12.13 Nuclear Spaces
Definition 12.13.1 (Nuclear Space).label Let $E$ be a separated locally convex space over $K \in \RC$, then the following are equivalent:
- (1)
There exists a fundamental system of convex and circled neighbourhoods $\fB \subset \cn_{E}(0)$ such that for each $U \in \fB$, the canonical projection $\pi_{U}: E \to \wh E_{U}$ is nuclear.
- (2)
For each Banach space $F$ and $T \in L(E; F)$, $T$ is nuclear.
- (3)
For each convex and circled neighbourhood $U \in \cn_{E}(0)$, there exists $V \in \cn_{E}(0)$ with $V \subset U$ such that the induced map $\wh E_{V} \to \wh E_{U}$ is nuclear.
If the above holds, then $E$ is a nuclear space.
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 12.12.4.
(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
- (a)
For each $x \in E$, $\pi_{U} x = \sum_{n = 1}^{\infty} \lambda_{n} y_{n} \dpn{x, \phi_n}{E}$.
- (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 12.12.4.$\square$
Theorem 12.13.2.label Let $E$ be a nuclear space over $K \in \RC$, $U \in \cn_{E}(0)$, and $p \in [1, \infty]$, then there exists $V \in \cn_{E}(0)$ with $V \subset U$ such that $\wh E_{V}$ is isometrically isomorphic to a subspace of $l^{p}(\natp; K)$.
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
- (a)
For each $x \in E$, $\pi_{U} x = \sum_{n = 1}^{\infty} \lambda_{n} y_{n} \dpn{x, \phi_n}{E}$.
- (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.$\square$
Corollary 12.13.3.label Let $E$ be a complete nuclear space over $K \in \RC$, then $E$ is a projective limit of Hilbert spaces over $K$. For any Fréchet space $F$, $F$ is nuclear if and only if it is the projective limit of a sequence $\seq{H_n}$ of Hilbert spaces such that the mapping $H_{m} \to H_{n}$ is nuclear for all $1 \le m < n < \infty$.
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