# Generalized mean of order $t$ [Zorich's book]

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!

• Shouldn't $f$ be strictly convex because $f''$ is $\gt 0$ for all $x\gt 0$?
– Koro
May 16, 2021 at 16:34
• @Koro, you mean for all $t>0$?
– RFZ
May 16, 2021 at 16:41
• $t\ne 0$ 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?
– Koro
May 16, 2021 at 17:03
• @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?
– RFZ
May 16, 2021 at 17:22
• 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:)
– Koro
May 30, 2021 at 4:01

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 $$p implies $$M_p(x, \alpha).

Case 1: $$0.

$$x\mapsto x^{q/p}$$ 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.

$$x\mapsto \ln x$$ 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 $$p and $$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.

• 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.
– RFZ
May 30, 2021 at 15:10
• @ZFR, my pleasure ;) Jun 1, 2021 at 16:48