatorname {Tr} }=\|N\|_{\operatorname {Tr} }}$. [9]

Let X and Y be Banach spaces and let ${\displaystyle N:X\to Y}$ be a continuous linear operator.

• If ${\displaystyle N:X\to Y}$ is a nuclear map then its transpose ${\displaystyle {}^{t}N:Y_{b}^{\prime }\to X_{b}^{\prime }}$ is a continuous nuclear map (when the dual spaces carry their strong dual topologies) and ${\displaystyle \|{}^{t}N\|_{\operatorname {Tr} }\leq \|N\|_{\operatorname {Tr} }}$.[10]

Let X and Y be Hilbert spaces and let N : X → Y be a continuous linear map whose absolute value is R : X → X. The following are equivalent:

1. N : X → Y is nuclear.
2. R : X → X is nuclear.[11]
3. R : X → X is compact and ${\displaystyle \operatorname {Tr} R}$ is finite, in which case ${\displaystyle \operatorname {Tr} R=\|N\|_{\operatorname {Tr} }}$.[11]
   • Here, ${\displaystyle \operatorname {Tr} R}$ is the trace of R and it is defined as follows: Since R is a continuous compact positive operator, there exists a (possibly finite) sequence ${\displaystyle \lambda _{1}>\lambda _{2}>\cdots }$ of positive numbers with corresponding non-trivial finite-dimensional and mutually orthogonal vector spaces ${\displaystyle V_{1},V_{2},\ldots }$ such that the orthogonal (in H) of ${\displaystyle \operatorname {span} \left(V_{1}\cup V_{2}\cup \cdots \right)}$ is equal to ${\displaystyle \operatorname {ker} R}$ (and hence also to ${\displaystyle \operatorname {ker} N}$) and for all k, ${\displaystyle R(x)=\lambda _{k}x}$ for all ${\displaystyle x\in V_{k}}$; the trace is defined as ${\displaystyle \operatorname {Tr} R:=\sum _{k}\lambda _{k}\operatorname {dim} V_{k}}$.
4. ${\displaystyle {}^{t}N:Y_{b}^{\prime }\to X_{b}^{\prime }}$ is nuclear, in which case ${\displaystyle \|{}^{t}N\|_{\operatorname {Tr} }=\|N\|_{\operatorname {Tr} }}$. [9]
5. There are two orthogonal sequences ${\displaystyle \left(x_{i}\right)_{i=1}^{\infty }}$ in X and ${\displaystyle \left(y_{i}\right)_{i=1}^{\infty }}$ in Y, and a sequence ${\displaystyle \left(\lambda _{i}\right)_{i=1}^{\infty }}$ in ${\displaystyle l^{1}}$ such that for all ${\displaystyle x\in X}$, ${\displaystyle N(x)=\sum _{i}\lambda _{i}\langle x,x_{i}\rangle y_{i}}$.[11]
6. N : X → Y' is an integral map.[12]

• Let W, X, Y, and Z be Hausdorff locally convex spaces, ${\displaystyle N:X\to Y}$ a nuclear map, and ${\displaystyle M:W\to X}$ and ${\displaystyle P:Y\to Z}$ be continuous linear maps. Then ${\displaystyle N\circ M:W\to Y}$, ${\displaystyle P\circ N:X\to Z}$, and ${\displaystyle P\circ N\circ M:W\to Z}$ are nuclear and if in addition W, X, Y, and Z are all Banach spaces then ${\displaystyle \|P\circ N\circ M\|_{\operatorname {Tr} }\leq \|P\|\|N\|_{\operatorname {Tr} }\|\|M\|}$.[13][14]
• If ${\displaystyle N:X\to Y}$ is a nuclear map between two Hausdorff locally convex spaces, then its transpose ${\displaystyle {}^{t}N:Y_{b}^{\prime }\to X_{b}^{\prime }}$ is a continuous nuclear map (when the dual spaces carry their strong dual topologies).[2]
  • If in addition X and Y are Banach spaces, then ${\displaystyle \|{}^{t}N\|_{\operatorname {Tr} }\leq \|N\|_{\operatorname {Tr} }}$.[9]
• If ${\displaystyle N:X\to Y}$ is a nuclear map between two Hausdorff locally convex spaces and if ${\displaystyle {\hat {X}}}$ is a completion of X, then the unique continuous extension ${\displaystyle {\hat {N}}:{\hat {X}}\to Y}$ of N is nuclear.[14]

Let X and Y be Hausdorff locally convex spaces and let ${\displaystyle N:X\to Y}$ be a continuous linear operator.

• The following are equivalent:

1. ${\displaystyle N:X\to Y}$ is nuclear.
2. (Definition) There exists a convex balanced neighborhood U of the origin in X and a bounded Banach disk B in Y such that ${\displaystyle N(U)\subseteq B}$ and the induced map ${\displaystyle {\overline {N}}_{0}:{\hat {X}}_{U}\to Y_{B}}$ is nuclear, where ${\displaystyle {\overline {N}}_{0}}$ is the unique continuous extension of ${\displaystyle N_{0}:X_{U}\to Y_{B}}$, which is the unique map satisfying ${\displaystyle N=\operatorname {In} _{B}\circ N_{0}\circ \pi _{U}}$ where ${\displaystyle \operatorname {In} _{B}:Y_{B}\to Y}$ is the natural inclusion and ${\displaystyle \pi _{U}:X\to X/p_{U}^{-1}(0)}$ is the canonical projection.[6]
3. There exist Banach spaces ${\displaystyle B_{1}}$ and ${\displaystyle B_{2}}$ and continuous linear maps ${\displaystyle f:X\to B_{1}}$, ${\displaystyle n:B_{1}\to B_{2}}$, and ${\displaystyle g:B_{2}\to Y}$ such that ${\displaystyle n:B_{1}\to B_{2}}$ is nuclear and ${\displaystyle N=g\circ n\circ f}$.[8]
4. There exists an equicontinuous sequence ${\displaystyle \left(x_{i}^{\prime }\right)_{i=1}^{\infty }}$ in ${\displaystyle X^{\prime }}$, a bounded Banach disk ${\displaystyle B\subseteq Y}$, a sequence ${\displaystyle \left(y_{i}\right)_{i=1}^{\infty }}$ in B, and a complex sequence ${\displaystyle \left(c_{i}\right)_{i=1}^{\infty }}$ such that ${\displaystyle \sum _{i=1}^{\infty }|c_{i}|<\infty }$ and ${\displaystyle N}$ is equal to the mapping:[8] ${\displaystyle N(x)=\sum _{i=1}^{\infty }c_{i}x_{i}^{\prime }(x)y_{i}}$ for all ${\displaystyle x\in X}$.

• If X is barreled and Y is quasi-complete, then N is nuclear if and only if N has a representation of the form ${\displaystyle N(x)=\sum _{i=1}^{\infty }c_{i}x_{i}^{\prime }(x)y_{i}}$ with ${\displaystyle \left(x_{i}^{\prime }\right)_{i=1}^{\infty }}$ bounded in ${\displaystyle X^{\prime }}$, ${\displaystyle \left(y_{i}\right)_{i=1}^{\infty }}$ bounded in Y and ${\displaystyle \sum _{i=1}^{\infty }|c_{i}|<\infty }$.[8]

The following is a type of Hahn-Banach theorem for extending nuclear maps:

• If ${\displaystyle E:X\to Z}$ is a TVS-embedding and ${\displaystyle N:X\to Y}$ is a nuclear map then there exists a nuclear map ${\displaystyle {\tilde {N}}:Z\to Y}$ such that ${\displaystyle {\tilde {N}}\circ E=N}$. Furthermore, when X and Y are Banach spaces and E is an isometry then for any ${\displaystyle \epsilon >0}$, ${\displaystyle {\tilde {N}}}$ can be picked so that ${\displaystyle \|{\tilde {N}}\|_{\operatorname {Tr} }\leq \|N\|_{\operatorname {Tr} }+\epsilon }$.[15]
• Suppose that ${\displaystyle E:X\to Z}$ is a TVS-embedding whose image is closed in Z and let ${\displaystyle \pi :Z\to Z/\operatorname {Im} E}$ be the canonical projection. Suppose all that every compact disk in ${\displaystyle Z/\operatorname {Im} E}$ is the image under ${\displaystyle \pi }$ of a bounded Banach disk in Z (this is true, for instance, if X and Z are both Fréchet spaces, or if Z is the strong dual of a Fréchet space and ${\displaystyle \operatorname {Im} E}$ is weakly closed in Z). Then for every nuclear map ${\displaystyle N:Y\to Z/\operatorname {Im} E}$ there exists a nuclear map ${\displaystyle {\tilde {N}}:Y\to Z}$ such that ${\displaystyle \pi \circ {\tilde {N}}=N}$.
  • Furthermore, when X and Z are Banach spaces and E is an isometry then for any ${\displaystyle \epsilon >0}$, ${\displaystyle {\tilde {N}}}$ can be picked so that ${\displaystyle \|{\tilde {N}}\|_{\operatorname {Tr} }\leq \|N\|_{\operatorname {Tr} }+\epsilon }$.[15]

Let X and Y be Hausdorff locally convex spaces and let ${\displaystyle N:X\to Y}$ be a continuous linear operator.

• Any nuclear map is compact.[2]
• For every topology of uniform convergence on ${\displaystyle L(X;Y)}$, the nuclear maps are contained in the closure of ${\displaystyle X^{\prime }\otimes Y}$ (when ${\displaystyle X^{\prime }\otimes Y}$ is viewed as a subspace of ${\displaystyle L(X;Y)}$).[6]