Tag: Power Mean

Prove connection between power mean and arithmetic mean of a function

by RoRi

September 6, 2015

Let $f$ be a continuous, strictly monotonic function on $\mathbb{R}_{>0}$ with inverse $g$ , and let $a_1 < a_2 < \cdots < a_n$ be given positive real numbers. Then define,

$M_f = g \left( \frac{1}{n} \sum_{i=1}^n f(a_i) \right).$

This $M_f$ is called the mean of $a_1, \ldots, a_n$ with respect to $f$ . (When $f(x) = x^p$ for $p \neq 0$ , this coincides with the $p$ th power mean from this exercise).

Show that

$f(M_f) = \frac{1}{n} \sum_{i=1}^n f(a_i).$

Proof. Since $g$ is the inverse of $f$ we know $f(g(x)) = x$ for all $x$ in the range of $f$ , i.e., for all $x$ such that there is some $c \in \mathbb{R}_{>0}$ such that $f(c) = x$ .

By the definition of $M_f$ then, we have that

$f(M_f) = f \left( g \left( \frac{1}{n} \sum_{i=1}^n f(a_i) \right) \right).$

So, if $\frac{1}{n} \sum_{i=1}^n f(a_i)$ is in the domain of $g$ then we are done. Since $g$ is the inverse of $f$ it’s domain is equal to the range of $f$ . We show that this value is in the range of $f$ using the intermediate value theorem.

Without loss of generality, assume $f$ is strictly increasing (the alternative assumption, that $f$ is strictly decreasing will produce an almost identical argument). Then, since $a_1 < a_2 < \cdots < a_n$ are all positive real numbers we have $f(a_1) < f(a_2) < \cdots < f(a_n)$ . (Here if we’d assumed that $f$ was strictly decreasing the roles inequalities would be reversed.) Then we have,

$f(a_1) = \frac{1}{n} \sum_{i=1}^n f(a_1) < \frac{1}{n} \sum_{i=1}^n f(a_i) < \frac{1}{n} \sum_{i=1}^n f(a_n) = f(a_n).$

Hence, by the intermediate value theorem, since

$\frac{1}{n} \sum_{i=1}^n f(a_i) \in [f(a_1), f(a_n)]$

there must be some $c \in \mathbb{R}_{>0}$ such that

$f(c) = \frac{1}{n} \sum_{i=1}^n f(a_i).$

Thus, $\frac{1}{n} \sum_{i=1}^n f(a_i)$ is in the domain of $g$ , so

$f(M_f) = f \left( g \left( \frac{1}{n} \sum_{i=1}^n f(a_i) \right) \right) = \frac{1}{n} \sum_{i=1}^n f(a_i). \qquad \blacksquare$

Use the pth power mean to prove an inequality

by RoRi

July 10, 2015

If $a^2 + b^2 + c^2 = 8$ for $a,b,c > 0$ , prove that

$a^4 + b^4 + c^4 \geq \frac{64}{3}.$

Proof. For $a,b,c \in \mathbb{R}$ not all equal, we may apply the previous exercise with $x_1 = a, \ x_2 = b, \ x_3 = c$ and with $p = 2$ , to obtain,

$\begin{align*} M_2 < M_4 \qquad \qquad &\implies& \left( \frac{a^2+b^2+c^}{3} \right)^{1/2} &< \left( \frac{a^4+b^4+c^4}{3} \right)^{1/4} \\ &\implies& \left( \frac{8}{3} \right)^2 &< \frac{a^4+b^4+c^4}{3} \\ &\implies& a^4 + b^4 + c^4 &> \frac{64}{3}. \end{align*}$

Further, if $a=b=c$ , then

$\begin{align*} a^2+b^2+c^2 = 8 &&\implies 3a^2 &= 8\\ &&\implies a^2 = b^2 = c^2 &= \frac{8}{3}\\ &&\implies a^4 = b^4 = c^4 &= \frac{64}{9} \\ &&\implies a^4 + b^4 + c^4 &= \frac{64}{3}. \end{align*}$

Thus, the requested inequality holds for any $a,b,c \in \mathbb{R}$ , and as a bonus, we prove equality holds if and only if $a=b=c. \qquad \blacksquare$

Prove the pth power mean is less than the (2p)th power mean

by RoRi

July 10, 2015

We recall the definition of the $p$ th power mean $M_p$ .

For $x_1, \ldots, x_n \in \mathbb{R}_{>0}$ , and $p \in \mathbb{Z}$ with $p \neq 0$ , we define the $p$ th power-mean $M_p$ as:

$M_p = \left( \frac{x_1^p + \cdots + x_n^p}{n} \right)^{1/p}.$

Now, for $p > 0$ , prove $M_p < M_{2p}$ for $x_1, \ldots, x_n$ not all equal.

Proof. From the Cauchy-Schwarz inequality we know that for real numbers $a_1, \ldots, a_n$ and $b_1, \ldots, b_n$ , we have

$\left( \sum_{k=1}^n a_k b_k \right)^2 \leq \left( \sum_{k=1}^n a_k^2 \right) \left( \sum_{k=1}^n b_k^2 \right)$

with equality if and only if there is some $y \in \mathbb{R}$ such that $a_k y + b_k = 0$ for all $k$ . Letting $a_k = x_k^p$ and $b_k = 1$ we have

$\left( \sum_{k=1}^n x_k^p \cdot 1 \right)^2 < \left( \sum_{k=1}^n x_k^{2p} \right) \left( \sum_{k=1}^n 1 \right).$

This inequality is strict since if equality held there would exist some $y \in \mathbb{R}$ such that $(x_k^p)y + 1 = 0$ for all $k$ , but this would imply $x_k = \left(-\frac{1}{p} \right)^{1/p}$ for all $k$ , contradicting our assumption that the $x_k$ are not all equal. Since $\sum_{k=1}^n 1 = n$ (see here), this implies

$\begin{align*} &&\sum_{k=1}^n x_k^p &< \left( \sum_{k=1}^n x_k^{2p} \right)^{1/2} n^{1/2} \\ \implies && \left( \sum_{k=1}^n x_k^p \right)^{1/p} &< \left( \sum_{k=1}^n x_k^{2p} \right)^{1/2p} \cdot n^{1/2p} &(\text{raising to } 1/p)\\ \implies && \left (\frac{1}{n} \right)^{1/p} \cdot \left( \sum_{k=1}^n x_k^p \right)^{1/p} &< \left( \sum_{k=1}^n x_k^{2p} \right)^{1/2p} \cdot \left( \frac{1}{n} \right)^{1/2p} & (\text{multiplying by } (1/n)^{1/p})\\ \implies && \left( \frac{\sum_{k=1}^n x_k^p}{n} \right)^{1/p} &< \left( \frac{\sum_{k=1}^n x_k^{2p}}{n} \right)^{1/2p} & (\text{combining terms})\\ \implies && M_p &< M_{2p}. \qquad \blacksquare \end{align*}$