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Let $f : A \to A \subseteq \mathbb R$ be a real function with a fixed point $a_0 = f(a_0)$ which is not attractive.

Let $f^k = f \circ f \circ ... \circ f$ be the $k^{th}$ iterate of $f$ (with the usual convention that $f^0 = \text{id}_A$).

Let $A_n(a) = \frac{1}{n}\sum_{k=0}^{n-1} f^k(a)$ be the average of the first $n$ iterates of $f$ starting at $a \in A$.

Are there known sufficient conditions for the limit to exist $lim_{n \to \infty} A_n(a) = a_0$?

The above quite obviously holds if $x_0$ is an attractive fixed point, or if $f^k(a)$ can be otherwise shown to converge. But in cases where $f^k(a)$ diverges, it is still possible that $A_n$ converges.
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The question is related to a post on MSE the other day (deleted by its author since) that noted that for $f(x) = \ln(|x|)$ the average of the first $n$ iterates appeared to converge to a constant limit, which (not unexpectedly) turned out to be $-\Omega = -W(1)$ the fixed point where $log(-x) = x$.

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You might like to check this out for starters

https://www.openstarts.units.it/dspace/bitstream/10077/6488/1/MassaRendMat09.pdf

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  • $\begingroup$ Thank you for the pointer. It makes sense that this would have come up in the context of fixed point approximations. $\endgroup$
    – dxiv
    Commented Sep 22, 2016 at 17:58
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This seems underspecified. For instance, is the fixed point unique? Or even, is the function continuous at the given fixed point? If not, the following construction may give you some insight.

Consider g(x) = 1.5x: The fixed point at 0 is non-attractive. Define f to be g on [-2/3, 2/3] and repetitions of the same everywhere else: f(x) = g(frac(x)), where frac denotes the fractional part of a Real number (and is antisymmetric: frac(-x) = -frac(x)).

This function's only fixed point is 0, and it is still non-attractive. Its values under iteration fill the range [-1,1] according to some distribution (which may depend on the starting point). The limiting mean of any sequence of iterates is obviously well defined, since the range is bounded. For some non-zero starting values (probably almost all, but I do not see any easy proof of that), the mean of these values is 0.

You (or others) may be able to improve this construction or proof. But I suspect the primary value may be to refine the question.

[Apologies for not using MathJax here.]

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  • $\begingroup$ The question was deliberately underspecified as to not exclude more general answers. Any answer to "are there known sufficient conditions" could of course include restrictions such as number of fixed points, continuity etc. Thanks for the proposed example. $\endgroup$
    – dxiv
    Commented Sep 22, 2016 at 18:06

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