Can averaged limits of sequences be realized as limits of sequences? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T03:49:32Z http://mathoverflow.net/feeds/question/68034 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/68034/can-averaged-limits-of-sequences-be-realized-as-limits-of-sequences Can averaged limits of sequences be realized as limits of sequences? deltuva 2011-06-17T07:46:25Z 2011-06-17T14:25:28Z <p>Let a summation method take a sequence $(x_n)$ to a net $(y_\alpha)$, where $\alpha$ runs over a partially ordered set, $y_\alpha=\sum c_{\alpha,n}x_n$ ($c_{\alpha,n}\geq 0$, $\sum_n c_{\alpha,n}=1$ for every $\alpha$ and $c_{\alpha,n}\to 0$ in $\alpha$ for every $n$). <em>Is it possible to find a sequence</em> $(\alpha_m)$ <em>of indices such that convergence of</em> $(y_\alpha)$ <em>imply convergence of</em> $(y_{\alpha_m})$?</p> <p>Take the Abel summation method as an example: the set of indices is $(0, 1)$, and convergence of the Abel means $(y_r)$ yields that for $(y_{r_m})$, where $(r_m)$ is an arbitrary sequence tending to 1.</p> http://mathoverflow.net/questions/68034/can-averaged-limits-of-sequences-be-realized-as-limits-of-sequences/68055#68055 Answer by Martin Sleziak for Can averaged limits of sequences be realized as limits of sequences? Martin Sleziak 2011-06-17T13:21:12Z 2011-06-17T14:25:28Z <p>If I understand your question correctly, you want the same sequence $(\alpha_m)$ for each sequence from the convergence field of your summation method.</p> <p>I'll try to show that <a href="http://en.wikipedia.org/wiki/Ultralimit" rel="nofollow">ultralimit</a> can be realized as a summability method in the way you described.</p> <p>Let $\mathcal F$ be any free ultrafilter. Let us define $$D=\{(A,n); n\in A, A\in\mathcal F\}$$ and $$(A,n)\le (B,m) \Leftrightarrow A\supseteq B.$$ Then $(D,\le)$ is a directed set. For $\alpha=(A,n)\in D$ we define $$c_{\alpha,k}= \begin{cases} 1;&amp;k=n,\\ 0;&amp;k\ne n. \end{cases}$$ which means that $y_\alpha=x_n$.</p> <p>Now $y_\alpha$ converges to $L$ if and only if $L$ is the $\mathcal F$-limit of the sequence $(x_n)$. (This is very similar to the usual correspondence between filters and nets in topological spaces, see e.g. Proposition 6.2. <a href="http://www.math.uga.edu/~pete/convergence.pdf" rel="nofollow">here</a>.) This implies that every bounded sequence is summable.</p> <p>However, if we choose any sequence $\alpha_m=(A_m,n_m)$ then the convergence of $(x_{\alpha_m})$ is in fact the convergence of $(x_{n_m})$. For any given sequence $(n_m)$ it is easy to exhibit an example of a bounded sequence $x$ such that $(x_{n_m})$ is not convergent.</p> <hr> <p>The above example shows that your claim is not true for arbitrary directed sets. However, if you work with a directed set which contains a <a href="http://en.wikipedia.org/wiki/Cofinal_%28mathematics%29" rel="nofollow">cofinal subset</a> of order type $(\mathbb N,\le)$, then it is obviously true. (Which is the case in the example you suggested.) I suspect that it should work even for directed sets having any countable cofinal subset. There is a post at math.stackexchange related to the nets of this type: <a href="http://math.stackexchange.com/questions/41634/ordered-sets-that-are-like-sequences/" rel="nofollow">http://math.stackexchange.com/questions/41634/ordered-sets-that-are-like-sequences/</a></p>