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

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 12.12.4.$\square$