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The Wikipedia article on nuclear spaces say the following:

"There are no Banach spaces that are nuclear, except for the finite-dimensional ones. In practice a sort of converse to this is often true: if a "naturally occurring" topological vector space is not a Banach space, then there is a good chance that it is nuclear."

Question: Can the converse part of this statement be made precise in some sense? Relevant references would be appreciated.

Update 1: A related interesting question from Math Overflow.

Update 2: Possibly the simplest setting that contains both nuclear and Banach spaces is Fréchet spaces that are defined by a countable family of increasing Hilbert space norms. Under such constraints, we cannot partition the spaces into Banach and nuclear. In particular, there is a non-nuclear non-Banach space of this type that is defined as an intersection of weighted sequence spaces.

Let $\ell_2(w)$ denote a weighted sequence space; $a\in \ell_2(w)$ if

\begin{equation} \sum_{n=1}^{\infty} \left(a(n)w(n)\right)^2 < \infty. \end{equation}

Considering the weights $w_{\epsilon}(n) = n^{\epsilon}$, the set

\begin{equation} \bigcap_{\epsilon>0} \ell_2(w_{\epsilon}), \end{equation} with the initial topology is neither Banach nor nuclear.

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