Timeline for Are weak and strong convergence of sequences not equivalent?
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| Mar 3, 2012 at 0:52 | comment | added | TaQ | I have now also observed that Schur's result is contained in Corollary 4.21.6 (p. 276) in R. E. Edwards' Functional Analysis. An example of a weakly but not in norm to zero convergent sequence in $L^p([0,1])$ for $1\le p<+\infty$ is $\langle[\langle \sin(2\pi(i+1)t):0\le t\le 1\rangle]:i\in\mathbb N_0\rangle$. Thanks again to Pietro Majer for providing refreshment to my memory. | |
| Mar 1, 2012 at 20:38 | comment | added | TaQ | Now that Schur has been mentioned, I found from Jarchow's Locally Convex Spaces, Theorem 10.5.2, proof pp. 205−206, which is the result of Schur Jochen Wengenroth referred to, however without the phrase "Schur property". | |
| Mar 1, 2012 at 19:35 | comment | added | TaQ | @Pietro Majer: In $\ell^1(\mathbb N_0)$, the sequence of standard unit vectors converges to zero pointwise but neither weakly, nor (globally) in measure. Anyway, thanks for the additional reference. en.wikipedia.org/wiki/Convergence_in_measure | |
| Mar 1, 2012 at 17:20 | comment | added | Pietro Majer | As to $L^1[0,1]$ and more generally $L^1(X,\mu)$ the complete statement is: $f_n$ converges strongly if and only if it converges weakly and in measure. (Of course in a discrete measure space such as $\mathbb{N}$, for a bounded sequence, weak convergence, convergence in measure and pointwise convergence all coincide). Check e.g. Dunford&Schwartz. | |
| Mar 1, 2012 at 16:27 | vote | accept | TaQ | ||
| Mar 1, 2012 at 15:39 | comment | added | Matthew Daws | See also Wikipedia: en.wikipedia.org/wiki/Schur%27s_property I can also recommend Albiac and Kalton's book. | |
| Mar 1, 2012 at 15:38 | history | answered | Jochen Wengenroth | CC BY-SA 3.0 |