Some days ago, when I again read the William Lowell Putnam Mathematical Competition (1979), I found this nice problem:

Let $p_{j}\in [0,1],j=1,2,\cdots,n$. Prove, that $$\inf_{0\le x\le 1}\sum_{j=1}^{n}\dfrac{1}{|x-p_{j}|}\le 8n\left(1+\dfrac{1}{3}+\cdots+\dfrac{1}{2n-1}\right)$$

This problem solution can split the inteval $[0,1]$ into $2n$ intervals of the same length, such $I_{k}=[\dfrac{k}{2n},\dfrac{k+1}{2n})$. Next, we choose x in an interval that does not contain any of the numbers $p_{j}$,then it not hard to prove it.

But I fell this inequality right constant $8$ can smaller, such $6$ is also hold.

Question 1:

$$\inf_{0\le x\le 1}\sum_{j=1}^{n}\dfrac{1}{|x-p_{j}|}\le 6n\left(1+\dfrac{1}{3}+\cdots+\dfrac{1}{2n-1}\right)$$

So I can think, the following problem maybe is also interesting:

Question 2

$$\inf_{0\le x\le 1}\sum_{j=1}^{n}\dfrac{1}{|x-p_{j}|}\le An\left(1+\dfrac{1}{3}+\cdots+\dfrac{1}{2n-1}\right)$$

Find the best constant of the $A$!

  • $\begingroup$ Interesting question. What have you tried so far? $\endgroup$ Sep 20, 2015 at 15:14

1 Answer 1


Given $p_1,\ldots,p_n\in[0,1]$ and some $r\in\left(0,\frac{1}{2}\right)$, let $f_k(x)$, for $k=1,\ldots,n$, the function supported on $E_k=[0,1]\cap \{x:|x-p_k|>r\}$ and given by: $$ f_k(x) = \frac{\mathbb{1}_{E_k}(x)}{|x-p_k|}.\tag{1} $$ It is straightforward to check that the mean value of $f_k$ over $E_k$ is at most: $$ \frac{-\log(2r)}{1-2r}\tag{2}$$ and if we restrict $f_k$ to a subset $E\subseteq E_k$, the mean value over the new set is at most $\frac{|E_k|}{|E|}$ times $(2)$.

The idea is so to take $E=[0,1]\setminus\bigcup_{k=1}^{n}E_k$ for some $r\leq\frac{1}{2n}$ and minimize: $$\frac{-\log(2r)\cdot n}{1-2rn}\tag{3}$$ that is an upper bound for the mean value of $f_1+\ldots+f_n$ over $E$. $E$ has a positive measure, hence for some $x\in E$ the value of $f_1+\ldots+f_n$ has to be $\leq $ than the mean value. If we take $r=\frac{1}{3n}$, for instance, we have that our $\inf$ is $\leq 3n\log\left(\frac{3n}{2}\right)$. Clearly $\frac{1}{1}+\frac{1}{3}+\ldots+\frac{1}{2n-1}$ behaves like $\frac{1}{2}\log(2n)$, so our argument proves that we may replace the constant $8$ given in the original statement with the conjectured (by Question $1.$) constant $\color{red}{6}$. Provided that $n$ is big enough, we may take $r=\frac{1}{4n}$ and get an upper bound that equals $2n\log(2n)$, so the previous constant can be improved up to $\color{red}{4}$. The same technique also shows that, if $n$ is big enough, we may replace the original constant $8$ with $\color{red}{2+\varepsilon}$.

By considering equi-spaced $p_k$s, we may also check that such a bound is optimal.

  • $\begingroup$ Nice,!+1 jack,But you it is straightforward to check that the mean value $(2)$,How to do have this result? $\endgroup$
    – math110
    Sep 21, 2015 at 11:12
  • 1
    $\begingroup$ @math110: you just have to compute the integral of $\frac{1}{|x-y|}$ over $x\in[0,1]\cap\{x:|x-y|>r\}$. The worst case is given by $y=\frac{1}{2}$. $\endgroup$ Sep 21, 2015 at 11:30

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