1. Vladimir Protasov (Moscow State University) Primitivity of matrix families and the problem of distribution of power random series

2. Random power series How to separate these two cases by a criterion ? x

3. A continuous monotone function is absolutely continuous iff it has a summable derivative 01 0 1 A continuous monotone function is purely singular iff A ``typical’’ monotone function is of mixed type, i.e. Any continuous monotone function can be decomposed in a unique way as

4. The ``simplest case’’. Bernoulli convolutions. 1 01 01 ? 0

8. In 1995 G.Derfel, N.Dyn, A.Levin formulated a ``dual’’ problem : The opposite case. The ``dual’’ problem. Erdos problem DDL problem Nothing is changed if we take t = 1/n, n >1 is an integer. Derfel, Dyn and Levin applied this problem to algorithms of extrapolation of functions by its values at integer points.

9. Theorem 1 (G.Derfel, N.Dyn, A.Levin, 1995, Y.Wang, 1995) This condition is not necessary ! Example 6. If P {h is even } = P { h is odd} = ½, then is absolutely continuous.  = 0, 1, 2, 3 P = 1/6, 1/3, 1/6, 1/3 P { h is even} = 1/3, P { h is odd} = 2/3 However, exists and, moreover, continuous. The density of distributionsatisfies the refinement equation The distribution is absolutely continuous if and only if How to check by the coefficients whether

10. To construct a system of compactly supported wavelets one needs to solve a refinement equation is a sequence of complex numbers sutisfying some constraints. This is a usual difference equation, but with the double contraction of the argument Refinement equations in the construction of wavelets

11. If it possesses a compactly supported solution such that What is known about refinement equations ? Conversely, if, then there exists a compactly supported solution, which is unique, up to multiplication by a constant and has its support on [0, N]. But only in the sense of distributions ! 0N then The refinable functions are never infinitely smooth

12. Example 1. Trivial: 01 Example 2. 02 Example 3. 0 3 The solution is unstable! A small perturbation of the coefficients may lead to the loss of absolute continuity: Example 4. The same with Special examples of refinement equations

13. Cavaretta, Dahmen and Micchelli (1991) Classified all refinable splines with integral nodes. Lawton, Lee and Shen (1995) Classified all refinable splines. For any N there are finitely many refinable splines of order N Berg and Plonka (2000), Hirn (2008) Protasov (2005) Classified all piecewise-smooth refinable functions. Thus, all piecewise-smooth refinable functions are splines. All of them are spanned by integral translates of the B-spline.

14. A “ typical ” refinable function and wavelet function Example 5 03 is the exponent of regularity (the Holder exponent) is not differentiable Nevertheless, it is differentiable almost everywhere the local exponent of regularity at the point x (very smooth) (very irregular) (breaks at all dyadic points) Fractal nature of refinable functions. Varying local regularity

15. How to check if I.Daubechies, D.Lagarias, 1991 A.Cavaretta, W.Dahmen, C.Micchelli, 1991 C.Heil, D.Strang, 1994 R.Q.Jia, 1995, K.S.Lau, J.Wang, 1995 Y.Wang, 1996 Example. How to compute the regularity of refinable functions ? The joint spectral radius of linear operators

16. The concept of primitive families. Definition. A family of matrices is called primitive if there exists at least one positive product.

17. Perron-Frobenius theorem (1912) A matrix A is not primitive if it is either reducible or one of the following equivalent conditions is satisfied:

18. A criterion of primitivity (conjectured in 2010 )

19. Now we can solve the problem by a criterion of absolute continuity The criterion is formulated in terms of roots of the characteristic function on a binary tree. Binary tree T: 1/2 3/4 1/4 1/8 1/16 7/8 15/16 5/83/8 9/16 5/16 13/163/16 11/16 7/16 a a/20.5 + a/2

20. 1/2 3/4 1/4 1/8 1/16 7/8 15/16 5/83/8 9/16 5/16 13/163/16 11/16 7/16 Definition 1. Examples. The sets A = {1/2}, A = {1/4, 3/4} are blocking A = {1/8, 5/16, 13/16, 3/4 } is not (it is non-symmetric). Theorem.. The distribution is absolutely continuous if and only if there is a blocking set A that consists of roots of the polynomial m(z), i.e., m(A) = 0. This condition can be checked within finite time.

21. 3/4 1/4 1/87/8 5/83/8 Example 7 (the case of Theorem 1 of DDL). P { h is even} = P { h is odd} = ½ iff A = {root} Example 8 (the case of Example 6).  = 0, 1, 2, 3 P = 1/6, 1/3, 1/6, 1/3 A = {1/4, 3/4} 1/2 The criterion is sharp, each case can be realized. A = {1/2}

22. m(1/2) =0 m(1/4) =0 m(3/4) =0 A = {1/4, 3/4} Thank you ! What is the set of all sets of probabilities for which the distribution is absolutely continuous ?