Let
be a continuous, strictly monotonic function on
with inverse
, and let
be given positive real numbers. Then define,
![Rendered by QuickLaTeX.com \[ M_f = g \left( \frac{1}{n} \sum_{i=1}^n f(a_i) \right). \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-5ac115f91e9c0ba480487dfa7b2c61fd_l3.png)
This
is called the mean of
with respect to
. (When
for
, this coincides with the
th power mean from this exercise).
Show that
![Rendered by QuickLaTeX.com \[ f(M_f) = \frac{1}{n} \sum_{i=1}^n f(a_i). \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-8e9059c7983f547203cd5311caf945d3_l3.png)
Proof. Since
is the inverse of
we know
for all
in the range of
, i.e., for all
such that there is some
such that
.
By the definition of
then, we have that
![Rendered by QuickLaTeX.com \[ f(M_f) = f \left( g \left( \frac{1}{n} \sum_{i=1}^n f(a_i) \right) \right). \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-a0cf5044abb131368c5a9de88275ffa7_l3.png)
So, if
is in the domain of
then we are done. Since
is the inverse of
it’s domain is equal to the range of
. We show that this value is in the range of
using the intermediate value theorem.
Without loss of generality, assume
is strictly increasing (the alternative assumption, that
is strictly decreasing will produce an almost identical argument). Then, since
are all positive real numbers we have
. (Here if we’d assumed that
was strictly decreasing the roles inequalities would be reversed.) Then we have,
![Rendered by QuickLaTeX.com \[ f(a_1) = \frac{1}{n} \sum_{i=1}^n f(a_1) < \frac{1}{n} \sum_{i=1}^n f(a_i) < \frac{1}{n} \sum_{i=1}^n f(a_n) = f(a_n). \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-ceb0d1cca9a8e2d4d424c3ecf11cf37f_l3.png)
Hence, by the intermediate value theorem, since
![Rendered by QuickLaTeX.com \[ \frac{1}{n} \sum_{i=1}^n f(a_i) \in [f(a_1), f(a_n)] \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-caa1877d1abb07caee5588f4f75f3509_l3.png)
there must be some
such that
![Rendered by QuickLaTeX.com \[ f(c) = \frac{1}{n} \sum_{i=1}^n f(a_i).\]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-71000e7680f85b902bb3ff529e3e0b29_l3.png)
Thus,
is in the domain of
, so
![Rendered by QuickLaTeX.com \[ f(M_f) = f \left( g \left( \frac{1}{n} \sum_{i=1}^n f(a_i) \right) \right) = \frac{1}{n} \sum_{i=1}^n f(a_i). \qquad \blacksquare\]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-a2d3d3139c737c3bc8fda20f8b08dac4_l3.png)