### $ \lambda(P)$-nuclearity of locally convex spaces having generalized bases

#### Abstract

It has been established that a $DF$-space having a fully-$\lambda(P)$-basis is $\lambda(P)$-nuclear wherein $P$ is a stable nuclear power set of infinite type. It is shown that a barrelled $G_1$-space $\lambda(Q)$ is uniformly $\lambda(P)$-nuclear iff $\{e_i,e_i\}$ is a fully-$\lambda(P)$-basis for $\lambda(Q)$. Suppose $\lambda$ is a $\mu$-perfect sequence space for a perfect sequence space $\mu$ such that there exist $u\in \lambda^\mu$ and $v\in \mu^x$ with $u_i\ge \varepsilon >0$ and $v_i\ge \iota >0$ for some $\varepsilon$ and $\iota$ and for all $i$. Then the following results are found to be true.

(i) A sequentially complete space having a fully-$(\lambda,\sigma \mu)$-basis is $\lambda(P)$-nuclear, provided $\mu$ is a $DF$-space in which $\{e_i,e_i\}$ is a semi-$\lambda(P)$-basis.

(ii) Suppose $\{e_i,e_i\}$ is a fully-$(\lambda, \sigma \mu)$-basis for a barrelled $G_i$-space $\lambda(Q)$. If $\mu$ is barrelled and $\{e_i,e_i\}$ is a semi-$\lambda(P)$-basis for $\mu$ then $\lambda(Q)$ is uniformly $\lambda(P)$-nuclear.

(iii) A $DF$-space with a fully-$(\lambda, \sigma\mu)$-basis is $\lambda(P)$-nuclear wherein $(\lambda,\sigma\mu)$ is barrelled in which $\{e_i,e_i\}$ is a semi-$\lambda(P)$-basis.

#### Full Text:

PDFDOI: http://dx.doi.org/10.5556/j.tkjm.31.2000.410

Sponsored by Tamkang University | ISSN 0049-2930 (Print), ISSN 2073-9826 (Online) |