Suppose $x_1\ge x_2\ge \cdots \ge x_n\ge 0$ and $y_1\ge y_2\ge\cdots\ge y_n\ge0$ be reals such that for any positive integer $p$, $$ \sum_{i=1}^n x_i^p \geq \sum_{i=1}^n y_i^p. $$

Question: Is there a constant $c\ge 1$ such that for all $x,y$ satisfying the above, we have $cx\succ_w y$.

Here $cx$ denotes the sequence $cx_1, cx_2,\ldots, cx_n$ and $a \succ_w b$ means that $a$ weakly majorizes $b$ from below.

Recall that a sequence $a_1\ge a_2\ge \cdots\ge a_n$ is said to weakly majorize from below another sequence $b_1\ge b_2\ge\cdots\ge b_n$ if for all $1\le k \le n$ $$ \sum_{i=1}^k a_i\ge \sum_{i=1}^k b_i. $$

Update: user35593's example $x=(4,1,1)$ and $y=(3,3,0)$ shows that if there is such a $c$, then $c\ge 1.2$.

Remark 1: If $\sum f(x_i)\ge \sum f(y_i)$ for all convex non-decreasing functions, then we have $x\succ_w y$, a result due to Karamata; Hardy, Littlewood and Pólya. In fact testing against all functions of the form $z\mapsto (z-t)^+$ is enough.

  • 1
    $\begingroup$ Choosing $x=(4,1,1)$ and $y=(3,3,0)$ we see that $c$ is at least $1.2$. $\endgroup$
    – user35593
    Mar 24, 2016 at 19:23

1 Answer 1


I think there is no such $c$. In fact, for $k$ and $\epsilon \gt 0$ given, there are $n \gg k$ and $x_1\ge x_2\ge \cdots \ge x_n\ge 0$ and $y_1\ge y_2\ge\cdots\ge y_n\ge0$ with $$\sum_{i=1}^n x_i^p \geq \sum_{i=1}^n y_i^p.$$ for all $p\ge 1$ but $$k\cdot \sum_{i=1}^k (x_i - \epsilon) \le\sum_{i=1}^k y_i$$ So $c$ had to be at least $k$.

The sequences are as follows: $$x_1 = 1 + \epsilon, x_2 = \cdots = x_n = \epsilon$$ $$ y_1 = \cdots = y_k = 1, y_{k+1} = \cdots = y_n = 0.$$ Now $$k\cdot \sum_{i=1}^k (x_i - \epsilon) = \sum_{i=1}^k y_i = k$$ So we only have to choose $n \gg k$ s. t. $$\sum_{i=1}^n x_i^p \geq \sum_{i=1}^n y_i^p = k \quad \forall p \ge 1$$ This is clearly possible, since $x_1^p \ge k$ for all but a finite number of $p$, and if $p$ is fixed, $n \ge 1 + \frac{k-1}{\epsilon^p}$ also does the job.


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