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Hi all,

I'm interested in a class of 'generalised tent maps' $f:[0,1]\to[0,1]$ for which

1) $f$ is strictly increasing on $[0, \frac{1}{2}]$, $f(0)=0$ and $f(\frac{1}{2})=1$

2) $f$ is symmetric about $\frac{1}{2}$, i.e. $f(x)=f(1-x)$.

3) $f$ is differentiable at 0 with $f'(0)>1$

4) $f$ is piecewise convex, but not strictly convex, on pieces $[0,1/2]$ and $[\frac{1}{2},1]$

5) $f$ is continuous.

Is it known that such functions f preserve absolutely continuous invariant probability measures?

I've seen a few papers proving the existence of acips for certain classes of piecewise convex functions, such as Lasota and Yorke (Trans AMS, 1982) and Bose et al (Studia Math 2003), but they always require that the function f is increasing on each of the pieces, which doesn't hold for the tent like constructions I'm interested in.

Thanks,

Tom

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  • $\begingroup$ What exactly do you mean by (4)? Is $f$ continuous at least? $\endgroup$
    – fedja
    Commented Nov 21, 2012 at 14:24
  • $\begingroup$ hi fedja, yes f is continuous, I've edited to include this. By 'convex but not strictly convex' I mean that, for $x,y \in [0,1/2]$, f(tx+(1−t)y)≤tf(x)+(1−t)f(y) for each t∈(0,1), but the inequality can't be made strict. The same holds for $x,y$ in [1/2,1]. $\endgroup$
    – Tom Kempton
    Commented Nov 21, 2012 at 14:41

1 Answer 1

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If $f$ is piecewise $C^2$, since your map is piecewise expanding it has an ACIM. See Lasota and York (Trans AMS, 1973).

See also the Chaos, Fractals, and Noise by A. Lasota and M. Mackey. One of the theorems in Ch. 6 might work for your maps.

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  • $\begingroup$ Thanks, I shall take another look at the book of Lasota and Mackey when I'm back in my home department, it's a really excellent book! Unfortunately the C^2 property is not ideal for me, I'm looking for a property that ensures that a generalised tent map has an acip and which is stable under perturbation. Since the limit of a convergent sequence of convex maps is convex I was tempted to try and use a convexity argument. $\endgroup$
    – Tom Kempton
    Commented Dec 4, 2012 at 8:37
  • $\begingroup$ You can get away with $C^{1+\alpha}$, but with less regularity, I'm not sure. $\endgroup$
    – Banach
    Commented Dec 4, 2012 at 12:06

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