Inequality of means for functions

Question

According to Wikipedia one can define the arithmetic mean of an integrable function $f: M \to \mathbb{R}$ over a relatively compact domain $M$ as:

$$A(f) = \frac{1}{\mu(M)}\int_M f$$

where $\mu(M)$ is the measure of $M$. Moreover, if $f > 0$ one can also define its geometric mean as:

$$G(f) = \exp\left(\frac{1}{\mu(M)}\int_M \log f\right)$$

The inequality of the arithmetic and geometric means is a well-known result in the discrete case. ¿Is it also true in this context? That is, is it true that:

$$A(f) \geq G(f)$$

for every positive integrable function $f$?

Answer 1

By Jensen's inequality you have that for a concave function $h$ (usually it is stated for convex functions with the other way round inequality)

$$h\left(\frac{1}{\mu(M)}\int_M f\right) \ge \frac{1}{\mu(M)}\int_M h(f)$$

As the $\log$ is concave you get that

$$\log \left(\frac{1}{\mu(M)}\int_M f\right) \ge \frac{1}{\mu(M)}\int_M \log(f)$$

As $e^x$ is increasing in $x$ the inequality is preserved when exponentiating, yielding to

$$e^{\log \left(\frac{1}{\mu(M)}\int_M f\right)} = \frac{1}{\mu(M)}\int_M f \ge \exp\left({\frac{1}{\mu(M)}\int_M \log(f)}\right)$$

Answer 2

Yes it is true: Use Jensen's inequality

And $x\mapsto \exp (x)$ is convex. Then you get,

$$G(f) = \exp\left(\frac{1}{\mu(M)}\int_M \log f\right)\le \frac{1}{\mu(M)}\int_M \exp(\log f) = A(f)$$

That is $$G(f) \le A(f)$$