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?


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.


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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.