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

  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

\[T: E \to l^{p}(\natp; K) \quad (Tx)_{n} = \lambda_{n}^{1/p}\dpn{x, \phi_n}{E}\]

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,

\[\normn{\pi_U x}_{\wh E_U}= \norm{\sum_{n = 1}^\infty \lambda_n y_n \dpn{x, \phi_n}{E}}_{\wh E_U}\le \sum_{n = 1}^{\infty} \lambda_{n} |\dpn{x, \phi_n}{E}|\]

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.$\square$

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