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 $\{b_k\}$.

  • $\begingroup$ 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. $\endgroup$
    – BigM
    Nov 25 '15 at 5:38
  • $\begingroup$ I made a correction. $\endgroup$
    – BigM
    Nov 25 '15 at 14:40
  • $\begingroup$ @BigM - see the edit. $\endgroup$ Nov 26 '15 at 12:22
  • $\begingroup$ Im slightly confused, since Karamata's inequality precisely guarantees that strictly convex function preserves the order of inequality. $\endgroup$
    – BigM
    Nov 26 '15 at 18:04
  • 1
    $\begingroup$ 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. $\endgroup$ Nov 26 '15 at 20:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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