# Weyl-Hardy-Littlewood-Polya inequality

Let $a_1,\cdots,a_n$ and $b_1,\cdots,b_n$ be two decreasing sequences of positive numbers such that $\sum_{k=1}^n a_k\leq \sum_{k=1}^n b_k$ and assume $F(x)$ is a convex function defined on $\mathbb R$ with $\displaystyle\lim_{x\to-\infty}F(x)=0$.

Does it follow (perhaps by Weyl-Hardy-Littlewood or Karamata inequality) that $$\sum_{k=1}^n F(a_k)\leq \sum_{k=1}^n F(b_k)$$
In particular does the inequality $\displaystyle\sum_{k=1}^n a^2_k\leq \sum_{k=1}^n b^2_k$ hold?

(the original question asked about concave $F$ and gave the particular case of whether $\sum \sqrt{a_k} \le \sum \sqrt{b_k}$. This post offered a counter-example to that.)

$0.25 + 0.24 < 0.5 + 0.01$ but $0.5 + 0.4898979... > 0.7071... + 0.1$

Though I note that your particular case does not satisfy the condition that $$\lim_{x\to-\infty} F(x) = 0$$

It took me a some time thinking about the new version before I realized I was overlooking an easy counter: Choose the two sequences so that their sum is equal. Then you can consider either one to be $\{a_k\}$ and the other to be $\{b_k\}$. Unless $F$ is linear, we can expect that in general $\sum F(a_k) \ne \sum F(b_k)$. So choose $\{a_k\}$ to be the sequence that gives the larger value of $\sum F(a_k)$, and your inequality fails.

Further, if $F$ is continuous for at least one of the $a_k$ or $b_k$, then we can make a minute adjustment to that value to obtain a counter-example with strict inequality.

Note also that the only properties of $F$ used are non-linearity and continuity at one point. So the problem here is not $F$, but rather the weakness of the conditions on $\{a_k\}$ and $\{b_k\}$.

• That inequality can be expressed for an interval without assumption at infinity. It seems I have the inequality backward since the function is concave, whereas most of those inequalities are stated for convex functions.
– BigM
Nov 25 '15 at 5:38
• I made a correction.
– BigM
Nov 25 '15 at 14:40
• @BigM - see the edit. Nov 26 '15 at 12:22
• Im slightly confused, since Karamata's inequality precisely guarantees that strictly convex function preserves the order of inequality.
– BigM
Nov 26 '15 at 18:04
• Karamata's inequality requires that $b$ majorizes $a$, which requires (1) that $\sum_{k=1}^n a_k \color{red}{=}\sum_{k=1}^nb_k$, and also that for all $m < n$, that $\sum_{k=1}^m a_k \le \sum_{k=1}^mb_k$. You didn't include either condition. Without them, the inequality fails. Nov 26 '15 at 20:16