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 applications of these results to determine the similarity of flat polygonal lines in contour 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 support on K [1, 5]. Hence any linear functional can be uniquely represented in the form
f ( z )   z  t  d  f  t ,
(2)
K
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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 ,
j 1
(3)
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  
K R 2

z m  t  d  f  t   Sm ,
(4)
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 produce 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:
 x, y  | x  y  r  r ,
  x, y  | r  x  y  r  r ,
  x, y  | x  y  r  r .
K1r1 
2
K 2r2 ,r3
K3r4
2
2
2
2
2
1
2
2
2
0
2
2
4
2
3
2
0
2
0
Then we define continuous linear functionals
f1r1 ( z ) 
*
 z  x, y  dL  C ( K ) ,
(6)
K1r1
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*
 z  x, y  dxdy  C ( K ) ,
f 2r2 ,r3 ( z ) 
(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
the scalar products
m
z1 , z2
z1 , z2
z1 , z2
1
2
3

 z1  x, y  z2  x, y  dL,
K1r1


z1  x, y  z2  x, y  dxdy ,
K2r2 , r3

 z1  x, y  z2  x, y  dxdy,
K1r4
So for all m  0 we have the zero moments
1
f1r1 ( z m )  Sm


z m dL  0,
K1r1

f 2r2 ,r3 ( z m )  Sm2 
z m dxdy  0,
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
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-
Then we fix 0  r1 
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 representations x  Re  z  
zz
z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
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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 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 ,  
k
j
j
j 1
under the
assumption that all the numbers z j are pairwise distinct, and  j  0 . In (10) we put
z 0  1 for all z  C .
Two sets S 
 z ,  
k
j
j
j 1
and S 
 z ,  
k
j
j
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
  z mj  j  z mj  j   0, m  0..M ,
k
j 1
(11)

 

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  ,
   
Denote by z j  z j the intersection of sets z j | j  1..k and z j | j  1..k . Now
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
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 z mj  j   z mj  j 
jJ z


jJ zz

jJ z
j

(12)
  h j   z mj  0, m  0..M .
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.
j 1
(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  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
 
Let’s consider the unclosed polyline Lk  R 2 with k nodes t j  x j , y j
 j 1 , and
k
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.
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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   j   z  Lkj      z j   z j 1  z j   ,
 j  arg  z j 1  z j  ,
z j  z t j  z ( x j , y j )  x j  iy j ,
and take a system of complex moments Sm with m  0..M  N 0
Sm 
z
m
Lk
k 1 1
dL    z mj    dz j    .
k
(14)
j 1 0
After simple calculations we obtain
1
Sm 
 ei1  z1m 1 
m 1

k 1

 e
j 1
i j 1
e
 i j

 z mj 1
e
ik 1
 zkm 1
(15)
.
Let sm  m  S m 1 , 1  ei1 , k  eik 1 ,  j  e
easy that
  j  ei    e
k 1
k
1
j 1
i j 1
e
i j
j 2
e
ik 1
i j 1
e
i j
with j  2..k  1 . It’s
 0,
and therefore (15) can be written as (10).
By assumption of polyline we have
z2  z1
1  ei1  
 0,
z2  z1
k  eik 1 
j  e
i j 1
e
zk  zk 1
zk  zk 1
 i j

(16)
 0.
z j  z j 1
z j  z j 1
(17)

z j 1  z j
 0,
z j 1  z j
(18)
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 .
points z j
k
k
 z ,  
k
It is easy to prove that the pairs
j
j
j 1
completely determine the nodes and seg-
ments of the polyline Lk . Indeed, 1 uniquely determines the value of ei1  11 and
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  from node t   Re  z  , Im  z  to the
  
thus the direction w   Re 11 ,  Im 11
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
exist in the node set t j
k
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 z k uniquely
identifies the segment in the polyline  tk 1 , tk  . As for the other pairs of values, we can
use the obvious equality e
 i j
j
n j
s 1
s 0
   s 
 ns
and consistently hold the previous
arguments, starting with one of the node t1 or t k .
A similar result holds for the closed polyline Lk
t   x , y 
k
j
j
j
j 1
defined by the nodes
with t1  tk . In this case, the analogue of (15) takes the form (10) if
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 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 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
z j 
1 k
1 k
zs   0  0  z j     0  0  zs  

k s 1
k s 1

1 k 
0   z j   zs  ,
k s 1 

and so without loss of generality, we assume that
k
k
j 1
j 1
 z j   z j  0
or the same
 0  0.
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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 ,
(19)
 zj m j   zj m   zj 1, zj , zj 1   sm ,
(20)
k
k
j 1
j 1
k
k
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
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.
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