12.12 Nuclear Operators
Definition 12.12.1 (Nuclear Operator Between Banach Spaces).label Let $E, F$ be Banach spaces, $E^{*}$ be the dual of $E$, equipped with the uniform topology, and $T \in L(E; F)$, then $T$ is nuclear if there exists $\seq{\phi_n}\subset E^{*}$ and $\seq{y_n}\subset F$ such that:
- (1)
For each $x \in E$, $Tx = \sum_{n = 1}^{\infty} y_{n} \dpn{x, \phi_n}{E}$.
- (2)
$\sum_{n \in \natp}\norm{y_n}_{F}\norm{\phi_n}_{E^*}< \infty$.
The set $N(E; F)$ is the space of nuclear operators from $E$ to $F$. For each $T \in N(E; F)$, let
then $\norm{\cdot}_{N(E; F)}$ is a norm on $N(E; F)$, and $N(E; F)$ is a Banach space.
Lemma 12.12.2.label Let $E, F$ be Banach spaces, $E^{*}$ be the dual of $E$, equipped with the uniform topology, then the mapping
extends continuously into a surjective linear map $E^{*} \tilde \otimes_{\pi} F \to N(E; F)$.
Definition 12.12.3 (Nuclear Operator).label Let $E, F$ be separated locally convex spaces over $K \in \RC$ and $T \in L(E; F)$, then the following are equivalent:
- (1)
There exists convex and circled sets $U \in \cn_{E}(0)$ and $B \in B(F)$ such that:
- (a)
The auxiliary space $F_{B}$ is a Banach space.
- (b)
$T(U) \subset B$.
- (c)
The induced map $\wh E_{U} \to F_{B}$ is nuclear.
- (2)
There exists an equicontinuous sequence $\seq{\phi_n}\subset E^{*}$, a convex, circled, and bounded subset $B \subset F$, $\seq{y_n}\subset B$, and $\seq{\lambda_n}\subset K$ such that
- (a)
The auxiliary space $F_{B}$ is a Banach space.
- (b)
For each $x \in E$, $Tx = \sum_{n = 1}^{\infty} \lambda_{n} y_{n} \dpn{x, \phi_n}{E}$.
- (c)
$\sum_{n \in \natp}|\lambda_{n}| < \infty$.
If the above holds, then $T$ is nuclear. The set $N(E; F)$ is the space of nuclear operators from $E$ to $F$.
Proof, [Theorem III.7.1, SW99]. (1) $\Rightarrow$ (2): Let $\pi: E \to E_{U}$ be the canonical projection map associated with $E_{U}$ and $\iota: F_{B} \to F$ be the canonical inclusion map associated with $F_{B}$. By assumption (1b), there exists an induced map $\hat T: E_{U} \to F_{B}$ such that the following diagram commutes:
By the linear extension theorem, $E_{U}^{*} = (\wh E_{U})^{*}$. Assume without loss of generality that $E_{U}$ is a Banach space, then (1c) implies that $\hat T \in L(E_{U}; F_{B})$ is a nuclear operator. By Lemma 12.12.2 and Theorem 12.11.3, there exists $\seq{\phi_n}\subset E_{U}^{*}$, $\seq{y_n}\subset F_{B}$, and $\seq{\lambda_n}\subset K$ such that:
- (i)
$\sum_{n \in \natp}|\lambda_{n}| < \infty$.
- (ii)
$\limv{n}\phi_{n} = 0$ and $\limv{n}y_{n} = 0$.
- (iii)
For each $x \in E_{U}$, $\hat Tx = \sum_{n = 1}^{\infty} \lambda_{n} y_{n} \dpn{x, \phi_n}{E_U}$.
By (ii), $\sup_{n \in \natp}\norm{\phi_n}_{E_U^*}< \infty$ and $\sup_{n \in \natp}\norm{y_n}_{F_B}< \infty$, so $\seq{\phi_n}$ is equicontinuous, and there exists $R > 0$ such that $\seq{y_n}\subset RB$. After rescaling, assume without loss of generality that $\seq{y_n}\subset B$. By unraveling the factorisation, (iii) shows that for each $x \in E$,
Therefore the decomposition using $\seq{\phi_n \circ \pi}\subset E^{*}$, $B \subset F$, $\seq{y_n}\subset B$, and $\seq{\lambda_n}\subset K$ given above satisfies (2).
(2) $\Rightarrow$ (1): Since $\seq{\phi_n}$ is equicontinuous, $U = \bigcap_{n \in \natp}\phi_{n}^{-1}(B_{K}(0, 1))$ is a convex and circled neighbourhood of $0$ in $E$.
(1b): Using assumption (2c) and rescaling, assume without loss of generality that $\sum_{n = 1}^{\infty} |\lambda_{n}| < 1$. Let $\rho: F_{B} \to [0, \infty)$ be the gauge of $B$, then for any $x \in U$,
so $\rho(Tx) < 1$ and $Tx \in B$. Therefore $T(U) \subset B$.
(1c): Let $\pi: E \to E_{U}$ be the canonical projection map associated with $E_{U}$ and $n \in \natp$. By construction, $U \subset \phi_{n}^{-1}(B_{K}(0, 1))$, so there exists $\hat \phi_{n} \in E_{U}^{*}$ such that the following diagram commutes:
Thus for each $x \in E$,
and the induced map $\hat T: \wh E_{U} \to F_{B}$ takes the form $\hat Tx = \sum_{n = 1}^{\infty} y_{n} \dpn{x, \lambda_n \hat \phi_n}{\wh E_U}$. Finally, for each $n \in \natp$, $U \subset \phi_{n}^{-1}(B_{K}(0, 1))$, and $\normn{\hat \phi_n}_{E_U^*}\le 1$. Similarly, since $y_{n} \in B$, $\norm{y_n}_{F_B}\le 1$ as well. Therefore
and $\hat T: \wh E_{U} \to F_{B}$ is nuclear.$\square$
Proposition 12.12.4.label Let $E, F, G, H$ be separated locally convex spaces and $S \in N(F; G)$, then:
- (1)
$S$ is compact.
- (2)
For any $T \in L(E; F)$, $S \circ T \in N(E; G)$.
- (3)
For any $R \in L(G; H)$, $R \circ S \in N(F; H)$.
- (4)
There exists a unique $\wh S \in L(\wh F; G)$ such that $\wh S|_{F}= S$. Moreover, $\wh S \in N(\wh F; G)$.
Proof, [Corollary III.7.1.1-III.7.1.3, SW99]. Let $\seq{\phi_n}\subset F^{*}$ be an equicontinuous sequence, $B \in B(G)$ be convex and circled, $\seq{y_n}\subset B$, and $\seq{\lambda_n}\subset K$ such that
- (a)
The auxiliary space $G_{B}$ is a Banach space.
- (b)
For each $x \in F$, $Sx = \sum_{n = 1}^{\infty} \lambda_{n} y_{n} \dpn{x, \phi_n}{F}$.
- (c)
$\sum_{n \in \natp}|\lambda_{n}| < \infty$.
(1): Let $U = \bigcap_{n \in \natp}\phi_{n}^{-1}(B_{K}(0, 1))$, then since $\seq{\phi_n}$ is equicontinuous, $U$ is a convex and circled neighbourhood of $0$ in $F$. Given that $G_{B}$ is complete, $S$ is the following composition of continuous maps:
By Tychonoff’s Theorem, $\overline{B_K(0,1)}^{\natp}$ is compact. Since $S(U)$ is contained in its image in the above diagram, $S(U)$ is relatively compact.
(2): Since $T \in L(E; F)$, $\seq{\phi_n \circ T}\subset E^{*}$ is equicontinuous. Thus for any $x \in E$,
and $S \circ T \in N(E; G)$.
(3): Using (1), assume without loss of generality that $B$ is also compact. In which case, $R(B)$ is a convex, circled, and compact set in $H$ containing $0$. Thus $H_{R(B)}$ is a Banach space. For each $x \in F$,
and $R \circ S \in N(F; H)$.
(4): By the linear extension theorem, such an extension exists and is unique. Moreover, $\seq{\phi_n}\subset F^{*}$ extend into an equicontinuous family $\bracsn{\wh \phi_n}_{1}^{\infty} \subset \wh F^{*}$. Since $G_{B}$ is complete, the extension $\wh S \in L(\wh F; G)$ takes the form
Therefore $\wh S \in N(\wh F; G)$.$\square$
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