ON A MOMENT PROBLEM FOR SETS OF POINTS IN THE COMPLEX PLANE

Abstract

This paper deals with a uniqueness problem of determination a set of points in the complex plane by the degree-like moments. We discuss applications of these results to determine the similarity of flat polygonal lines in contour analysis method. f (z)   z t  d f t ,

Keywords

• moment problem
• contour analysis
• image processing

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Introduction A moment problem is closely associated with many questions of functional analysis [1], integral geometry [2, 3], problems of interpolations, and function classifications [4]. Several problems concerning the uniqueness of solutions of operator equations can be reformulated in terms of determining continuous linear functional f : X  C by the known values on a system of basic elements zmmI N0 of a normed linear space X : f (zm )  sm .

In applications X is a space of complex-valued functions that are continuous on compact set K . Such space with the uniform norm is denoted as C (K ) . By C(K )* we denote the linear space that topologically conjugates to C (K ) . So C(K )* contains all continuous linear functionals f : C(K )  C . Based on the Riesz-Radon theorem we know that C(K )* is isometric to the space of Radon measures with compact support on K [1, 5]. Hence any linear functional can be uniquely represented in the form

K (1) (2) here  f is a measure with compact support K f  K that is uniquely defined by the functional f .

In this paper we will be interested in the case of (1) or (2), where K  R2 , zm t   zm (x, y)  zm   x  iy m , i2  1, and  f t  is a function of bounded variation with compact support K f  K  R2 . The integral in (2) is understood as an integral of Lebesgue-Stieltjes.

If K f  t j | t j   x j , y j  , j 1..k is a finite set of points, then the integral in (2) is reduced to the finite sum

k k f (z)   z t j    f t j    z j   fj ,

j1 j1 f (zm )  or where z j  x j  iy j , and  fj  C . So (1) takes one of the forms

 zm t  d f t   K R2

 K f R2

zm t  d f t   Sm , k f (zm )   z mj   fj  sm. (5) j1 We will show the uniqueness of a linear functional f (and so K f ) that is determined from (5) by a known finite number of values sm . Then we extend this result to the special case of (4), when the compact support K f is a polyline with a finite number of segments, and the integral is understood as a line integral along a plane curve. Note that the moment problem (5) arises in a contour analysis based on the integral representations for Gaussian beams [6].

It is easy to give an example of different compact subsets K f  K  R2 , which produce an equal moments Sm to a corresponding functionals f , even in the case m  I  N0 .

As a compact K  R2 we consider a circle K   x, y  | x2  y2  r02 and define a family of subcompacts: K1r1   x, y  | x2  y2  r12  r02, K2r2 ,r3   x, y  | r22  x2  y2  r32  r02, K3r4   x, y  | x2  y2  r42  r02.

Then we define continuous linear functionals f1r1 (z)   z  x, y  dL C(K )*,

K1r1 (3) (4) (6) f2r2,r3 (z)   z  x, y  dxdy C(K )*,

K2r2 ,r3 (7) f3r4 (z)   z  x, y  dxdy C(K )*, (8)

K1r4 where the integral in (6) is understood as а line integral.

For all m  N0 functions zm (x, y)  zm   x  iym form an orthogonal system over the scalar products z1, z2 1   z1  x, y  z2  x, y  dL,

K1r1 z1, z2 2   z1  x, y  z2  x, y  dxdy,

K2r2 ,r3 z1, z2 3   z1  x, y  z2  x, y  dxdy,

K1r4 So for all m  0 we have the zero moments f1r1 (zm )  Sm1   zm dL  0,

K1r1 f2r2 ,r3 (zm )  Sm2   zm dxdy  0,

K2r2,r3 f3r4 (zm )  Sm3   zm dxdy  0,

K1r4 and non zero moments S01  2 r1 , S02    r32  r22  , S03   r42 , for m  0 . Then we fix 0  r1  r02 and set r4  2 r1 , r32  r42  r22 . It's clear that for all 2 0  r22  r02  2 r1 we obtain an infinite number of different continuous linear functionals f1r1 , f2r2 ,r3 , f3r1 with compact supports K1r1 , K2r2 ,r3 , K3r4 , and equal moments f1r1 (zm ) , f2r2 ,r3 ( zm ) , f3r1 (zm ) .

This is explained by the fact that the set of functions zm (x, y)  zm   x  iym is not a dense set in C (K ) . According to the Weierstrass theorem, a dense set is formed by the system of polynomials pm,n ( x, y)  xm yn , m, n  N0 . Based on the representations x  Re  z   z 2 z , y  Im  z   z 2 z , we can say that the system of functions Pm,n ( x, y)  zm z n has the same quality.

Thus any continuous linear functionals f C(K )* , including those that are defined by (6)-(8) will be uniquely determined by the system of values If we consider the linear functionals defined by integrals over plain domains or rectifiable curves, then the system of moments (9) will uniquely determine bounded curve or plain domain. 2

Moment problem for a finite set of points Let’s consider a moment problem in the following form. From a given system of equations k  z mj   j  sm , m  0..M , M  N, (10) j1 we want to determine a set of pairs of complex numbers S   z j ,  j  k j1 under the assumption that all the numbers z j are pairwise distinct, and  j  0 . In (10) we put z0  1 for all z C .

Two sets S   z j ,  j  k j1 and S   z j ,  j  k j1 will be considered as equivalent S  S , if there exists a total bijection h :1..k 1..k that is z j  zh( j) and  j  h( j) (in other words, if they match up to a permutation).

We are interested in the uniqueness of the solution for the problem above, which is understood as equivalent in the above sense.

Suppose that there are two sets S and S that satisfies (10). Therefore we have the following system of equations k   z mj   j  z mj   j   0, m  0..M , (11) j1 Denote by z j   z j  the intersection of sets z j | j 1..k and z j | j 1..k . Now we define the sets of indices J z z   j | z j z j  z j , J z z   j | z j z j  z j , J z  1..k \ J z z , J z  1..k \ J z z .

Let s  J z z  J z z  k . Without loss of generality, we assume that J z z  J z z  1..s ( s  0 ), and J z z  Jz z   ( s  0 ). It is clear that in this case J z  J z  s  1..k ( s  k ) and J z  J z   ( s  k ).

Denote by h : J z z  J z z the total bijection that takes each j1 to j2 in accordance with the rule z j1  z j2 . Then the system of equalities (11) takes the form

, h jk s, k 1  j  2 k  s  zij1, 1  j  k, 1  i  M 1 i j  zhi1jk s , k 1  j  2 k  s, 1  i  M 1 , and rewrite (12) in the form 2k s  i j   j  0, i 1..M 1. j1 (12) (13) We note that the leading principal minors of the matrix i j  are Vandermonde determinants composed by powers of distinct numbers. Thus the rank of this matrix is equal to min  M 1, 2 k  s . Let us consider (13) as a system of linear equations with variables  j . If M 1  2 k  s , the system (13) can have only the trivial solution. In this case J z  J z   , J z z  Jz z  1..k , and for all j 1..k we obtain z j  zh j ,  j  h j.

It’s mean that S  S .

Obviously that M 1  2 k  s for all s  0 , if M 1  2 k . Hence, if M  2 k 1 , then the moment problem (10) can have only one solution. The converse may be proved in such a way as is done in [7]. 3

Moment problem for polyline with a finite number of segments Let’s consider the unclosed polyline Lk  R2 with k nodes t j   x j , y j  j1 without self-intersections. We assume that associated nodes t j1 , t j , t j1 does not lie k , and on the same straight line, and each node can belong to no more than two segments of a polyline. So polyline has k 1 segments. As before, we define the map z : R 2  C in accordance with the rule z  z t   z(x, y)  x  iy . For each of k 1 segments we consider the parameteriza

Lkj    t j  t j1  t j   . tions Lkj : 0,1  R2 that given by  Let’s denote z j  z t j   z(x j , y j )  x j  iy j , z j   j   z  Lkj    z j   z j1  z j   ,  j  arg  z j1  z j  , and take a system of complex moments Sm with m  0..M  N0 Sm   zm dLk  k1 1 z mj   d z j   .

Lk j1 0 After simple calculations we obtain Sm  m11   ei1  z1m1 

k 1  ei j1  ei j   z mj1  eik1  zkm1 .

j1 Let sm  m  Sm1 , 1  ei1 , k  eik1 ,  j  ei j1  ei j with j  2..k  1 . It’s easy that k  j  ei1  k1 ei j1  ei j   eik1  0, j1 j2 and therefore (15) can be written as (10).

By assumption of polyline we have 1  ei1   z2  z1  0,

z2  z1 k  eik1  zk  zk 1  0.

zk  zk 1  j  ei j1  ei j  z j  z j1 z j  z j1  z j1  z j  0,

z j1  z j with j  2..k  1 .

From the results of the previous section it follows that no exists more than one sets of points z j kj 1 and the weights  j kj 1 that satisfy (14) with m  0..2k . It is easy to prove that the pairs  z j ,  j  k j1 completely determine the nodes and segments of the polyline Lk . Indeed, 1 uniquely determines the value of ei1  11 and (14) (15) (16) (17) (18) thus the direction w   Re 11 ,  Im11  from node t1  Re z1 , Im z1  to the node t2 . Thus t2  t1  r  w for some real value r  0 . At least one such point must exist in the node set t j kj1 If there are several nodes that satisfy the relations t j  t1  rj  w with some rj  0 , then we should choose the one which corresponds to the minimum value of ri . This follows from our assumption that the points t j1 , t j , t j1 should not lie on the same straight line and each node can belong to no more than two segments of a polyline. The same applies for k , which together with zk uniquely identifies the segment in the polyline tk1, tk  . As for the other pairs of values, we can j n j use the obvious equality ei j   s   ns and consistently hold the previous s1 s0 arguments, starting with one of the node t1 or tk . t j   x j , y j 

j1 A similar result holds for the closed polyline Lk

k we substitute k 1 for k and put 1  eik1  ei1 .

with t1  tk . In this case, the analogue of (15) takes the form (10) if defined by the nodes 4

Applications to the recognition of similarity for planar contours A moment problem has a relation to the image analysis [8], including analysis of similarity discrete planar contours [6, 9].

We say that two sets M1  C and M 2  C are called similar, if there exist 0 , 0  C such that M2  z | z  0  0  z, z  M1 . Here 0 describes parallel transport of points M1 , 0 is equal to the coefficient of similarity, and arg  0  corresponds to the angle of rotation ( 0  0 ).

If the similar sets M1 , M 2 are finite, then zj  1 k

 zs  0  0  z j  k s1 1 k

 0  0  zs   k s1  0   z j   1 k 

 zs  , k s1  Let’s consider two unclosed polyline Lk , L k  R 2 with the same number of segments. As follows from the previous section, each of them is uniquely determined by a system of moments k k  z mj   j   z mj    z j1, z j , z j1   sm , (19) j1 j1 k k  zjm  j   zjm    zj1, zj , zj1   sm , (20) j1 j1 with m  0..2 k 1 and   z j1, z j , z j1  that satisfy (16)-(18) Given that   0  z j1, 0  z j , 0  z j1   

0   z j1, z j , z j1 , 0 we finally obtain the similarity criterion for polyline Lk , L k in the following form

Criterion (21) also holds for closed polylines. (21)

References