Subadditive Kingmans theorem for lattices. - MathOverflow most recent 30 from http://mathoverflow.net2013-05-25T10:52:32Zhttp://mathoverflow.net/feeds/question/74873http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/74873/subadditive-kingmans-theorem-for-latticesSubadditive Kingmans theorem for lattices. Piotr Miłoś2011-09-08T12:51:48Z2011-09-22T15:22:12Z
<p>I am looking for a multidimensional version of Kingman's subadditive theorem. I found <a href="http://www.jstor.org/stable/2243289" rel="nofollow">this</a> but it is not exactely what I need. </p>
<p>I would rather have something like that:</p>
<p>Let us consider $\mathbb{Z}^2_+$ and a family $(X)$ of random variables indexed by pairs of points in $\mathbb{Z}^2$ i.e. $X_{z_1, z_2}$ is a random variable associated with subgrid of $\mathbb{Z}^2$ "starting from" $z_1$ and ending on $z_2$. Assume that for any rectangular subgrid $\Lambda \subset \mathbb{Z}^2_+$ and point $x\in \Lambda $ we have </p>
<p>$X_{\Lambda}\leq X_{\Lambda_1} + X_{\Lambda_2}+X_{\Lambda_3}+X_{\Lambda_4},$</p>
<p>where $\Lambda_1,\Lambda_2,\Lambda_3,\Lambda_4$ is $\Lambda$ split in point $x$ into four subgrids. I suppose that this together with some "usual" ergodic theorem conditions should imply that </p>
<p>$X_{(0,0),(n,n)}/n^2$ converges in $L^1$ and a.s. </p>
<p>May be some one you could give me some references. </p>
<p>Being in the topic of subadditivity. I found a multidimensional version of <a href="http://arxiv.org/abs/0707.3903" rel="nofollow">Fekete lemma</a>. It is surprising for me that it was not done before 2007. But may be it was. Again I will be happy to know any.</p>
http://mathoverflow.net/questions/74873/subadditive-kingmans-theorem-for-lattices/74883#74883Answer by R W for Subadditive Kingmans theorem for lattices. R W2011-09-08T14:31:47Z2011-09-08T14:31:47Z<p>Have a look at the paper by Nguyen, Xuan-Xanh <i>Ergodic theorems for subadditive spatial processes</i>, Z. Wahrsch. Verw. Gebiete 48 (1979), no. 2, 159–176, MR0534842 (82c:60056). The setup there is somewhat different, but it seems that the main result there should contain your claim. </p>