## Weighted Power Means

Let $x,\alpha\in\mathbb{R}^{n}$ have positive components and satisfy $\sum_{k=1}^{n}\alpha_{k}=1$. We define the weighted power mean of order $t$ by

$\displaystyle M_{t}(x;\alpha):=\left(\sum_{k=1}^{n}\alpha_{k}x_{k}^{t}\right)^{\frac{1}{t}}$

I claim that

$\displaystyle\lim_{t\rightarrow 0^{+}}M_{t}(x;\alpha)=\prod_{k=1}^{n}x_{k}^{\alpha_{k}}$

Proof. Observe that

$\displaystyle t\sum_{k=1}^{n}\alpha_{k}\log\left|x_{k}\right|=\sum_{k=1}^{n}\alpha_{k}\log\left|x_{k}^{t}\right|\leq\log\left(\sum_{k=1}^{n}\alpha_{k}x_{k}^{t}\right)$,

by the concavity of $\log$. Hence,

$\displaystyle\exp\left(t\sum_{k=1}^{n}\alpha_{k}\log\left|x_{k}\right|\right)\leq\sum_{k=1}^{n}\alpha_{k}x_{k}^{t}$

Taking the $\frac{1}{t}$-root of both sides, we obtain that

$\displaystyle\prod_{k=1}^{n}x_{k}^{\alpha_{k}}=\exp\left(\sum_{k=1}^{n}\alpha_{k}\log\left|x_{k}\right|\right)\leq\left(\sum_{k=1}^{n}\alpha_{k}x_{k}^{t}\right)^{\frac{1}{t}}$

Hence, $\liminf_{t\rightarrow 0^{+}}M_{t}(x;\alpha)\geq\prod_{k=1}^{n}x_{k}^{\alpha_{k}}$. This establises a convenient lower bound. To prove the desired equality, we use l’hospital’s rule:

$\begin{array}{lcl}\displaystyle\lim_{t\rightarrow 0^{+}}\log\left(\sum_{k=1}^{n}\alpha_{k}x_{k}^{t}\right)^{\frac{1}{t}}=\lim_{t\rightarrow 0^{+}}\dfrac{\log\left(\sum_{k=1}^{n}\alpha_{k}x_{k}^{t}\right)}{t}&=&\displaystyle\lim_{t\rightarrow 0^{+}}\dfrac{\sum_{k=1}^{n}\alpha_{k}\log(x_{k})x_{k}^{t}}{\sum_{k=1}^{n}\alpha_{k}x_{k}^{t}}\\[.7 em]&=&\displaystyle\dfrac{\sum_{k=1}^{n}\alpha_{k}\log(x_{k})}{\sum_{k=1}^{n}\alpha_{k}}\\[.7 em]&=&\displaystyle\sum_{k=1}^{n}\alpha_{k}\log(x_{k})\end{array}$

By the continuity of the exponential, we obtain that

$\displaystyle\lim_{t\rightarrow 0^{+}}\left(\sum_{k=1}^{n}\alpha_{k}x_{k}^{t}\right)^{\frac{1}{t}}=\exp\left(\sum_{k=1}^{n}\alpha_{k}\log(x_{k})\right)=\prod_{k=1}^{n}x_{k}^{\alpha_{k}}$

$\Box$

Observe that for $t>0$,

$\begin{array}{lcl} \displaystyle M_{-t}(x;\alpha)=\left(\sum_{k=1}^{n}\alpha_{k}x_{k}^{-t}\right)^{-\frac{1}{t}}=\left(\sum_{k=1}^{n}\alpha_{k}(x_{k}^{-1})^{t}\right)^{-\frac{1}{t}}&=&\displaystyle M_{t}(x^{-1};\alpha)^{-1}\\[.7 em]&\leq&\displaystyle\left(\prod_{k=1}^{n}x_{k}^{-\alpha_{k}}\right)^{-1}\\[.7 em]&=&\displaystyle\prod_{k=1}^{n}x_{k}^{\alpha_{k}}\\[.7 em]&=&\displaystyle M_{0}(x;\alpha)\end{array}$

For all $s\leq t$, $M_{s}(x;\alpha)\leq M_{t}(x;\alpha)$.

Proof. First, assume that $s\leq t$ and $s,t \neq 0$. Since the function $x^{\frac{t}{s}}$ is convex, we have by Jensen’s inequality that

$\displaystyle\left(\sum_{k=1}^{n}\alpha_{k}x_{k}^{s}\right)^{\frac{t}{s}}\leq\sum_{k=1}^{n}\alpha_{k}(x_{k}^{s})^{\frac{t}{s}}=\sum_{k=1}^{n}\alpha_{k}x_{k}^{t}\Longrightarrow \left(\sum_{k=1}^{n}\alpha_{k}x_{k}^{s}\right)^{\frac{1}{s}}\leq\left(\sum_{k=1}^{n}\alpha_{k}x_{k}^{t}\right)^{\frac{1}{t}}$

We’ve already proven the cases $s=0$ and $t>0$ and $s<0$ and $t=0$, so we’re done. $\Box$

Other properties of the weighted power means include that the function $t\mapsto t\log M_{t}(x;\alpha)$ is convex on $\mathbb{R}$. Since $t\mapsto t\log M_{t}(x;\alpha)$ is continuous, it suffices to show that (see my post for why)

$\displaystyle (t+h)\log M_{t+h}(x;\alpha)+(t-h)\log M_{t-h}(x;\alpha)-2t\log M_{t}(x;\alpha)\geq 0$

for all $t\in\mathbb{R}$ and $h>0$. Observe that this expression is equal to

$\begin{array}{lcl}&=&\displaystyle\log\left(\sum_{k=1}^{n}\alpha_{k}x_{k}^{t+h}\right)+\log\left(\sum_{k=1}^{n}\alpha_{k}x_{k}^{t-h}\right)-2\log\left(\sum_{k=1}^{n}\alpha_{k}x_{k}^{t}\right)\\[1.1 em]&=&\displaystyle\log\left(\sum_{k=1}^{n}\sum_{j=1}^{n}\alpha_{k}\alpha_{j}x_{k}^{t+h}x_{j}^{t-h}\right)-\log\left(\sum_{k=1}^{n}\sum_{j=1}^{n}\alpha_{k}\alpha_{j}(x_{k}x_{j})^{t}\right)\\[1.1 em]&=&\displaystyle\log\left(\dfrac{\sum_{k=1}^{n}\sum_{j=1}^{n}\alpha_{k}\alpha_{j}(x_{k}x_{j})^{t}(x_{k}x_{j}^{-1})^{h}}{\sum_{k=1}^{n}\sum_{j=1}^{n}\alpha_{k}\alpha_{j}(x_{k}x_{j})^{t}}\right)\end{array}$

Since

$\displaystyle\left(\dfrac{x_{j}}{x_{k}}\right)^{h}+\left(\dfrac{x_{k}}{x_{j}}\right)^{h}=\dfrac{x_{j}^{2h}+x_{k}^{2h}}{(x_{k}x_{j})^{h}}\geq \dfrac{2x_{j}^{h}x_{k}^{h}}{(x_{j}x_{k})^{h}}=2$,

we conclude that

$\displaystyle\log\left(\dfrac{\sum_{k=1}^{n}\sum_{j=1}^{n}\alpha_{k}\alpha_{j}(x_{k}x_{j})^{t}(x_{k}x_{j}^{-1})^{h}}{\sum_{k=1}^{n}\sum_{j=1}^{n}\alpha_{k}\alpha_{j}(x_{k}x_{j})^{t}}\right)> \log(1)=0$

So far, we have not said anything about the limiting behavior of $M_{t}(x;\alpha)$ at $t=\pm\infty$. We shall redress this omission now. Define $M_{-\infty}(x;\alpha):=\inf_{1\leq k\leq n}x_{k}$ and $M_{\infty}(x;\alpha):=\sup_{1\leq k\leq n}x_{k}$. I claim that

$\displaystyle M_{\infty}(x;\alpha)=\lim_{t\rightarrow\infty}M_{t}(x;\alpha),\indent M_{-\infty}(x;\alpha)=\lim_{t\rightarrow-\infty}M_{t}(x;\alpha)$

Proof. Let $j,j'$ be integers in $\left\{1,\cdots,n\right\}$ such that $x_{j}=M_{\infty}(x;\alpha)$ and $x_{j'}=M_{-\infty}(x;\alpha)$. Then for $t>0$,

$\displaystyle\alpha_{j}^{\frac{1}{t}}M_{\infty}(x;\alpha)=\left(\alpha_{j}x_{j}^{t}\right)^{\frac{1}{t}}\leq M_{t}(x;\alpha)\leq\left(\sum_{k=1}^{n}\alpha_{k}M_{\infty}(x;\alpha)^{t}\right)^{\frac{1}{t}}=M_{\infty}(x;\alpha)$

and for $t<0$,

$\displaystyle\alpha_{j'}^{\frac{1}{t}}M_{-\infty}(x;\alpha)=\left(\alpha_{j'}x_{j'}^{t}\right)^{\frac{1}{t}}\leq M_{t}(x;\alpha)\leq\left(\sum_{k=1}^{n}\alpha_{k}M_{-\infty}(x;\alpha)^{t}\right)^{\frac{1}{t}}=M_{-\infty}(x;\alpha)$

Since $\alpha_{j}^{\frac{1}{t}}\rightarrow 1$ as $t\rightarrow\infty$ and $\alpha_{j'}^{\frac{1}{t}}\rightarrow 1$ as $t\rightarrow-\infty$, an application of the squeeze theorem completes the proof. $\Box$