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Vladimir Protasov (Moscow State University)
Primitivity of matrix families and
the problem of distribution of power random series
Random power series
 


k 1
t k k ,
{ k } are i.i.d.
k has the same distribution as  for all k . F (x) = P {   x}.
A simple fact: If E 


x d F   , then the series  =

t
k 1
k
k
converges with probability 1 for every t  (0,1).
A less obvious fact: The sum  is always a continuous random value of ``pure type'', i.e.,
1) the distribution function F is continuous;
2) it is either absolutely continuous (there exists the density F '  L1 (R))
or purely singular ( F ' ( x)  0 for almost all x).
(O.Zakusilo, 1975-76, G.Derfel, 1989).
F ( x )
x
How to separate these two cases by a criterion ?
Having t and the distribution F0 one needs to determine
whether F is absolutely continuous or purely singular ?
The problem has found applications in
 number theory (P.Erdos, 1939-40; A.M.Garsia; 1962; R.Salem, 1963)
 fractals and dynamical systems (B.Solomyak, Yu.Peres, W.Schlag, 1995-2000)
 PDE and mathematical physics (G.Derfel, 1989; R.Schilling; J.Moravec, 2001)
 approximations and algorithms (G.Derfel, N.Dyn, A.Levin, 1995; Y.Wang, 1995-96)
A continuous monotone function is absolutely continuous iff it has a summable derivative
f ( x)
f ( x)
f '( x)  L 1 [0,1]
f '( x)  0 a.e.
x
0
1
f ( x)   f '(t ) d t
0
1
0
A continuous monotone function is purely singular iff f  const ,
f '( x)  0 a.e.
Any continuous monotone function can be decomposed in a unique way as
f  p f cont  q fsing ,
A ``typical’’ monotone function is of mixed type, i.e.
pq  1
p, q  0
The distribution function F of  

t
k
k 1
 k is always of pure type.
How to determine whether F is absolutely continuous of purely singular
by t and the distribution of 0 ?
The ``simplest case’’. Bernoulli convolutions.
In 1939-40 P.Erdos sdudied the special case 0  1 with probabilities 1/2
 

t
k
(Bernoulli convolutions). For which t this is singular ? Still unsolved.
k 1
F ( x )
 For t  0.5 we have a uniform distribution
-1
0
1
F ( x )
 For t  0.5 the function F is a piecewise-constant a.e.,
-1
and therefore it is purely singular.
 What can be said about t  (0.5 , 1) ?
Is it true that F is absolutely continuous for all such t ?
F ( x )
-1
0
?
0
1
1
The first examples of numbers t  (0.5, 1), for which the distribution of  

 t
k
k=1
is purely singular were found by P.Erdos in 1940.
If t   1, where  is a Pisot number, then  is singular.
Pisot numbers:  is an algebraic integer such that | |  1 and all its algebraic conjugate numbers
are smaller than 1 by modulus.
Pn ( z )  z n  an 1 z n 1
 a0
, z1 ,
, zn are roots. | z1 |  1 , | z j |  1, j  2,
  z1 is a Pisot number.
There are infinitely many Pisot numbers on (1 , 2) (so that  -1  (0.5 , 1)).
For example,  = ( 5  1) / 2 , P(z) = z 2  z  1. They are all irrational.
P.Erdos showed that for t   1 the function F is purely singular.
,n
Idea. Pisot numbers possess the following (characteristic ?) property: dist ( k , Z)  0 as k  ,
where dist ( k , Z) is the distance to the closest integer.
Indeed,
z1k  z2k 
 znk  Z, but z2k 
 znk  0, therefore dist ( z1k , Z)  dist ( k , Z)  0.
In particular cos 2  k  1 with an exponential rate.
Let  = F and ˆ ( y)    ( x) e
We have  ( x) 

2
2 ixy
R
 ( x -1) 

2
d x , where F is the distribution of  

k 1
 ( x  1)

Then

ˆ ( y ) 
ˆ ( y)



cos (2  1 y) ˆ ( 1 y)
cos 2   k y
k 1
For all integers n we have ˆ ( n )  C  0 uniformly for n  N .
Taking y   n , we obtain ˆ ( y )  0 as y  , and hence   L1 (R).
k
.
 F is absolutely continuous for almost all t  (0.5 , 1).
(B.Solomyak, 1995)
 Concrete examples: F is absolutely continuous for Salem numbers and for Garsia numbers.
 F is singular for uncountably many t  (0.5 , 1).
No concrete examples of such t are known thus far.
 Nothing is known for all rational t  (0.5 , 1).
Yu. Peres, 1997.
The opposite case. The ``dual’’ problem.
In 1995 G.Derfel, N.Dyn, A.Levin formulated a ``dual’’ problem:
Problem. Let t  1/ 2,  is an arbitrary integer-valued random value: P{  k}  pk ,
p
k
 1.

For which sequences {p k } the value  =  2  k k has density?
k =1
Erdos problem
η =  1 with equal probabilities.
t is arbitrary
DDL problem
η is an arbitrary integer-valued random value
t = 0.5 is fixed
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.
The density of distribution
  ( x)
 ( x) 
satisfies the refinement equation
N
2p
k 0
k
 (2 x  k )
The distribution is absolutely continuous if and only if
How to check by the coefficients whether
  L1 (R)
  L1 (R) ?
  L 1 (R)   1 (T0 |V , T1 |V )  1
1/ m
 m

 1 (T0 , T1 )  lim  2  Td 1 T d m 
m 
d1 , , d m


T0 , T1 are N  N  matrices, (Ti ) j k  2 p2 j  k 1i
Theorem 1 (G.Derfel, N.Dyn, A.Levin, 1995, Y.Wang, 1995)
If P { is even}  P { is odd} = ½, then F is absolutely continuous.
For N  4
 p0

p
T0  2  2
 p4

0
0
p1
p3
0
p0
p2
0
p4
0

0
p1 

p3 
 p1

p
T1  2  3
0

0
p0
p2
p4
0
0
p1
p3
0
0

p0 
p2 

p4 
This condition is not necessary !
Example 6.   0 ,
1 ,
2 , 3
P = 1/6 , 1/3 , 1/6 , 1/3
P { is even} = 1/3, P { is odd} = 2/3
However,  exists and, moreover, continuous.
Refinement equations in the construction of wavelets
To construct a system of compactly supported wavelets one needs to solve a refinement equation
 ( x) 
N
c
k 0
c0 ,
, cN
k
 (2 x  k ) ,
is a sequence of complex numbers sutisfying some constraints.
cN  (2 x  N )
( x)
c0  (2 x)
c1  (2 x  1)
.......
This is a usual difference equation, but with the double contraction of the argument
What is known about refinement equations ?
 ( x) 
N
c
k 0
k
 (2 x  k ) ,
  ( x) d x
If it possesses a compactly supported solution such that
N
Conversely, if
c
k 0
k
 0
N
then
c
k 0
k
 2
 2 , then there exists a compactly supported solution, which is unique,
up to multiplication by a constant and has its support on [0, N].
( x)
But only in the sense of distributions !
0
ˆ ( )

m( / 2) ˆ ( / 2) ,
ˆ ( )


1 N
m( ) 
c k e  2  i k

2 k 0
m (2
j
)
j 1
The refinable functions are never infinitely smooth
  C N ( R)
N
Special examples of refinement equations
Example 1.
c0  c1  1
 ( x)   (2 x)   (2 x  1)
Example 2.
Example 3.
c0 
c0 

Trivial:

0
1
1
, c1  1 , c 2 
2
2


1
3
3
1
, c1  , c 2  , c 3 
4
4
4
4


1
0
2
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
c0 
1

2
1

c0 
4
c1  1  
3
c1   
4
c2 
c2 
1
2
 is purely singular.
3
4
c3 
1
4
 is purely singular.
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.
Theorem. If there exists   0 such that the smoothness of a refinable function  on (0,  )
exceeds its smoothness on R , then  is a refinable spline with integral nodes.
Thus, all piecewise-smooth refinable functions are splines.
All of them are spanned by integral translates of the B-spline.
A “ typical ” refinable function and wavelet function
Example 5
 ( x)

1
2
1
2
 (2 x)   (2 x  1)   (2 x  2)   (2 x  3)
3
3
3
3
 ( x)  log 2 3  1.58...
(very smooth)
 ( x)
 ( x)  log 2 1.5  0.58...
(very irregular)
0

 sup {   0 |
3
(breaks at all dyadic points)
|  ( x  h)   ( x) |  C h  for all x, h }
is the exponent of regularity (the Holder exponent)
  log 2 1.5  0.58...


is not differentiable
Nevertheless, it is differentiable almost everywhere
 ( x)  sup {   0 |
|  ( x  h)   ( x ) |  C h 
}
the local exponent of regularity at the point x
At almost all points  ( x)  log 2 2.25  1.17...
Hence  '( x)  0 a.e.
Fractal nature of refinable functions. Varying local regularity
How to compute the regularity of refinable functions ?
The joint spectral radius of linear operators
How to check if
  C ( R) ?   L 1 ( R) ?
 C (R)    (T0 |V , T1 |V )  1
I.Daubechies, D.Lagarias, 1991
A.Cavaretta, W.Dahmen, C.Micchelli, 1991
C.Heil, D.Strang, 1994
Moreover,
  L 1 (R)   1 (T0 |V , T1 |V )  1
R.Q.Jia, 1995,
K.S.Lau, J.Wang, 1995
Y.Wang, 1996
Example.

 1 ( A0 , A1 )  lim  2 k  Ad 1
k 
d1 , , d k

  ( A0 , A1 )  lim
k 
Ad k



d1 , , d k
Ad 1
N  4, c0 , c1 , c2 , c3 , c4
1/ k
1/ k
max
   log 2   (T0 |V , T1 |V )
Ad k
T0 , T1 are N  N  matrices, (Ti ) j k  c2 j  k 1i
 c0

c2

T0 
 c4

0
 c1

c
T1   3
0

0
0
c4
0

0
c1 

c3 
c0
c2
c4
0
0
c1
c3
0
0

c0 
c2 

c4 
0 0
c1 c0
c3 c2
The concept of primitive families.
Definition. A family of matrices is called primitive if there exists at least one positive product.
Theorem. If all coefficients of the refinement equation are nonnegatve, then   L1
if and only if the pair of matrices {T0 , T1} is primitive and possesses a common
invariant affine subspace not passing through the origin.
Perron-Frobenius theorem (1912) A matrix A is not primitive if it is either reducible
or one of the following equivalent conditions is satisfied:
1) there are r  2 eigenvalues of A, counting with multiplicities, equal by modulo
to its spectral radius ( A).
They are all different and equal to  k =  e
2 ik
r
, k  0,... , r  1.
2) (Romanovsky, 1933) All cycles of the graph of the matrix A have g.c.d. = r.
3) There is a partition of the set   {1,..., d } into r sets  1 ,...,  r ,
on which A acts as a cyclic permutation.
0

B1
4) On the basis corresponding to that partition A has the form: A  


0
The number r is the imprimitivity index of the matrix A.
0
0
B r 1
Br 

0


0 
A criterion of primitivity
Let us have a family { A1 ,..., Am } of nonnegative matrices. We make the following two assumptions:
(a)
The family is irreducible, i.e., the matrices A1 ,..., Am do not share a common invariant coordinate plane.
(b)
All matrices of the family have neither zero rows nor zero columns.
Theorem 1 [P., Voynov, 2012]. If a family of matrices satisfies (a) and (b), then it either possesses
a positive product, or there is a partition of the set   {1,..., d } into r  2 sets  1 ,...,  r ,
on which every matrix A j acts as a permutation.
(conjectured in 2010)
Remark 1. This means that there is a permutation of the basis of R d d, after which every matrix Aj
gets the block form:
Aj
 0

0
 


 Br
0
B2
0
0
B1 




0
Remark 2. The permutations are not necessarily cyclic ! (In contrast to the P-F theorem)
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.
Consider first the case when  can take finitely many values.
Without loss of generality assume   {0, 1, .... , N}. Thus P{  k}  pk , k  0, ...., N ,
Binary tree T:
1/2
1/8
1/16
9/16
5/8
5/16
a
3/4
1/4
3/8
7/8
13/16 3/16 11/167/16
15/16
a/2
0.5 + a/2
N

k 0
pk  1.
1/2
3/4
1/4
1/8
1/16
9/16
5/8
5/16
3/8
7/8
13/16 3/16 11/167/16
15/16
Definition 1.
A subset of vertices A  T is blocking if it is symmetric and it is a minimal cut set
(any infinite path from the root contains precisely one element of A).
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).
N
m( z )   pk e  2 i k z is a characteristic function of .
k 0
Theorem.. The distribution F 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.
Example 7 (the case of Theorem 1 of DDL). P { is even} = P { is odd} = ½
Example 8 (the case of Example 6). m( z ) 
A = {root}
1
1
2 e 6 iz  e 4 iz  2e 2 iz  1 
2e 2 iz  1  e 4 iz  1


6
6
  0 ,
1 ,
2 , 3
P = 1/6 , 1/3 , 1/6 , 1/3
1/2
A = {1/2}
3/4
1/4
1/8
iff
5/8
3/8
A = {1/4 , 3/4}
7/8
The criterion is sharp, each case can be realized.
What is the set of all sets of probabilities ( p0 , p1 ,
, pN )  R N 1
for which the distribution is absolutely continuous ?
A = {1/2}
m(1/2) =0
p
2k

p
2 k 1
p1
A = {1/4, 3/4}
m(1/4) =0
m(3/4) =0
p
p
4k

4 k 1
p
 p
4k 2
pN
4 k 3
p0
Thank you !