Logarithm of AM/GM ratio: $\sqrt{\log((x+y)/(2\sqrt{xy}))}$ - MathOverflow

Logarithm of AM/GM ratio: $\sqrt{\log((x+y)/(2\sqrt{xy}))}$

2010-09-28T16:02:00Z

Recently, while playing around with infinite-divisibility, i arrived at the following metric:

$$d(x,y) := \sqrt{\log\left(\frac{x+y}{2\sqrt{xy}}\right)},$$

defined for positive reals $x$ and $y$. Proving that $d$ is a metric is trivial, except for the triangle-inequality. However, we can bypass a direct proof by appealing to Schoenberg's theorem (I. J. Schoenberg. Metric spaces and positive-definite functions, TAMS, 1938), from which the metricity follows easily because $-\log(x+y)$ is a conditionally positive-definite kernel.

However, i have been searching for following:

Applications / situations where this metric shows up?

An elementary proof of $d(x,y)$ being a metric.

Remarks

a. A google search on "ratio arithmetic geometric mean" yields some applications of the ratio alone;

b. An elementary proof should exist, but my initial attempts have not been that successful, especially as i stubbornly did not want to use differential calculus.

c. Notice that while proving $$d(x,y) \le d(x,z) + d(y,z),$$ we may assume wlog $x < 1$ and $y > 1$ and $z=1$, as proving the other cases ranges from very-trivial to trivial.

Answer by Willie Wong for Logarithm of AM/GM ratio: $\sqrt{\log((x+y)/(2\sqrt{xy}))}$

2010-09-29T17:40:24Z

Take the coordinate transformation from $\mathbb{R}$ to $\mathbb{R}_+$ by the exponential map. Then $(x+y) / \sqrt{xy} = e^{a-b} + e^{b-a}$ where $x = e^{2a}$ and $y = e^{2b}$. So we re-write

$$ d(e^{2a}, e^{2b}) = \sqrt{\log \cosh (a-b) } $$

So it suffices to consider the function $q(x) = \sqrt{ \log \cosh x}$, which is an even function with a unique minimum at $x = 0$, where $q(0) = 0$. Triangle inequality reduces then to checking that $q$ is sub-additive. If you were willing to use calculus, then this follows from the fact that $x q'(x)$ is negative except at $0$, where the derivative is not well defined.

In any case, a simple lemma for situations like this is

Lemma Let $f(x)$ be an increasing function on the positive real line with $f(0) = 0$. Then $f$ is sub-additive (i.e. $f(s+t) \leq f(s) + f(t)$) if $f(x)/x$ is decreasing.

Thus the triangle inequality boils down to checking $\log \cosh x / x^2$ is a decreasing function of (positive) $x$. At this point, I am lost as to how to check this fact without using any calculus at all. (Somewhere you will need to delve into properties of the natural logarithm or hyperbolic cosine, and calculus is the most natural tool there.)

As to where this metric comes up: the function $q(x)^2$ behaves like $x^2/2$ near the origin and $|x| - \log 2$ for large values. This makes it a Huber function*, which has found many uses in the literature of applied mathematics, especially with image processing (related to fMRI and things like that) and in robust statistics. (My knowledge here is only superficial, so you are probably better off asking a local expert in those fields about the $\log \cosh$ function.)

*Huber, P. J., 1973, Robust regression: Asymptotics, conjectures, and Monte Carlo: Ann. Statist., 1, 799-821

Answer by fedja for Logarithm of AM/GM ratio: $\sqrt{\log((x+y)/(2\sqrt{xy}))}$

2010-09-29T21:14:56Z

This is just the $L^2(\frac{dt}{t})$ distance between $e^{-xt}$ and $e^{-yt}$ (Frullani integrals) up to some positive factor. I'm not sure whether you'll call this "elementary" though.