Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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

share|improve this question
    
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
add comment

1 Answer

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.

share|improve this answer
    
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. –  Tom Kempton Dec 4 '12 at 8:37
    
You can get away with $C^{1+\alpha}$, but with less regularity, I'm not sure. –  Banach Dec 4 '12 at 12:06
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.