# Nuclear operator

In mathematics, nuclear operators are an important class of linear operators introduced by Alexander Grothendieck in his doctoral dissertation. Nuclear operators are intimately tied to the projective tensor product of two topological vector spaces (TVSs).

## Preliminaries and notation

Throughout let X,Y, and Z be topological vector spaces (TVSs) and L : XY be a linear operator (no assumption of continuity is made unless otherwise stated).

• The projective tensor product of two locally convex TVSs X and Y is denoted by ${\displaystyle X\otimes _{\pi }Y}$  and the completion of this space will be denoted by ${\displaystyle X{\widehat {\otimes }}_{\pi }Y}$ .
• L : XY is a topological homomorphism or homomorphism, if it is linear, continuous, and ${\displaystyle L:X\to \operatorname {Im} L}$  is an open map, where ${\displaystyle \operatorname {Im} L}$ , the image of L, has the subspace topology induced by Y.
• If S is a subspace of X then both the quotient map XX/S and the canonical injection SX are homomorphisms.
• The set of continuous linear maps XZ (resp. continuous bilinear maps ${\displaystyle X\times Y\to Z}$ ) will be denoted by L(X, Z) (resp. B(X, Y; Z)) where if Z is the underlying scalar field then we may instead write L(X) (resp. B(X, Y)).
• Any linear map ${\displaystyle L:X\to Y}$  can be canonically decomposed as follows: ${\displaystyle X\to X/\operatorname {ker} L\;{\overset {L_{0}}{\rightarrow }}\;\operatorname {Im} L\to Y}$  where ${\displaystyle L_{0}\left(x+\operatorname {ker} L\right):=L(x)}$  defines a bijection called the canonical bijection associated with L.
• X* or ${\displaystyle X^{\prime }}$  will denote the continuous dual space of X.
• To increase the clarity of the exposition, we use the common convention of writing elements of ${\displaystyle X^{\prime }}$  with a prime following the symbol (e.g. ${\displaystyle x^{\prime }}$  denotes an element of ${\displaystyle X^{\prime }}$  and not, say, a derivative and the variables x and ${\displaystyle x^{\prime }}$  need not be related in any way).
• ${\displaystyle X^{\#}}$  will denote the algebraic dual space of X (which is the vector space of all linear functionals on X, whether continuous or not).
• A linear map L : HH from a Hilbert space into itself is called positive if ${\displaystyle \langle L(x),X\rangle \geq 0}$  for every ${\displaystyle x\in H}$ . In this case, there is a unique positive map r : HH, called the square-root of L, such that ${\displaystyle L=r\circ r}$ .[1]
• If ${\displaystyle L:H_{1}\to H_{2}}$  is any continuous linear map between Hilbert spaces, then ${\displaystyle L^{*}\circ L}$  is always positive. Now let R : HH denote its positive square-root, which is called the absolute value of L. Define ${\displaystyle U:H_{1}\to H_{2}}$  first on ${\displaystyle \operatorname {Im} R}$  by setting ${\displaystyle U(x)=L(x)}$  for ${\displaystyle x=R\left(x_{1}\right)\in \operatorname {Im} R}$  and extending ${\displaystyle U}$  continuously to ${\displaystyle {\overline {\operatorname {Im} R}}}$ , and then define U on ${\displaystyle \operatorname {ker} R}$  by setting ${\displaystyle U(x)=0}$  for ${\displaystyle x\in \operatorname {ker} R}$  and extend this map linearly to all of ${\displaystyle H_{1}}$ . The map ${\displaystyle U{\big \vert }_{\operatorname {Im} R}:\operatorname {Im} R\to \operatorname {Im} L}$  is a surjective isometry and ${\displaystyle L=U\circ R}$ .
• A linear map ${\displaystyle \Lambda :X\to Y}$  is called compact or completely continuous if there is a neighborhood U of the origin in X such that ${\displaystyle \Lambda (U)}$  is precompact in Y.[2]
• In a Hilbert space, positive compact linear operators, say L : HH have a simple spectral decomposition discovered at the beginning of the 20th century by Fredholm and F. Riesz:[3]
There is a sequence of positive numbers, decreasing and either finite or else converging to 0, ${\displaystyle r_{1}>r_{2}>\cdots >r_{k}>\cdots }$  and a sequence of nonzero finite dimensional subspaces ${\displaystyle V_{i}}$  of H (i = 1, 2, ${\displaystyle \ldots }$ ) with the following properties: (1) the subspaces ${\displaystyle V_{i}}$  are pairwise orthogonal; (2) for every i and every ${\displaystyle x\in V_{i}}$ , ${\displaystyle L(x)=r_{i}x}$ ; and (3) the orthogonal of the subspace spanned by ${\displaystyle \cup _{i}V_{i}}$  is equal to the kernel of L.[3]

### Notation for topologies

• 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).

### A canonical tensor product as a subspace of the dual of Bi(X, Y)

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 XY then it can be shown that XY together with 𝜒 forms a tensor product of X and Y (where xy := 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(XY; Z) → Bi(X, Y; Z) given by uu𝜒 is an isomorphism of vector spaces. In particular, this allows us to identify the algebraic dual of XY with the space of bilinear forms on X × Y.[4] Moreover, if X and Y are locally convex topological vector spaces (TVSs) and if XY 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 XY 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]

## Nuclear operators between Banach spaces

There is a canonical vector space embedding ${\displaystyle I:X^{\prime }\otimes Y\to L(X;Y)}$  defined by sending ${\displaystyle z:=\sum _{i}^{n}x_{i}^{\prime }\otimes y_{i}}$  to the map

${\displaystyle x\mapsto \sum _{i}^{n}x_{i}^{\prime }(x)y_{i}}$ .

Assuming that X and Y are Banach spaces, then the map ${\displaystyle I:X_{b}^{\prime }\otimes _{\pi }Y\to L_{b}(X;Y)}$  has norm ${\displaystyle 1}$  (to see that the norm is ${\displaystyle \leq 1}$ , note that ${\displaystyle \|I(z)\|=\sup _{\|x\|\leq 1}\|I(z)(x)\|=\sup _{\|x\|\leq 1}\left\|\sum _{i=1}^{n}x_{i}^{\prime }(x)y_{i}\right\|\leq \sup _{\|x\|\leq 1}\sum _{i=1}^{n}\left\|x_{i}^{\prime }\right\|\|x\|\left\|y_{i}\right\|\leq \sum _{i=1}^{n}\left\|x_{i}^{\prime }\right\|\left\|y_{i}\right\|}$  so that ${\displaystyle \|I(z)\|\leq \|z\|_{\pi }}$ ). Thus it has a continuous extension to a map ${\displaystyle {\hat {I}}:X_{b}^{\prime }{\widehat {\otimes }}_{\pi }Y\to L_{b}(X;Y)}$ , where it is known that this map is not necessarily injective.[6] The range of this map is denoted by ${\displaystyle L^{1}(X;Y)}$  and its elements are called nuclear operators.[7] ${\displaystyle L^{1}(X;Y)}$  is TVS-isomorphic to ${\displaystyle \left(X_{b}^{\prime }{\widehat {\otimes }}_{\pi }Y\right)/\operatorname {ker} {\hat {I}}}$  and the norm on this quotient space, when transferred to elements of ${\displaystyle L^{1}(X;Y)}$  via the induced map ${\displaystyle {\hat {I}}:\left(X_{b}^{\prime }{\widehat {\otimes }}_{\pi }Y\right)/\operatorname {ker} {\hat {I}}\to L^{1}(X;Y)}$ , is called the trace-norm and is denoted by ${\displaystyle \|\cdot \|_{\operatorname {Tr} }}$ . Explicitely,[clarification needed explicitly or especially?] if ${\displaystyle T:X\to Y}$  is a nuclear operator then ${\displaystyle \left\|T\right\|_{\operatorname {Tr} }:=\inf _{z\in {\hat {I}}^{-1}\left(T\right)}\left\|z\right\|_{\pi }}$ .

### Characterization

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]

### Properties

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]

## Nuclear operators between Hilbert spaces

Nuclear automorphisms of a Hilbert space are called trace class operators.

Let X and Y be Hilbert spaces and let N : XY be a continuous linear map. Suppose that ${\displaystyle N=U\circ R}$  where R : XX is the square-root of ${\displaystyle N^{*}\circ N}$  and U : XY is such that ${\displaystyle U{\big \vert }_{\operatorname {Im} R}:\operatorname {Im} R\to \operatorname {Im} L}$  is a surjective isometry and ${\displaystyle N=U\circ R}$ . Then N is a nuclear map if and only if R is a nuclear map; hence, to study nuclear maps between Hilbert spaces it suffices to restrict one's attention to positive linear operators.[10]

### Characterizations

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

1. N : XY is nuclear.
2. R : XX is nuclear.[11]
3. R : XX 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 : XY' is an integral map.[12]

## Nuclear operators between locally convex spaces

Suppose that U is a convex balanced closed neighborhood of the origin in X and B is a convex balanced bounded Banach disk in Y with both X and Y locally convex spaces. Let ${\displaystyle p_{U}(x)=\inf _{r>0,x\in rU}r}$  and let ${\displaystyle \pi :X\to X/p_{U}^{-1}(0)}$  be the canonical projection. One can define the auxiliary Banach space ${\displaystyle {\hat {X}}_{U}}$  with the canonical map ${\displaystyle {\hat {\pi }}_{U}:X\to {\hat {X}}_{U}}$  whose image, ${\displaystyle X/p_{U}^{-1}(0)}$ , is dense in ${\displaystyle {\hat {X}}_{U}}$  as well as the auxiliary space ${\displaystyle F_{B}=\operatorname {span} B}$  normed by ${\displaystyle p_{B}(y)=\inf _{r>0,y\in rB}r}$  and with a canonical map ${\displaystyle \iota :F_{B}\to F}$  being the (continuous) canonical injection. Given any continuous linear map ${\displaystyle T:{\hat {X}}_{U}\to Y_{B}}$  one obtains through composition the continuous linear map ${\displaystyle {\hat {\pi }}_{U}\circ T\circ \iota :X\to Y}$ ; thus we have an injection ${\displaystyle L\left({\hat {X}}_{U};Y_{B}\right)\to L(X;Y)}$  and we henceforth use this map to identify ${\displaystyle L\left({\hat {X}}_{U};Y_{B}\right)}$  as a subspace of ${\displaystyle L(X;Y)}$ .[7]

Definition: Let X and Y be Hausdorff locally convex spaces. The union of all ${\displaystyle L^{1}\left({\hat {X}}_{U};Y_{B}\right)}$  as U ranges over all closed convex balanced neighborhoods of the origin in X and B ranges over all bounded Banach disks in Y, is denoted by ${\displaystyle L^{1}(X;Y)}$  and its elements are call nuclear mappings of X into Y.[7]

When X and Y are Banach spaces, then this new definition of nuclear mapping is consistent with the original one given for the special case where X and Y are Banach spaces.

### Sufficient conditions for nuclearity

• 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]

### Characterizations

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]

### Properties

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]

## References

1. ^ Trèves 2006, p. 488.
2. ^ a b c Trèves 2006, p. 483.
3. ^ a b Trèves 2006, p. 490.
4. ^ Schaefer & Wolff 1999, p. 92.
5. ^ a b Schaefer & Wolff 1999, p. 93.
6. ^ a b c Schaefer & Wolff 1999, p. 98.
7. ^ a b c Trèves 2006, pp. 478-479.
8. Trèves 2006, pp. 481-483.
9. ^ a b c Trèves 2006, p. 484.
10. ^ a b Trèves 2006, pp. 483-484.
11. ^ a b c Trèves 2006, pp. 492-494.
12. ^ Trèves 2006, pp. 502-508.
13. ^ Trèves 2006, pp. 479-481.
14. ^ a b Schaefer & Wolff 1999, p. 100.
15. ^ a b Trèves 2006, p. 485.

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