Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Can anyone prove this statement? It seems true, but I'm finding it tricky to give a concise proof.

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, $\tilde r \le \alpha r$.

share|improve this question
add comment

1 Answer 1

up vote 1 down vote accepted

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.

share|improve this answer
    
Very nice, thanks much. –  Jeff Nov 21 '12 at 18:20
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.