Show the mean of a strictly monotonic function lies in an interval

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

$a_1 < M_f < a_n.$

Proof. Since $f$ is strictly monotonic on the positive real axis and $a_1 < a_2 < \cdots < a_n$ are $n$ positive reals, we know $f$ is strictly increasing or strictly decreasing, and correspondingly we have,

$f(a_1) < f(a_2) < \cdots < f(a_n) \qquad \text{or} \qquad f(a_1) > f(a_2) > \cdots > f(a_n).$

First, assume $f$ is striclty increasing, then

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

Since $f$ is strictly increasing so is its inverse $g$ (by Apostol’s Theorem 3.10); thus, we have

$\begin{align*} &g \left( \frac{1}{n} \sum_{i=1}^n f(a_1) \right) < g \left( \frac{1}{n} \sum_{i=1}^n f(a_i) \right) < g \left( \frac{1}{n} \sum_{i=1}^n f(a_n) \right) \\[8pt] &\implies \quad g(f(a_1)) < M_f < g(f(a_n)) \\ &\implies \quad a_1 < M_f < a_n. \end{align*}$

If $f$ is strictly decreasing then

$\begin{align*} & f(a_1) > \cdots > f(a_n) \\[8pt] \implies & \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 \\[8pt] \implies & g\left( \frac{1}{n} \sum_{i=1}^n f(a_1) \right) < g \left( \frac{1}{n} \sum_{i=1}^n f(a_i) \right) < g \left( \sum_{i=1}^n f(a_n) \right) \\[8pt] \intertext{(where the inequalities reverse since $f$ decreasing implies its inverse, $g$, is also decreasing)} \implies & a_1 < M_f < a_n. \qquad \blacksquare \end{align*}$

Stumbling Robot

Show the mean of a strictly monotonic function lies in an interval

Related

Point out an error, ask a question, offer an alternative solution (to use Latex type [latexpage] at the top of your comment): Cancel reply