Suppose a bounded sequence $(x_n)$ converges to $x$ in the Cesaro sense (i.e., $\frac{1}{n}(x_1 + x_2 + \dots + x_n)\rightarrow x$) in a separable Hilbert space $H$. How to prove that some subsequence $(x_{n_k})$ converges weakly to $x$?
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asked Jul 19, 2010 at 14:24
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1$\begingroup$ Reminds me of Banach-Saks theorem, which goes the other way. $\endgroup$– Gjergji ZaimiCommented Jul 19, 2010 at 15:07
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$\begingroup$ Indeed, looks like the exercise might've been intended the other way. For the record: the proof of Banach-Saks can be found in books.google.com/… $\endgroup$– Kestutis CesnaviciusCommented Jul 19, 2010 at 16:53
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1$\begingroup$ Whatever the corrected question is supposed to be, it almost certainly is not research level. $\endgroup$– Zen HarperCommented Jul 19, 2010 at 18:13
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If we take $x_n = (-1)^n x$ then $x_n$ converges to $0$ in Cesaro sence. But no subsequence of $x_n$ converges weakly to $0$. $x_n$ is also a bounded sequence. Hence your statements seems wrong.
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$\begingroup$ Thanks. It was an exercise that I couldn't solve, so I assumed it should hold. $\endgroup$ Commented Jul 19, 2010 at 15:13
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5$\begingroup$ @Kestutis Cesnavicius: A good habit to be in is that if you cannot prove a statement, try to break it. Either you succeed in falsifying the statement, or you gain more understanding of why the statement should be true. $\endgroup$ Commented Jul 19, 2010 at 17:57
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2$\begingroup$ Kestutis, may be it was asked (or they wanted to ask) to prove that some subsequence weakly converges, but not necessary to $x$? Which is not very deep, too, but it is true at least. $\endgroup$ Commented Jul 19, 2010 at 18:06