I am trying to show the following:

Denote with $|\cdot|_p$ the p-norm of a vector. Given $x_1, x_2 \in R^n$ so that all components of $x_1, x_2$ are strictly positive, $|x_1|_1=|x_2|_1=1$ and $|x_1|_2 \leq |x_2|_2$. Does this imply that $|x_1|_p \leq |x_2|_p$ for $p>2$?

Any help is much appreciated!

P-norms are non-increasing with p so I tried taking derivatives of the above to show that one side decreases faster, without success. This problem feels connected with Rényi entropy but I found nothing useful there. I found examples where this is true and no counter example.

  • $\begingroup$ Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or closed. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers. $\endgroup$ Jun 10, 2022 at 7:34
  • $\begingroup$ Thank you, I added some of the things I tried. I stumbled on this problem in the context of higher moments of quadratic forms in random variables, which I don't feel is very relevant to the discussion... $\endgroup$
    – mrav
    Jun 10, 2022 at 8:16
  • $\begingroup$ I doubt it is true, and I would expect to be able to produce a counterexample in $\mathbb R^3$ $\endgroup$
    – Henry
    Jun 10, 2022 at 8:47
  • $\begingroup$ This is the reason why I think this is true: the Rényi entropy is the same (up to sign and irrelevant constant) as the Rényi divergence wrt the uniform distribution, i.e. the information content of the distribution. Both sequences are probability distributions (positive and sum one). At order two one of the two distributions contains more information than the other and if the inequality did not hold for any order I could find an order for which the other contains more information, which feels wrong... $\endgroup$
    – mrav
    Jun 10, 2022 at 9:45

1 Answer 1


As a simple counter-example consider

  • $x_1=(0.15,0.25,0.6)$ with $|x_1|_1=1$, $|x_1|_2\approx 0.667$, $|x_1|_3\approx 0.617$
  • $x_2=(0.05,0.45,0.5)$ with $|x_2|_1=1$, $|x_2|_2\approx 0.675$, $|x_2|_3\approx 0.600$

or in terms of integers

  • $(3,5,12)$ with sum $20$, sum of squares $178$, sum of cubes $1880$
  • $(1,9,10)$ with sum $20$, sum of squares $182$, sum of cubes $1730$

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