Means and inequalities for functions

A post on Monday looked at means an inequalities for a lists of non-negative numbers. This post looks at analogous means and inequalities for non-negative functions. We go from means defined in terms of sums to means defined in terms of integrals.

Let p(x) be a non-negative function that integrates to 1. (That makes p(x) a probability density, but you don’t have to think of this in terms of probability.) Think of p(x) as being a weighting function. The dependence on p(x) will be implicit. For a non-negative function f(x) and real number r ≠ 0, define M_r( f ) by

$M_r(f) = \left(\int p(x) \left( f(x) \right)^r \, dx \right)^{1/r}$

If r > 0 and the integral defining M_r( f ) diverges, we say M_r( f ) = ∞ and M_–r( f ) = 0.

Define the geometric mean of the function f by

$G(f) = \exp\left( \int p(x) \log f(x)\, dx \right)$

There are a few special cases to consider when defining G(f). See Inequalities for details.

First we give several limiting theorems.

As r → –∞, M_r( f ) → min( f )
As r → +∞, M_r( f ) → max( f )
As r → 0⁺, M_r( f ) → G( f )

And now for the big theorem: If r ≤ s, then M_r( f ) ≤ M_s( f ).The conditions under which equality hold are a little complicated. Again, see Inequalities for details.

We could derive analogous results for infinite sums since sums are just a special case of integrals.

The assumption that the weight function p(x) has a finite integral is critical. We could change the definition of M_r( f ) slightly to accommodate the case that the integral of p(x) is finite but not equal to 1, and all the conclusions above would remain true. But if we allowed p(x) to have a divergent interval, the theorems do not hold. Suppose p(x) is constantly 1, and our region of integration is (0, ∞). Then M_r( f ) might be more or less than M_s( f ) depending on f. For example, let f(x) = b exp( – bx ) for some b > 0. M₁( f ) = 1, but M_∞( f ) = b. Then M₁( f ) is less than or greater than M_∞( f ) depending on whether b is less than or greater than 1.