most recent 30 from http://mathoverflow.net 2013-05-23T23:12:59Z http://mathoverflow.net/feeds/question/114017 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/114017/length-of-intersection-of-intervals Jeff 2012-11-21T03:48:53Z 2012-11-21T16:19:51Z

<p>Can anyone prove this statement? It seems true, but I'm finding it tricky to give a concise proof.</p> <p>Fix $\alpha\in[0,1]$. Let $\mu$ be Lebesgue measure. Define $B(c,r)\equiv[c-r,c+r]$, where $[\cdot, \cdot]$ denotes an interval. For $i=1,\ldots,n$, fix $r_1,\ldots,r_n\in[0,\infty)$, and $c_1,\ldots,c_n\in\mathbb R$. Let $\bar c, \bar r, \tilde c$, and $\tilde r$ satisfy $B(\bar c, \bar r)=\bigcap_{i=1}^{n}B(c_{i},r_{i})$ and $B(\tilde c, \tilde r)=\bigcap_{i=1}^{n}B(c_{i},\alpha r_{i})$. Then, $\tilde r \le \alpha r$.</p>

http://mathoverflow.net/questions/114017/length-of-intersection-of-intervals/114021#114021 Alexandre Eremenko 2012-11-21T05:15:36Z 2012-11-21T05:15:36Z

<p>The problem is reduced to the case $n=2$. Indeed, let $\cap B(c_i,r_i)=(a,b)$. Changing the numeration we can assume $a=c_1-r_1$. Then $b=c_k+r_k$ for some $k$. Now it is clear that if $\cap B(c_i\alpha r_i)=(a',b')$ then $a'\geq c_1-\alpha r_1$ and $b'\leq c_k+\alpha r_k$. So it is enough to prove that $$c_k-c_1+\alpha r_k+\alpha r_1\leq \alpha(c_k-c_1+r_k+r_1),$$ that is $c_1\geq c_k$. But this last inequality is easy, just draw a picture of intersection of two intervals.</p>