Let $x=(x_1,\dots,x_n)$ and $\alpha=(\alpha_1,\dots,\alpha_n)$, where $x_i\geq 0, \alpha_i>0$ for $i=1,\dots,n$ and $\sum \limits_{i=1}^{n}\alpha_i=1$. For any number $t\neq 0$ we consider the mean of order $t$ of the numbers $x_1,\dots,x_n$ with weights $\alpha_i$: $$M_t(x,\alpha)=\left(\sum \limits_{i=1}^{n}\alpha_i x_i^t\right)^{1/t}$$

Show that a) $\lim \limits_{t\to 0} M_t(x,\alpha)=x_1^{\alpha_1}\dots x_n^{\alpha_n}$; b)$\lim \limits_{t\to +\infty} M_t(x,\alpha)=\max \limits_{1\leq i\leq n} x_i$; c)$\lim \limits_{t\to -\infty} M_t(x,\alpha)=\min \limits_{1\leq i\leq n} x_i$;

d) $M_t(x,\alpha)$ is a nondecreasing function of $t$ on $\mathbb{R}$ and is strictly increasing if $n>1$ and the numbers $x_i$ are all nonzero.

This is actually a problem from Zorich's book but I guess there is a flaw in the question.

  1. I guess we need to assume that all $x_i>0$ because $x_i^t$ is undefined for $t<0$ and $x_i=0$. Am I right?

  2. I solved parts a)-c) quite easily but I have some issues with d).

Let's take the derivative of $M_t(x,\alpha)$ with respect to $t$ for $t\neq 0$:

$$M'_t(x,\alpha)=\frac{M_t(x,\alpha)}{t^2(\alpha_1x_1^t+\dots+\alpha_n x_n^t)} \left[ \alpha_1x_1^t \ln x_1^t+\dots +\alpha_nx_n^t \ln x_n^t-(\alpha_1x_1^t+\dots+\alpha_n x_n^t)\ln(\alpha_1x_1^t+\dots+\alpha_n x_n^t)\right];$$

So I will assume that all $x_i>0$ (I've explained above why I need to do this).

Easy to see that $\dfrac{M_t(x,\alpha)}{t^2(\alpha_1x_1^t+\dots+\alpha_n x_n^t)}>0$.

Since the function $f:(0,\infty)\to \mathbb{R}$ defined by $x\mapsto x\ln x$ is convex then by Jensen's inequality it follows that $$f(\alpha_1x_1^t+\dots+\alpha_nx_n^t)\leq \alpha_1 f(x_1^t)+\dots+\alpha_n f(x_n^t)$$ or equivalently $$ \alpha_1x_1^t \ln x_1^t+\dots +\alpha_nx_n^t \ln x_n^t-(\alpha_1x_1^t+\dots+\alpha_n x_n^t)\ln(\alpha_1x_1^t+\dots+\alpha_n x_n^t)\geq 0.$$ Hence $M'_{t}\geq 0$ which means that $M_t$ is nondecreasing function.

But I have issues to prove the second part in d), i.e. the case when it is strictly increasing. I am trying to use this question but it did not work out.

I'd be thankful for your help!

  • $\begingroup$ Shouldn't $f$ be strictly convex because $f''$ is $\gt 0$ for all $x\gt 0$? $\endgroup$
    – Koro
    May 16, 2021 at 16:34
  • $\begingroup$ @Koro, you mean for all $t>0$? $\endgroup$
    – RFZ
    May 16, 2021 at 16:41
  • $\begingroup$ $t\ne 0$ doesn’t have anything to do with $f$. The mapping f is $x\to \ln x$. And as you mentioned in your post, $x_i’s$ are positive. Right? $\endgroup$
    – Koro
    May 16, 2021 at 17:03
  • $\begingroup$ @Koro, Oh I see. I forgot that $t$ has not nothing in common with $f$. Yes I think that $f''(x)=\frac{1}{x}>0$. So what do you want to say? $\endgroup$
    – RFZ
    May 16, 2021 at 17:22
  • 1
    $\begingroup$ Hi @ZFR :) How are you? I hope you’re doing well. I was involved in some other question so couldn’t respond sooner. I have a class in 20 mins. I’ll take a look again as my earlier comment didn’t solve your question. Meanwhile, it will be great if you’d let me know why my comment didn’t solve your question. Thanks:) $\endgroup$
    – Koro
    May 30, 2021 at 4:01

1 Answer 1


Here is a step-by-step proof for the point d) using strict Jensen's inequality. But before that, we need to reword a bit the proposition:

  1. $M_t(x, \alpha)$ is not formally defined at $t=0$. However, it can be trivially defined by continuity: $M_0(x, \alpha)\ \widehat=\ \lim_{t\to0} M_t(x, \alpha) = \prod_i x_i^{\alpha_i}$, as demonstrated in a). Doing so, $M_t(x, \alpha)$ is defined $\forall t\in\mathbb{R}$.
  2. If all the $x_i$ are equal, then $M_t(x, \alpha)=x_1$. Therefore, for $M_t(x, \alpha)$ to be strictly increasing with $t$, all the $x_i$ can not be equal; being not all zero is not enough.

We show now that $p<q$ implies $M_p(x, \alpha)<M_q(x, \alpha)$.

Case 1: $0<p<q$.

$x\mapsto x^{q/p}$ being strictly convex, we can use strict Jensen's inequality: $$ \left(\sum_i \alpha_i y_i\right)^{q/p} < \sum_i \alpha_i y_i^{q/p} $$ Posing $y_i=x_i^p$, it comes $$ \left(\sum_i \alpha_i x_i^p\right)^{q/p} < \sum_i \alpha_i x_i^q. $$ $x\mapsto x^{1/q}$ being strictly increasing for $q>0$, we find immediately $$ \left(\sum_i \alpha_i x_i^p\right)^{1/p} < \left(\sum_i \alpha_i x_i^q\right)^{1/q}, $$ that is, $M_p(x, \alpha) < M_q(x, \alpha)$.

Case 2: $0<p$.

$x\mapsto \ln x$ being strictly concave, we can use also the strict Jensen's inequality for strictly concave functions: $$ \ln\left(\sum_i\alpha_i y_i\right) > \sum_i\alpha_i \ln y_i $$ Posing $y_i=x_i^p$, this becomes $$ \ln\left(\sum_i\alpha_i x_i^p\right) > \sum_i\alpha_i \ln x_i^p = \ln \prod_i x_i^{\alpha_i p } = \ln \left(\prod_i x_i^{\alpha_i}\right)^p $$

$\ln$ being strictly increasing, we obtain $$ \sum_i\alpha_i x_i^p > \left(\prod_i x_i^{\alpha_i}\right)^p, $$ $x\mapsto x^{1/p}$ being strictly increasing for $p>0$, we finally obtain $$ \left(\sum_i\alpha_i x_i^p\right)^{1/p} > \prod_i x_i^{\alpha_i}, $$ that is, $M_p(x, \alpha) > M_0(x, \alpha)$.

The demonstration for the two other cases, that is $p<q<0 \Rightarrow M_p(x, \alpha)<M_q(x, \alpha)$ and $p<0 \Rightarrow M_p(x, \alpha) < M_0(x, \alpha)$, are similar. Just be careful that $x\mapsto x^{1/p}$ and $x\mapsto x^{1/q}$ are now decreasing functions, which leads to a change of the inequalities during the derivation of the proof.

  • $\begingroup$ Thanks a lot for your help! Indeed, when I was trying to solve it a while ago I noticed that some condition is missing. And you pointed out that not all of the $x_i$ are equal. Indeed, in order to apply Jensen's inequality for strictly convex functions in a correct way we need the following: we are given positive numbers $\lambda_1,\dots,\lambda_n$ whose sum is 1 and we are given points $x_1,\dots,x_n\in (a,b)$ where at least two are not equal. $\endgroup$
    – RFZ
    May 30, 2021 at 15:10
  • $\begingroup$ @ZFR, my pleasure ;) $\endgroup$ Jun 1, 2021 at 16:48

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