Absolutely Continuous Invariant Measures for Piecewise Convex Maps

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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What exactly do you mean by (4)? Is $f$ continuous at least? –  fedja Nov 21 '12 at 14:24
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]. –  Tom Kempton Nov 21 '12 at 14:41

If $f$ is piecewise $C^2$, since your map is piecewise expanding it has an ACIM. See Lasota and York (Trans AMS, 1973).
You can get away with $C^{1+\alpha}$, but with less regularity, I'm not sure. –  Banach Dec 4 '12 at 12:06