13.14 Nuclear Spaces

Proof. (1) $\Rightarrow$ (2): Let $U = T^{-1}(B_{F}(0, 1))$, then there exists $V \in \fB$ with $V \subset U$. In which case, there exists $\wh T \in L(\wh E_{V}; F)$ such that the following diagram commutes:

Since $\pi_{V} \in N(E; \wh E_{V})$, $T = \wh T \circ \pi_{V}$ is nuclear by Proposition 13.13.4.

(2) $\Rightarrow$ (3): Let $U \in \cn_{E}(0)$, then the canonical map $\pi_{U}: E \to \wh E_{U}$ is nuclear. Thus there exists an equicontinuous sequence $\seq{\phi_n}\subset E^{*}$, $\seq{y_n}\subset B_{\wh E_U}(0, 1)$, and $\seq{\lambda_n}\subset K$ such that

1. (a)

   For each $x \in E$, $\pi_{U} x = \sum_{n = 1}^{\infty} \lambda_{n} y_{n} \dpn{x, \phi_n}{E}$.

2. (b)

   $\sum_{n \in \natp}|\lambda_{n}| < \infty$.

Let $V = U \cap \bigcap_{n \in \natp}\phi_{n}^{-1}(B_{K}(0, 1))$, then by equicontinuity of $\seq{\phi_n}$, $V \in \cn_{E}(0)$. Moreover, for each $n \in \natp$, there exists $\wh \phi_{n} \in \wh E_{V}^{*}$ such that the following diagram commutes:

As $V \subset \phi_{n}^{-1}(B_{K}(0, 1))$, $\normn{\widehat \phi_n}_{\wh E_V^*}\le 1$. Thus the induced map $\widehat \pi_{U}: \wh E_{V}\to \wh E_{U}$ takes the form

with

Therefore $\wh \pi_{U}$ is nuclear.

(3) $\Rightarrow$ (1): Let $U \in \cn_{E}(0)$ be convex and circled, then there exists a convex circled neighbourhood $V \in \cn_{E}(0)$ such that the induced map $\wh \pi_{U}: \wh E_{V} \to \wh E_{U}$ is nuclear. In which case, the canonical map $\pi_{U}: E \to \wh E_{U}$ is the composition of $\pi_{V}$ and $\wh \pi_{U}$. Thus $\pi_{U}: E \to \wh E_{U}$ is nuclear by Proposition 13.13.4.$\square$

Proof, [III.7.3, SW99]. Assume without loss of generality that $U$ is convex and circled, and the canonical projection $\pi_{U}: E \to \wh E_{U}$ is nuclear. In which case, there exists an equicontinuous sequence $\seq{\phi_n}\subset E^{*}$, $\seq{y_n}\subset B_{\wh E_U}(0, 1)$, and $\seq{\lambda_n}\subset K$ such that

1. (a)

   For each $x \in E$, $\pi_{U} x = \sum_{n = 1}^{\infty} \lambda_{n} y_{n} \dpn{x, \phi_n}{E}$.

2. (b)

   $\sum_{n \in \natp}|\lambda_{n}| < \infty$.

By rescaling, further assume without loss of generality that $\sum_{n \in \natp}|\lambda_{n}| = 1$ and $\lambda_{n} > 0$ for all $n \in \natp$. Under the convention that $1/\infty = 0$, define

then for each $x \in E$, $\norm{Tx}_{l^p(\natp; K)}\le \norm{\pi_U x}_{\wh E_U}$, so $T$ is continuous.

On the other hand,

Let $q \in [1, \infty]$ be the Hölder conjugate of $p$. By Hölder’s inequality applied to $\bracsn{\lambda_n^{1/p}\dpn{x, \phi_n}{E}}_{1}^{\infty}$ and $\bracsn{\lambda_n^{1/q}}_{1}^{\infty}$, $\normn{\pi_U x}_{\wh E_U}\le \norm{Tx}_{l^p(\natp; K)}$.

Finally, let $V = T^{-1}(B_{l^p(\natp; K)})$, then $V \subset U$, and $\wh E_{V}$ is isomorphic to $\ol{T(E)}$, with equal norms.$\square$

Proof. See Proposition 13.14.6, Proposition 13.14.5, Proposition 13.14.7, and Proposition 13.14.8.$\square$

Proof, [Theorem III.7.4, SW99]. Firstly, a setup about auxiliary spaces and subspaces is required. Let $U \in \cn_{E}(0)$ be convex and circled, then the composition of the inclusion map $\iota: F \to E$ and the canonical projection $\pi_{U}: E \to E_{U}$ factors through $F_{U \cap F}$ as follows:

where $\widehat \pi_{U}$ is an isometric embedding. As a result, the factored map $\widehat \pi_{U}: F_{U \cap F}\to E_{U}$ extends to an isometric embedding on the completions:

which enables identifying $\widehat F_{U \cap F}$ as a closed subspace of $\widehat E_{U}$.

To start the proof, let $U \in \cn_{E}(0)$ be a given convex and circled neighbourhood. Since $E$ is nuclear, there exists a convex and circled neighbourhood $V \in \cn_{E}(0)$ with $V \subset U$ such that the induced map $\widehat \pi_{U}: \widehat E_{V} \to \widehat E_{U}$ is nuclear. By prior discussion, the following diagram commutes:

Thus the induced map $\widehat \pi_{U \cap F}: \widehat F_{V \cap F}\to \widehat F_{U \cap F}$ corresponds to the restriction of $\widehat \pi_{U}$ to $\widehat F_{V \cap F}$. Since $\widehat \pi_{U}$ is nuclear, there exists $\seq{\phi_n}\subset E_{V}^{*}$ and $\seq{y_n}\subset \widehat E_{U}$ such that

and $\sum_{n \in \natp}\norm{y_n}_{\widehat E_U}\norm{\phi_n}_{E_V^*}< \infty$.

Now, using Theorem 13.14.2, further assume without loss of generality that $\widehat E_{U}$ is a Hilbert space. Let $P: \widehat E_{U}\to \widehat F_{U \cap F}$ be the orthogonal projection of $\widehat E_{U}$ onto $\widehat F_{U \cap F}$, then

with

Therefore the induced map $\widehat \pi_{U \cap F}$ is nuclear, and $F$ is a nuclear space.$\square$

Proof, [Theorem III.7.4, SW99]. Firstly, a setup about auxiliary spaces and quotients is required. Let $p: E \to E/F$ be the canonical projection and $U \in \cn_{E}(0)$ be a convex and circled neighbourhood, then the composition of maps $E \to E/F \to (E/F)_{p(U)}$ factors through $E_{U}$ as follows:

This extends through the completion

and yields that $\widehat{(E/F)}_{p(U)}$ is a quotient space of $\widehat E_{U}$.

To begin the proof, let $U \in \cn_{E}(0)$ be a convex and circled neighbourhood. Since $E$ is nuclear, there exists a convex and circled neighbourhood $V \in \cn_{E}(0)$ with $V \subset U$ such that the induced map $\widehat \pi_{U}: \widehat E_{V} \to \widehat E_{U}$ is nuclear. The composition of maps $\wh E_{V} \to \wh E_{U} \to \wh{(E/F)}_{p(U)}$ then factors through $\widehat{(E/F)}_{p(V)}$ as $\widehat \pi_{p(U)}$:

Since $\wh \pi_{U}: \wh E_{V} \to \wh E_{U}$ is nuclear, there exists $\seq{\phi_n}\subset E_{V}^{*}$ and $\seq{y_n}\subset \wh E_{U}$ such that

and $\sum_{n \in \natp}\norm{y_n}_{\wh E_U}\norm{\phi_n}_{E_V^*}< \infty$.

Now, using Theorem 13.14.2, further assume without loss of generality that $\wh E_{V}$ is a Hilbert space. Identify $(\widehat{E/F})_{p(V)}$ as a closed subspace of $\widehat E_{V}$, and let $P: \widehat E_{V} \to (\widehat{E/F})_{p(V)}$ be the orthogonal projection of $\widehat E_{V}$ onto $(\widehat{E/F})_{p(V)}$. This allows rewriting

where

Therefore $\widehat \pi_{p(U)}$ is nuclear, and $E/F$ is a nuclear space.$\square$

Proof, [Theorem III.7.4, SW99]. For each $n \in \natp$, identify $E_{n}$ as a subspace of $\bigoplus_{n = 1}^{\infty} E_{n}$. Let $F$ be a Banach space and $T \in L(\bigoplus_{n = 1}^{\infty} E_{n}; F)$. For each $n \in \natp$, $E_{n}$ is a nuclear space, so $T|_{E_n}: E_{n} \to F$ is a nuclear operator, and there exists $\bracsn{\phi_{n, k}}_{k = 1}^{\infty} \subset E_{n}^{*}$ equicontinuous, $\bracsn{y_{n, k}}_{k = 1}^{\infty} \subset B_{F}(0, 1)$, and $\bracsn{\lambda_{n, k}}_{k = 1}^{\infty} \subset K$ such that $\sum_{k \in \natp}|\lambda_{n, k}| \le 2^{-n}$ and

for all $x \in E_{n}$. Thus for any $x \in \bigoplus_{n = 1}^{\infty} E_{n}$,

where $\sum_{n \in \natp}\sum_{k \in \natp}|\lambda_{n, k}| \le \sum_{n \in \natp}2^{-n}< \infty$ and $\bracsn{y_{n, k}|n, k \in \natp}\subset B_{F}(0, 1)$.

Finally, for each $n \in \natp$, let $U_{n} = \bigcap_{k \in \natp}\phi_{n, k}^{-1}(B_{K}(0, 1))$, then $U_{n} \in \cn_{E_n}(0)$ by equicontinuity of $\bracsn{\phi_{n, k}}_{k = 1}^{\infty} \subset E_{n}^{*}$. Let $U = \aconv(\bigcup_{n \in \natp}U_{n})$, then $U \in \cn_{\bigoplus_{n = 1}^\infty E_n}(0)$ and $U \subset \bigcap_{n \in \natp}\bigcap_{k \in\natp}(\phi_{n, k}\circ \pi_{n})^{-1}(B_{K}(0, 1))$. Hence $\bracsn{\phi_{n, k} \circ \pi_n|n, k \in \natp}$ is equicontinuous, and $T$ is a nuclear operator.$\square$

Proof, [Theorem III.7.4, SW99]. Let $F$ be a Banach space and $T \in L(\prod_{i \in I}E_{i}; F)$, then there exists $J \subset I$ finite and $\wh T \in L(\prod_{j \in J}E_{j}; F)$ such that the following diagram commutes:

By Proposition 13.8.3, $\prod_{j \in J}E_{j} = \bigoplus_{j \in J}E_{j}$. By Proposition 13.14.7, $\bigoplus_{j \in J}E_{j}$ is a nuclear space, so $\widehat T: \bigoplus_{j \in J}E_{j} \to F$ is a nuclear operator. As the composition of a continuous operator and a nuclear operator, $T$ is nuclear by Proposition 13.13.4. Therefore $\prod_{i \in I}E_{i}$ is a nuclear space.$\square$