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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 pth 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*}

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