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A question about the meaning from following excerpt from german wikipedia adressing interesting crucial feature of nuclear spaces opposing them from Banach spaces (transl.):

While normed spaces, especially Banach or Hilbert spaces, are generalizations of finite-dimensional vector spaces (over $\Bbb R$ or $\Bbb C$) maintaining the existence of norm but losing compactness properties, the focus in the case of nuclear spaces, which cannot be normalized in the infinite-dimensional case, is on the compactness properties.

Could it be made precise in which sense nuclear spaces can be regarded as those spaces maintaining the compactness properties (contrasting their "nature" radically from Banach spaces)?

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  • $\begingroup$ Not sure what that means, but Smith spaces seem to more likely to have "compactness properties". $\endgroup$
    – Z. M
    Commented Sep 30 at 15:08

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See https://en.wikipedia.org/wiki/Nuclear_space

In a nuclear space, every closed bounded set is compact.

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    $\begingroup$ This is not true without an additional completeness property: Take any non-reflexive Banach space $X$. Then $(X,\sigma(X,X^*)$ is nuclear (the local Banach spaces corresponding to weak neighbourhoods are finite-dimesnional), the unit ball $B$ of $X$ is $\sigma(X,X^*)$-bounded but not weakly compact. It is true that bounded subsets of nuclear spaces are precompact. $\endgroup$
    – Jochen Wengenroth
    Commented Oct 1 at 8:15
  • $\begingroup$ hmm, it seems that the term "compactness properties" from above is phrased a bit misleading, even if replace it by "precompact". It seems that the seeked feature is that these spaces have a weakened version of Heine Borel property. But, is there maybe also some connection to concept of compact operators with resp. induced canonical seminorm maps? $\endgroup$
    – user267839
    Commented Oct 1 at 16:32
  • $\begingroup$ let me also add this script adressing Jochen Wengenroth's point, see 7 ff $\endgroup$
    – user267839
    Commented Oct 1 at 16:41
  • $\begingroup$ @JochenWengenroth A necessary and sufficient condition is quasicompleteness, that every closed bounded set is complete. $\endgroup$
    – Robert Furber
    Commented Oct 7 at 10:43
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    $\begingroup$ Wikipedia has a bit of a problem in certain places with people copying things from textbooks and papers without understanding. Unfortunately there is also the requirement of "no original research" that means you can't necessarily use your own expertise to fix it. $\endgroup$
    – Robert Furber
    Commented Oct 7 at 10:50

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