User nonameisfinetoo - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T23:41:43Z http://mathoverflow.net/feeds/user/14179 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/64052/multiple-ergodic-averages-with-varying-number-of-terms Multiple ergodic averages with varying number of terms nonameisfinetoo 2011-05-05T20:52:11Z 2011-05-06T06:14:07Z <p>Hi. I've been stuck on the following question for some time.</p> <p>Consider a sequence of functions $\left( f_n \right)$ from an ergodic space $\left( \mathsf{X}, \mathsf{S}, \mu \right)$ to $\left[ 0,1 \right]$ such that $f_{a+b} \leq f_a \mathsf{S}^a \left( f_b \right)$ for all integers $a, b \geq 0$, where, classically, $\mathsf{S} \left( g \right)$ is the map $x \longmapsto g \left( \mathsf{S} \left( x \right) \right)$.</p> <p>Obviously $f_n \leq f_{n+1}$ so that the sequence $\left( f_n \right)$ decreases at each point to a function $f$. Under the hypothesis $\int_{\mathsf{X}} f_1 d\mu &lt; 1$ I was able to prove that $f = 0$ almost everywhere. Here's how I dealt with the problem:</p> <p>For fixed $n$ and integers $k$ and $\alpha$ such that $k \alpha \leq n$, the inequality $$f_n \leq f_1 \mathsf{S}^k \left( f_1 \right) \ldots \mathsf{S}^{k\alpha} \left( f_1 \right)$$ holds, so that, taking averages, one gets $$f_n \leq \frac{1}{\lfloor n/ \alpha \rfloor} \sum_{k=0}^{\lfloor n/ \alpha \rfloor} f_1 \mathsf{S}^k \left( f_1 \right) \ldots \mathsf{S}^{k\alpha} \left( f_1 \right)$$ Taking integrals, and using the dominated convergence theorem and the Furstenberg-Katznelson theorem on multiple ergodic averages, one gets $$\int_{\mathsf{X}} f \leq lim_{n \longrightarrow \infty} \frac{1}{\lfloor n/ \alpha \rfloor} \sum_{k=0}^{\lfloor n/ \alpha \rfloor} \int_{\mathsf{X}} f_1 \mathsf{S}^k \left( f_1 \right) \ldots \mathsf{S}^{k\alpha} \left( f_1 \right) = \left( \int_{\mathsf{X}} f_1 \right)^\alpha$$ For all integers $\alpha$, which allows me to conclude.</p> <p>Now the question is: does the series $\sum f_n$ converge? That seems plausible considering the seemingly exponential decreasing of $f_n$ to $f$, but trying to use the same techniques leads to multiple ergodic averages for a varying number of terms - making the integer $\alpha$ dependent on $n$ that is; is there any way to deal with those? All suggestions are welcome.</p> http://mathoverflow.net/questions/61294/uniqueness-of-analytic-continuation-on-a-domain-of-cn Uniqueness of analytic continuation on a domain of C^n. nonameisfinetoo 2011-04-11T12:40:13Z 2011-04-11T16:40:03Z <p>Hi. I have been struggling with this question for a while now.</p> <p>Given a domain $\Omega \subset \mathbb{C}^n$, $\Omega^\prime = \Omega \cap \mathbb{R}^n$, and an analytic function $f: \Omega \longrightarrow \Omega$ such that $f_{|\Omega^\prime}$ is a (real) constant, under which assumptions on $\Omega$ and/or $\Omega^\prime$ can one conclude that $f$ is constant on the whole domain $\Omega$?</p> <p>The case $n = 1$ is quite classical. For $n>1$, assuming $\Omega$ is a pseudoconvex domain does not really help, and I'm not sure of what can be said about real domains of the zeros of an analytical function of several complex variables.</p> <p>Thanks for any clue about this.</p> http://mathoverflow.net/questions/60834/convexity-of-a-cone-in-cxc Convexity of a cone in CxC nonameisfinetoo 2011-04-06T16:41:57Z 2011-04-06T16:41:57Z <p>Hi. I have trouble deciding if the set of couples $\left( \xi, \zeta \right) \in \mathbb{C}^2$ with $Re \left( \xi \text{ } \overline{\zeta} \right) > 0$ is convex. It is a (real) cone, but is it a convex one? If not, could you provide me with a counter-example?</p> <p>thx.</p> http://mathoverflow.net/questions/64052/multiple-ergodic-averages-with-varying-number-of-terms/64088#64088 Comment by nonameisfinetoo nonameisfinetoo 2011-05-06T09:38:23Z 2011-05-06T09:38:23Z Overcomplicating indeed. Thank you. http://mathoverflow.net/questions/64052/multiple-ergodic-averages-with-varying-number-of-terms Comment by nonameisfinetoo nonameisfinetoo 2011-05-05T22:49:35Z 2011-05-05T22:49:35Z You're perfectly right, weak-mixing is the good working hypothesis. As for more expedient ways to prove that $f=0$ a.e.: For integers $n$ and $p$, $f_{n+p} \leq f_n S^n (f_p)$ so that when $p \rightarrow \infty$ and considering that $f_n \leq f_1$, the inequality $f \leq f_1 S^n (f)$ holds for all $n$. Taking averages and integrals, one finds that $\int f \leq 1/n \sum_{k=0}^n \int f_1 S^k (f) \rightarrow \int f_1 \int f$ so that $\int f \leq \int f_1 \int f$ which proves that $f=0$ almost everywhere. But this does not help for the convergence of $\sum f_n$ unless I am missing something? http://mathoverflow.net/questions/60834/convexity-of-a-cone-in-cxc Comment by nonameisfinetoo nonameisfinetoo 2011-04-06T17:38:53Z 2011-04-06T17:38:53Z Indeed. thank you.