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So I have this definition of Kadec norm

$\textbf{Definition:}$ Let $(X,\|\|)$ be a Banach space. The norm $\|\|$ is said to be a Kadec norm if $x_n \xrightarrow{w} \bar{x}$ and $\|x_n\|\to \|\bar{x}\|$ implies $x_n\to \bar{x}.$

I was wondering why is it necessary in this definition that $X$ to be a Banach space. I though that it is not at all, and that then the condition would imply that $X$ is Banach. But I was not able to prove that statement. Any thoughts?

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No need for $X$ to be a Banach space. For example in Megginson's An introduction to Banach space theory, the Kadets-Klee property is defined for normed spaces (Definition 2.5.26, pg. 220). The same page also provides examples of non Banach spaces which have the Kadets-Klee property:

By a Theorem of Kadets (1959) which is mentioned without proof in the end of this page, every seperable normed space has an equivalent locally uniformly rotund norm. A theorem by Vyborny (1956) states that locally uniformly rotund spaces have the Kadets-Klee property (see Theorem 5.3.7, pg. 463 in the same book). Combining these two theorems, you have that every seperable normed space has an equivalent Kadets norm.

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I am not sure if I understand your question - are you asking whether the definition makes sense for normed spaces? Certainly it does. The real question is what are you going to do with it?

I think it is unlikely that it'll imply completeness as the condition mentions only sequences that already have a weak limit so any subspace should inherit this property.

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  • $\begingroup$ The main question is: Why do we need to assume that $X$ is Banach? The property makes sense even without this assumption. I thought that this case was similar to reflexivity, where the condition already implies the completeness of the space $\endgroup$ – John D Jul 31 '17 at 1:51

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