2 Update on conjecture

I think I've essentially solved the problem for Mobius transformations (rational maps of degree 1). The situation can basically be summarized as "nothing to see here, folks" except for the one interesting case brought up by Michael Lugo. I think I know what happens in that case, but I haven't proved all of it. For brevity, I'm not going to prove everything rigorously because there are many cases.

In everything that follows, f(x) = (ax + b)/(cx + d), and {an(x)}n is the sequence of averages starting at x.

First a few obvious things:

(1) If x is a fixed point of f, then clearly an(x) = x for all n and thus converges to x trivially. A non-identity Mobius transformation typically has two fixed points, or one fixed point if f is parabolic (conjugate to a translation).

(2) Infinity has to be treated specially because if it appears anywhere in the sequence, then the sequence from that point on is constant at infinity and thus converges trivially to infinity. Thus, if x = f^(-n) (infinity) for any n, then an(x) converges to infinity.

The behavior of an(x) for general x depends on the classification of f as parabolic, hyperbolic/loxodromic, or elliptic. Also, whether or not infinity is one of the fixed points can be important.

(3) If f is parabolic, having a unique fixed point a: if a = infinity, then f is a translation, and it's easy to see that an(x) must diverge to infinity. If a ≠ infinity, then for all x in C, f^n(x) converges to a, and thus an(x) also converges to a (except as noted in point (2)).

(4) If f is loxodromic or hyperbolic, having two fixed points: if one of them is infinity, then conjugating by a Euclidean similarity, f(x) = rx for some complex r, where |r| ≠ 1. Then

an(x) = x(r^(n+1)-1)/((n+1)*(r-1)),

which converges to 0 for all x if |r| < 1 or diverges to infinity for all x if |r| > 1 (except as noted in point (1)). If infinity is not one of the fixed points, then f has two fixed points in the complex plane, one of them attracting and one of them repelling. f^n(x) and an(x) then converge to the attracting fixed point for all x (except as noted in points (1) and (2)).

(5) If f is elliptic, having two fixed points: if one of them is infinity, then f is a Euclidean rotation, and it's easy to see that an(x) converges to the center of rotation for all x. If infinity isn't a fixed point, then conjugating by a Euclidean similarity, the two fixed points are +/- i, so

f(x) = (cos(t)x - sin(t))/(sin(t)x + cos(t))

for some fixed t. If t is a rational multiple of π, then f has finite order, and an(x) actually converges to a rational function of x. Otherwise, let r = |(x - i)/(-ix + 1)|. If r ≠ 1, then an(x) converges to the integral

(1/(2π))∫0(re^(iθ) + i)/(ire^(iθ) + 1)dθ

which can be computed without too much difficulty (say, by residue calculus) to be i if r < 1 (if x is in the upper half-plane) or -i if r > 1 (if x is in the lower half-plane). If r = 1, then x is real. Then things get interesting, because {f^(-n)(infinity)} is a countable dense subset in R. However, if x is real and not in this set, then I believe (but haven't proved) that an(x) does not converge or diverge to infinity. My thinking: the distribution of the iterates f^n(x) can be shown to converge to the Cauchy distribution, with density 1/(π(1+x^2)) for x∈R. This distribution has pathological properties, such as not being integrable. In fact, if X1, ..., Xm are independent variables picked from this distribution, then (X1 + ... + Xm)/m actually has the same distribution. Based on that, a reasonable conjecture might be that the distribution of the averages an(x) also converges to the same Cauchy distribution, which would certainly preclude any kind of convergence of the values of the sequence. [edit: Actually, I can't support that conjecture numerically; it might be false, or the distribution might just converge very slowly.]

1

I think I've essentially solved the problem for Mobius transformations (rational maps of degree 1). The situation can basically be summarized as "nothing to see here, folks" except for the one interesting case brought up by Michael Lugo. I think I know what happens in that case, but I haven't proved all of it. For brevity, I'm not going to prove everything rigorously because there are many cases.

In everything that follows, f(x) = (ax + b)/(cx + d), and {an(x)}n is the sequence of averages starting at x.

First a few obvious things:

(1) If x is a fixed point of f, then clearly an(x) = x for all n and thus converges to x trivially. A non-identity Mobius transformation typically has two fixed points, or one fixed point if f is parabolic (conjugate to a translation).

(2) Infinity has to be treated specially because if it appears anywhere in the sequence, then the sequence from that point on is constant at infinity and thus converges trivially to infinity. Thus, if x = f^(-n) (infinity) for any n, then an(x) converges to infinity.

The behavior of an(x) for general x depends on the classification of f as parabolic, hyperbolic/loxodromic, or elliptic. Also, whether or not infinity is one of the fixed points can be important.

(3) If f is parabolic, having a unique fixed point a: if a = infinity, then f is a translation, and it's easy to see that an(x) must diverge to infinity. If a ≠ infinity, then for all x in C, f^n(x) converges to a, and thus an(x) also converges to a (except as noted in point (2)).

(4) If f is loxodromic or hyperbolic, having two fixed points: if one of them is infinity, then conjugating by a Euclidean similarity, f(x) = rx for some complex r, where |r| ≠ 1. Then

an(x) = x(r^(n+1)-1)/((n+1)*(r-1)),

which converges to 0 for all x if |r| < 1 or diverges to infinity for all x if |r| > 1 (except as noted in point (1)). If infinity is not one of the fixed points, then f has two fixed points in the complex plane, one of them attracting and one of them repelling. f^n(x) and an(x) then converge to the attracting fixed point for all x (except as noted in points (1) and (2)).

(5) If f is elliptic, having two fixed points: if one of them is infinity, then f is a Euclidean rotation, and it's easy to see that an(x) converges to the center of rotation for all x. If infinity isn't a fixed point, then conjugating by a Euclidean similarity, the two fixed points are +/- i, so

f(x) = (cos(t)x - sin(t))/(sin(t)x + cos(t))

for some fixed t. If t is a rational multiple of π, then f has finite order, and an(x) actually converges to a rational function of x. Otherwise, let r = |(x - i)/(-ix + 1)|. If r ≠ 1, then an(x) converges to the integral

(1/(2π))∫0(re^(iθ) + i)/(ire^(iθ) + 1)dθ

which can be computed without too much difficulty (say, by residue calculus) to be i if r < 1 (if x is in the upper half-plane) or -i if r > 1 (if x is in the lower half-plane). If r = 1, then x is real. Then things get interesting, because {f^(-n)(infinity)} is a countable dense subset in R. However, if x is real and not in this set, then I believe (but haven't proved) that an(x) does not converge or diverge to infinity. My thinking: the distribution of the iterates f^n(x) can be shown to converge to the Cauchy distribution, with density 1/(π(1+x^2)) for x∈R. This distribution has pathological properties, such as not being integrable. In fact, if X1, ..., Xm are independent variables picked from this distribution, then (X1 + ... + Xm)/m actually has the same distribution. Based on that, a reasonable conjecture might be that the distribution of the averages an(x) also converges to the same Cauchy distribution, which would certainly preclude any kind of convergence of the values of the sequence.