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Let $a_i > 0$ forall $1 \leq i \leq n$ and let $M(x) := \big(\frac{\sum_{i = 1}^n a_i^x}{n} \big)^{\frac{1}{x}}$ be the power means function. It is well known that the power means function is non-decreasing.

I am interested in the concavity properties of $M(x)$ but checking if $M''(x) \geq 0$ seems to be too difficult.

I was able to compute $M''(x)$. Let $M(x) = f(x)^{\frac{1}{x}}$ where $f(x) = \frac{\sum_{i=1}^n a_i^x}{n}$. Then I was able to find: $$ \frac{M''(x) f^2(x)}{\big(\frac{M(x)}{x^4} \big)} = (xff' - f \ln f)^2 + x^3(ff'' - f'^2) - 2x(xff' - f^2 \ln f) $$ where $'$ denotes derivative and number denotes powers. Checking the sign of $M''(x)$ is equivalent to the checking the sign of the RHS. This terms seem too hard to check for positivity. Can anyone help? I know for a fact that $M(x)$ is convex and concave on different intervals, but I can't find these intervals. Can anyone help?

Anon
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  • What do you mean by canonical answer? – Some Guy Apr 08 '21 at 04:21
  • In the equation for $M''$, the first term in RHS should be $(xf' - f \ln f)^2$. It seems difficult to find the exact intervals? Can we prove that $M''(x) < 0$ for $x > x_0$ for some $x_0\ge 1$? – River Li Apr 08 '21 at 05:48
  • @SomeGuy I meant just find the answer for upto the 3 variable case (unfortunately MSE gives a list of options which I can tick from, and the nearest to "find the answer" was "find the canonical answer", hence I ticked that. – Anon Apr 08 '21 at 05:49
  • @RiverLi If you can give me the value of $x_0$, that would helpful (I would accept it). However an answer like "$\exists x_0$, where .... " wouldn't be so helpful. – Anon Apr 08 '21 at 05:50
  • One possible way to find the intervals could be to equate the triple derivative to 0, to get the inflection points. – Anon Apr 08 '21 at 05:59
  • @Kaind Why did you choose the tag [contest-math] ? I'd guess it makes sense to replace the former by [convex-analysis] and possibly [convexity-inequality] . – Hanno Apr 09 '21 at 07:52
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    A good source might also be Ladislav Matejíčka: Short note on convexity of power mean https://www.researchgate.net/publication/285385599_Short_note_on_convexity_of_power_mean – Andreas Apr 09 '21 at 10:19
  • @Hanno If you find $M(x)$ to be convex/ concave on some interval, applying Jensen's to it, solves nearly half of all contest maths problems in one shot. – Anon Apr 09 '21 at 14:37

1 Answers1

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All this is complicated.

Too complicated $-$ consider alone $M''(x)=\ldots\;$
And paired with the resulting nescience, this excludes a canonical answer.

It is known that

  • if $n=2$ and $a_1\neq a_2$, then $M(x)$ possesses one inflection point (not necessarily at $x=0$). On the left of it, $M(x)$ is convex. And concave to the right of it.
    See Nam & Minh: "Proof for a Conjecture on General means"
    (J of Inequalities in Pure & Applied Math., Vol. 9, 2008)
  • if $n=3$, then more inflection points may pop up.
    A concrete instance having three inflection points (albeit in the weighted context) was proposed by Grant Keady , cf the first paragraph on page 2 of "Conjecture on General means"
    (J of Inequalities in Pure & Applied Math., Vol. 7, 2006).

The questions are as intriguing as old, you may also read H. Shniad, published 73 years ago
(Bulletin Amer. Math. Soc. 54, pp 770$-$776, 1948).

Hanno
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