=Paper=
{{Paper
|id=Vol-1638/Paper50
|storemode=property
|title=On a moment problem for sets of points in the complex plane
|pdfUrl=https://ceur-ws.org/Vol-1638/Paper50.pdf
|volume=Vol-1638
|authors=Victor P. Tsvetov
}}
==On a moment problem for sets of points in the complex plane ==
https://ceur-ws.org/Vol-1638/Paper50.pdf
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 mI 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
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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 , (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
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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-
zz zz
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
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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 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
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z mj j z mj j
jJ z jJ z
z 0, m 0..M .
(12)
j h j
m
j
jJ zz
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 ij1 , 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.
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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 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
ei1 z1m 1
(15)
k 1
e
j 1
i j 1
e
i j
z mj 1 eik 1 zkm 1 .
i j 1 i j
Let sm m S m 1 , 1 ei1 , k eik 1 , j e e with j 2..k 1 . It’s
easy that
j ei e e
k k 1
i j 1 i j ik 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 ei1 0, (16)
z2 z1
zk zk 1
k eik 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 ei1 11 and
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thus the direction w Re 11 , Im 11 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 eik 1 ei1 .
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.
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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 ,
k k
(19)
j 1 j 1
zj m j zj m zj 1, zj , zj 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.
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