Image Processing, Geoinformatics and Information Security ON A MOMENT PROBLEM FOR SETS OF POINTS IN THE COMPLEX PLANE V.P. Tsvetov Samara National Research University, Samara, Russia 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 applica- tions of these results to determine the similarity of flat polygonal lines in con- tour analysis method. Keywords: moment problem, contour analysis, image processing. Citation: Tsvetov VP. On a moment problem for sets of points in the complex plane. CEUR Workshop Proceedings, 2016; 1638: 411-418. DOI: 10.18287/1613-0073-2016-1638-411-418 1 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  N of a normed linear 0 space X : f ( z m )  sm . (1) 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 sup- port on K [1, 5]. Hence any linear functional can be uniquely represented in the form f ( z )   z  t  d  f  t , (2) K Information Technology and Nanotechnology (ITNT-2016) 411 Image Processing, Geoinformatics and Information Security Tsvetov VP… 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  R 2 , zm  t   zm ( x, y)  z m   x  iy  , i 2  1, and  f  t  is a function of bounded m variation with compact support K f  K  R 2 . 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 , (3) j 1 j 1 where z j  x j  iy j , and  fj  C . So (1) takes one of the forms f (zm )   z m t  d  f t    z m  t  d  f  t   Sm , (4) K R 2 K f R 2 or k f ( z m )   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  R 2 , which pro- duce an equal moments Sm to a corresponding functionals f , even in the case m  I  N0 .  As a compact K  R 2 we consider a circle K   x, y  | x 2  y 2  r02 and define a  family of subcompacts:  K1r1   x, y  | x 2  y 2  r12  r02 ,   K 2r2 ,r3   x, y  | r22  x 2  y 2  r32  r02 ,  K3r4   x, y  | x  y 2 2  r42  r02 . Then we define continuous linear functionals f1r1 ( z )   z  x, y  dL  C ( K )* , (6) K1r1 Information Technology and Nanotechnology (ITNT-2016) 412 Image Processing, Geoinformatics and Information Security Tsvetov VP… f 2r2 ,r3 ( z )   z  x, y  dxdy  C ( K ) , * (7) K2r2 , r3 f 3r4 ( z )   z  x, y  dxdy  C ( K )* , (8) K1r4 where the integral in (6) is understood as а line integral. For all m  N 0 functions zm ( x, y)  z m   x  iy  form an orthogonal system over m 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 ( z m )  Sm 1   z m dL  0, K1r1 f 2r2 ,r3 ( z m )  Sm2   z dxdy  0, m r ,r K 22 3 f3r4 ( z m )  Sm3   z m dxdy  0, K1r4   and non zero moments S01  2 r1 , S02   r32  r22 , S03   r42 , for m  0 . r02 Then we fix 0  r1  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 function- r r ,r r r r ,r r als f1 1 , f 2 2 3 , f 3 1 with compact supports K11 , K 22 3 , K 34 , and equal moments f1r1 ( z m ) , f 2r2 ,r3 ( z m ) , f 3r1 ( z m ) . This is explained by the fact that the set of functions zm ( x, y)  z m   x  iy  is not m 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 )  x m y n , m, n  N 0 . Based on the representa- zz zz tions x  Re  z   , y  Im  z   , we can say that the system of functions 2 2 Pm,n ( x, y )  z m 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 Information Technology and Nanotechnology (ITNT-2016) 413 Image Processing, Geoinformatics and Information Security Tsvetov VP… f ( z m z n )  Sm,n , m, n  N 0 . (9) If we consider the linear functionals defined by integrals over plain domains or recti- fiable 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  z ,   k we want to determine a set of pairs of complex numbers S  j j under the j 1 assumption that all the numbers z j are pairwise distinct, and  j  0 . In (10) we put z 0  1 for all z  C .  z ,    z ,   k k Two sets S  j j and S  j j will be considered as equivalent j 1 j 1 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   z mj  j  z mj  j   0, m  0..M , k (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  J z 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 Information Technology and Nanotechnology (ITNT-2016) 414 Image Processing, Geoinformatics and Information Security Tsvetov VP…  z mj  j   z mj  j  jJ z jJ z        z  0, m  0..M . (12)   j h j m j jJ zz Further denote  j   h j  , 1  j  s   j   j , s 1  j  k ,   h j  k  s , k  1  j  2 k  s  z ij1 , 1  j  k , 1  i  M  1  i j   i 1 ,  zh j  k  s  , k  1  j  2 k  s, 1  i  M  1  and rewrite (12) in the form 2 k s  i j   j  0, i 1..M  1. (13) j 1   We note that the leading principal minors of the matrix i j are Vandermonde deter- minants 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 varia- bles  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  J z 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    j 1 , and k Let’s consider the unclosed polyline Lk  R 2 with k nodes t j  x j , y j without self-intersections. We assume that associated nodes t j 1 , t j , t j 1 does not lie 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. Information Technology and Nanotechnology (ITNT-2016) 415 Image Processing, Geoinformatics and Information Security Tsvetov VP… 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- tions Lkj : 0,1  R 2 that given by   Lkj     t j  t j 1  t j  . 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  N 0 k 1 1 Sm   z m dL    z mj    dz j    . k (14) Lk j 1 0 After simple calculations we obtain Sm  1 m 1   ei1  z1m 1  (15)   k 1  e j 1 i j 1 e  i j  z mj 1  eik 1  zkm 1 . i j 1 i j Let sm  m  S m 1 , 1  ei1 , k  eik 1 ,  j  e e with j  2..k  1 . It’s easy that   j  ei    e e k k 1 i j 1 i j ik 1 1 e  0, j 1 j 2 and therefore (15) can be written as (10). By assumption of polyline we have z2  z1 1  ei1    0, (16) z2  z1 zk  zk 1 k  eik 1   0. (17) zk  zk 1 i j 1  i j z j  z j 1 z j 1  z j j  e e    0, (18) z j  z j 1 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   j 1 and the weights  j  j 1 that satisfy (14) with m  0..2k . k k points z j  z ,   k It is easy to prove that the pairs j j completely determine the nodes and seg- j 1 ments of the polyline Lk . Indeed, 1 uniquely determines the value of ei1  11 and Information Technology and Nanotechnology (ITNT-2016) 416 Image Processing, Geoinformatics and Information Security Tsvetov VP…    thus the direction w   Re 11 ,  Im 11   from node t   Re  z  , Im  z  to the 1 1 1 node t 2 . Thus t2  t1  r  w for some real value r  0 . At least one such point must   j 1 If there are several nodes that satisfy the relations k exist in the node set t j 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 z k uniquely identifies the segment in the polyline  tk 1 , tk  . As for the other pairs of values, we can j n j  i j use the obvious equality e    s    n  s and consistently hold the previous s 1 s 0 arguments, starting with one of the node t1 or t k . A similar result holds for the closed polyline Lk defined by the nodes t   x , y  k j j j with t1  tk . In this case, the analogue of (15) takes the form (10) if j 1 we substitute k  1 for k and put 1  eik 1  ei1 . 4 Applications to the recognition of similarity for planar contours A moment problem has a relation to the image analysis [8], including analysis of simi- larity 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 M 2   z  | z   0  0  z, z  M1 . Here  0 describes parallel transport of points M 1 , 0 is equal to the coefficient of similarity, and arg  0  corresponds to the angle of rotation (  0  0 ). If the similar sets M 1 , M 2 are finite, then 1 k 1 k z j   zs   0  0  z j     0  0  zs   k s 1 k s 1  1 k  0   z j   zs  ,  k s 1  k k and so without loss of generality, we assume that  z j   z j  0 or the same j 1 j 1  0  0. Information Technology and Nanotechnology (ITNT-2016) 417 Image Processing, Geoinformatics and Information Security Tsvetov VP… 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  z mj  j   z mj   z j 1, z j , z j 1   sm , k k (19) j 1 j 1  zj m j   zj m   zj 1, zj , zj 1   sm , k k (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    0  z j 1 , 0  z j , 0  z j 1  0    z j 1 , z j , z j 1 , we finally obtain the similarity criterion for polyline Lk , L k in the following form sm  0  0m 1  sm , m  0..2 k  1. (21) Criterion (21) also holds for closed polylines. References 1. Kantorovich LV, Akilov GP. Functional analysis. Moscow: “Nauka” Publisher; 1984. [In Russian] 2. Lavrentiev MM, Saveliev LYa Operator theory and ill posed problems. 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[In Russian] Information Technology and Nanotechnology (ITNT-2016) 418