most recent 30 from http://mathoverflow.net 2013-05-23T05:31:13Z http://mathoverflow.net/feeds/user/17271 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/67372/distributing-points-with-respect-to-a-concave-function/73290#73290 Trenton Osborn 2011-08-20T18:35:24Z 2011-08-20T19:07:59Z

<p>You can define the function $f(t)\cdot min_i|x_i-t|$ piecewise so that $f(t)\cdot min_i|x_i-t|=f(t)\cdot|x_i-t|$ when $t$ is on the interval $\left[\frac{x_{i-1}+x_i}{2},\frac{x_i+x_{i+1}}{2}\right]$. Here we have to define $x_0=-x_1$ (so $\frac{x_{0}+x_1}{2}=0$) and $x_{n+1}=2-x_n$ (so $\frac{x_n+x_{n+1}}{2}=1$). This makes the integral \begin{equation} F(x_1,...,x_n)=\sum_{i=1}^n\int_\frac{x_{i-1}+x_i}{2}^\frac{x_i+x_{i+1}}{2}f(t)\cdot|x_i-t|dt \end{equation} Now, $f$ isn't $0$ at the endpoints of these intervals and the intervals don't have length $1$, but they are concave and their average value is $\alpha/n$. You can consider f on each interval as the limit of a sequence of concave functions defined on that interval which are each $0$ at the endpoints, so you should be able to put a lower bound on each integral in the same way that you did it for $\int_0^1f(t)\cdot|x_1-t|dt$. I can't say what that lower bound might be, because I don't know how (or if) you used the length of the interval $[0,1]$ when you found the lower bound for $\int_0^1f(t)\cdot|x_1-t|dt$. If you didn't use the length of the interval at all, the integrals will have to be on average greater than or equal to $\alpha/6n$, implying that $F(x_1,...,x_n)$ is bounded below by $\alpha/6$, which would be kind of interesting. One more thing to note is that unlike your example when $n=1$, for the integrals in the sum above you have the additional condition that $f(x_i)$ is bounded away from $0$ by the value of $f$ at at least one of the endpoints of the intervals. Maybe this will make it possible to put stricter lower bounds on each of these integrals. I hope all that was clear (and correct). </p>