Sufficient conditions to the existence of a weakly convergent subsequence from a Cauchy sequence in a (merely) normed space - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T05:21:12Z http://mathoverflow.net/feeds/question/75321 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/75321/sufficient-conditions-to-the-existence-of-a-weakly-convergent-subsequence-from-a Sufficient conditions to the existence of a weakly convergent subsequence from a Cauchy sequence in a (merely) normed space Salvo Tringali 2011-09-13T15:41:28Z 2011-09-13T19:54:51Z <p>Bonsoir/bonjour à toutes et à tous.</p> <p>The title has it all, but... We know (as a consequence of the Eberlein-Šmulian theorem) that any bounded sequence, <code>$\{x_n\}_{n \;\! \in \;\! \mathbb{N}}$</code>, in a (real or complex) reflexive normed (and hence Banach) space, <code>$\mathbf{X} \equiv (X, \|\cdot\|)$</code>, contains a <em>weakly</em> convergent subsequence. Now, drop the assumption that $\mathbf{X}$ is complete and strengthen the hypotheses on <code>$\{x_n\}_{n \;\! \in \;\! \mathbb{N}}$</code> by replacing boundedness with cauchyness. Then the question comes:</p> <blockquote> <p><strong>Question.</strong> What are <em>sufficient</em> conditions to the existence of a weakly convergent subsequence from a Cauchy sequence in a (real or complex) normed space (which is <em>not</em> supposed to be complete)? Of course, let's rule out trivially tautological conditions such as "the existence of a weakly convergent subsequence", "the (strong) convergence of the sequence", ...</p> </blockquote> <p>I'm aware that the question may sound a little weird at face value - especially looking at the case in which $\mathbf{X}$ is a dense <em>proper</em> (normed) subspace of a Banach space, $\mathbf{Y}$, and <code>$\{x_n\}_{n \;\! \in \;\! \mathbb{N}}$</code> is a sequence in $X$ which is convergent in <code>$\mathbf{Y}$</code> but not in <code>$\mathbf{X}$</code>. Nevertheless, the particular problem (*) on which I am working, implies additional conditions on <code>$\{x_n\}_{n \;\! \in \;\! \mathbb{N}}$</code> that may still force, as I hope, the existence of a weakly convergent subsequence - and hence, in my <em>very particular</em> case, the (strong) convergence of the sequence -, and this is basically why I am posing the question.</p> <hr> <p>(*) Let me give some details on the problem, in the case that they may be useful to know. These include the existence of a bounded linear transformation, <code>$T: \mathbf{X} \to \mathbf{X}$</code>, such that </p> <ol> <li><code>$\{Tx_n\}_{n\;\! \in \;\! \mathbb{N}}$</code> is (strongly) convergent to some <code>$y \in X$</code>;</li> <li>for all <code>$\varphi \in X^\prime$</code>, there is a unique <code>$\psi \in X^\prime$</code> for which <code>$\varphi = \psi \circ T$</code>.</li> </ol> <p>Here, as you can guess, $X^\prime$ is the continuous dual of $\mathbf{X}$. Also, equipping $X^\prime$ with its usual norm, <code>$\|\cdot\|_{X^\prime}: X^\prime \to \mathbb{R}: \varphi \mapsto \sup_{x \;\! \in \;\! X, \|x\| \;\! \le \;\! 1} |\varphi(x)|$</code>, and letting $\mathbf{X}^\prime \equiv (X^\prime, \|\cdot\|_{X^\prime})$, the function, $\Psi: X^\prime \to X^\prime$, mapping any given $\varphi \in X^\prime$ to the unique $\psi \in X^\prime$ such that condition 2 above is satisfied, is actually a <strong>bi-Lipschitz isomorphism</strong> <code>$\mathbf{X}^\prime \to \mathbf{X}^\prime$</code>, so $\lim_n \varphi(x_n) = \Psi(\varphi)(y)$ for all $\varphi \in X^\prime$.</p> http://mathoverflow.net/questions/75321/sufficient-conditions-to-the-existence-of-a-weakly-convergent-subsequence-from-a/75334#75334 Answer by Bill Johnson for Sufficient conditions to the existence of a weakly convergent subsequence from a Cauchy sequence in a (merely) normed space Bill Johnson 2011-09-13T17:15:46Z 2011-09-13T17:15:46Z <p>I don't understand your question. If a norm Cauchy sequence has a weak cluster point, then it norm converges to the weak cluster point. Put another way, a norm Cauchy sequence is weakly Cauchy and hence weak* converges in the bidual to something; if that something is back in the space, the sequence is weakly convergent to it and, since the sequence is norm Cauchy, must norm converge to it (since e.g. tails of the sequence are contained in arbitrarily small balls and the weak limit must be in the closed convex hull of every tail of the sequence).</p> http://mathoverflow.net/questions/75321/sufficient-conditions-to-the-existence-of-a-weakly-convergent-subsequence-from-a/75347#75347 Answer by Bill Johnson for Sufficient conditions to the existence of a weakly convergent subsequence from a Cauchy sequence in a (merely) normed space Bill Johnson 2011-09-13T19:49:24Z 2011-09-13T19:54:51Z <p>Your condition (2) is that <code>$T^*$</code> is a surjective isomorphism, so $T$ induces a surjective isomorphism on the completion of $X$. For a counterexample, let $T$ be the right shift on $\ell_2(Z)$ restricted to an appropriate dense subspace. What subspace? Well, it must be dense, so throw in the unit vector basis. Throw in some natural vector $y$ which you want to be the limit of $Tx_n$; $y=\sum_{k=1}^\infty 2^{-k!} e_k$ should be fine. You need for $T$ to map the subspace back into itself, so throw in $T^k y$ for $k=1,2,...$. Let $X$ be the linear span of all the vectors thrown in. Set $x_n= \sum_{k=0}^n 2^{-(k+1)!} e_k$. Then $x_n$ converges to a point not in $X$ but $Tx_n$ converges to $y$.</p>