2
$\begingroup$

So I know that weak convergence does imply strong convergence if the dimension of the space is finite, and that in general it does not in infinite dimension. But I was wondering if there were any examples or specific cases where there is an infinite dimensional space where weak convergence would happen to imply strong convergence. That is, is it possible to construct an infinite dimensional norm space such that weak would somehow imply strong? Just curious.

$\endgroup$
6
$\begingroup$

You are talking about the Schur Property of normed vector spaces. Let $(X, \|\cdot\|)$ be a normed vector space. This normed vector space is said to have the Schur Property if for any weakly convergent sequence $x_n \rightarrow x$, we have $\|x_n - x\| \rightarrow 0$.

Many standard spaces, like $L^p[0,1]$, do not have this property, but some do, for instance $\ell^1$, defined to be the set of all infinite sequences of real (or complex) numbers $Z = (z_n)_{n \in \mathbb{N}}$ for which $$ \sum_n |z_n| < \infty$$. This vector space possesses the "obvious" scalar multiplication and addition, and is endowed with the norm $$ \|Z\| = \sum_{n = 1}^{\infty} |z_n| $$ I found an exposition of the proof of this fact here.

$\endgroup$
  • $\begingroup$ Thanks! That clears that up $\endgroup$ – Fractal20 Aug 10 '13 at 17:58

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.