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A sequence $(x_n)$ in a Banach space $X$ is called weakly Cauchy if for every $\ell \in X'$ the sequence $(\ell(x_n))$ is Cauchy in the scalar field.

I want to show that weakly Cauchy sequences are not necessarily weakly convergent.

This seems to be the case for Hilbert spaces On the limits of weakly convergent subsequences, whats the difference?

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    $\begingroup$ For a counterexample, consider the sequence $(y_n)$ defined by $y_n=e_1+e_2+\cdots+e_n$ in $c_0$, where $e_i$ is the $i$'th standard unit vector. Spaces where weakly Cauchy sequences are weakly convergent are called weakly sequentially complete. Reflexive spaces are such spaces. $\endgroup$ – David Mitra Nov 27 '12 at 17:22
  • $\begingroup$ Thanks, Is it obvious that this is cauchy? $\ell(y_m) - \ell(y_n) = (0,0,...0,-1-1...0,...,0)$ or something? Where we have min(n,m) zeros in the beginning..? And why is it not weakly convergent? I guess $y = (1,1,1,1,1...1)$ what is the functional that makes $lim_{n\rightarrow \infty}\ell(y_n) \rightarrow \ell(y)$ false? $\endgroup$ – Johan Nov 27 '12 at 17:34
  • $\begingroup$ For $m\ge n$ and $l=(l_1,l_2,\cdots)$, $l(y_m)-l(y_n)$ would have the form $\sum_{i=n+1}^m\ell_i$. But $l$ is an element of $\ell_1$; thus absolutely summable... $\endgroup$ – David Mitra Nov 27 '12 at 17:38
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    $\begingroup$ For the second part of your comment, note that $(1,1,\ldots)$ (which is the only candidate for a weak limit point) is not an element of $c_0$. $\endgroup$ – David Mitra Nov 27 '12 at 17:41
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    $\begingroup$ Yes, it should be obvious that $(y_n)$ is weakly Cauchy. In fact, it's obvious that $(l(y_n))$ is a convergent sequence in $\Bbb K$ for any $l$ in $\ell_1$; since $l(y_n)$ is just the sum of the first $n$ coordinates of $l$. To address your second question: when one talks about a "limit point" of a sequence, that point needs to be in the same space as the elements of the sequence. Here, we are talking about a sequence in the space $c_0$. Any limit point (of any type) must be an element of $c_0$ as well. $\endgroup$ – David Mitra Nov 27 '12 at 18:12
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In the link, it's actually true for a reflexive space.

So to get a counter-example, we have to think on a non-reflexive Banach space.

Take $X:=C[0,1]$ endowed with the supremum norm and $x_n(t):=t^n$. This sequence doesn't converge weakly, but for all $l\in X'$, we can represent it by a Radon measure, and $\left\{\int_{[0,1]}t^d\mu(t)\right\}$ is convergent by monotone convergence theorem applied to the positive and negative parts of $\mu$.

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  • $\begingroup$ I have not seen the Radon measures before. So since we can represent it by a Radon measure we have a weakly Cauchy sequence? $\endgroup$ – Johan Nov 27 '12 at 17:39
  • $\begingroup$ That right (actually, as $\Bbb R$ and $\Bbb C$ are complete, we have for all linear continuous functional that $\{L(x_n)\}$ is convergent. $\endgroup$ – Davide Giraudo Nov 27 '12 at 19:39

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