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Prove connection between power mean and arithmetic mean of a function

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

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


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

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

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