On the uniqueness of representation by linear superpositions
Let $Q$ be a set such that every function on $Q$ can be represented by linear superpositions. This representation is, in general, not unique. However, for some sets, it may be unique provided that the initial values of the representing functions are prescribed at some point of $Q$. We study the prop...
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| author | Ismailov, V. E. Ісмаілов, В. Є. |
| author_facet | Ismailov, V. E. Ісмаілов, В. Є. |
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| description | Let $Q$ be a set such that every function on $Q$ can be represented by linear superpositions. This representation is, in general,
not unique. However, for some sets, it may be unique provided that the initial values of the representing functions are
prescribed at some point of $Q$. We study the properties of these sets. |
| first_indexed | 2026-03-24T02:15:48Z |
| format | Article |
| fulltext |
UDC 517.5
V. E. Ismailov (Inst. Math. and Mech. Nat. Acad. Sci. Azerbaijan, Baku)
ON THE UNIQUENESS OF REPRESENTATION BY LINEAR SUPERPOSITIONS*
ПРО ЄДИНIСТЬ ЗОБРАЖЕННЯ ЧЕРЕЗ ЛIНIЙНI СУПЕРПОЗИЦIЇ
Let Q be a set such that every function on Q can be represented by linear superpositions. This representation is, in general,
not unique. However, for some sets, it may be unique provided that the initial values of the representing functions are
prescribed at some point of Q. We study the properties of these sets.
Нехай Q — така множина, що кожну функцiю на Q можна зобразити в термiнах лiнiйних суперпозицiй. У
загальному випадку таке зображення не є єдиним. Проте для деяких множин воно може бути єдиним, якщо
початковi значення функцiй з цього зображення задано в деякiй точцi Q. Вивчаються деякi властивостi таких
множин.
1. Introduction. Let X,X1, . . . , Xr be sets and hi : X \rightarrow Xi, i = 1, . . . , r, be arbitrarily fixed
mappings. Consider the set
\scrL = \scrL (h1, . . . , hr) =
\Biggl\{
r\sum
i=1
gi(hi(x)) : x \in X, gi : Xi \rightarrow \BbbR , i = 1, . . . , r
\Biggr\}
.
Members of this set will be called linear superpositions (see [12]). Linear superpositions were
begun to be systematically studied after the famous result of A. N. Kolmogorov [6] on Hilbert’s 13th
problem. The result states that for the unit cube \BbbI d, \BbbI = [0, 1], d \geq 2, there exists 2d+ 1 functions
\{ sq\} 2d+1
q=1 \subset C(\BbbI d) of the form
sq(x1, . . . , xd) =
d\sum
p=1
\varphi pq(xp), \varphi pq \in C(\BbbI ), p = 1, . . . , d, q = 1, . . . , 2d+ 1, (1.1)
such that each function f \in C(\BbbI d) admits the representation
f(x) =
2d+1\sum
q=1
gq(sq(x)), x = (x1, . . . , xd) \in \BbbI d, gq \in C(\BbbR ). (1.2)
This surprising and deep result was improved and generalized in several directions. It was first
observed by G. G. Lorentz [7] that the functions gq can be replaced with a single continuous function
g. D. A. Sprecher [9] showed that the theorem can be proven with constant multiples of a single
function \varphi and translations. Specifically, \varphi pq in (1.1) can be chosen as \lambda p\varphi (xp + \varepsilon q), where \varepsilon and
\lambda are some positive constants. B. L. Fridman [1] succeeded in showing that the functions \varphi pq can be
constructed to belong to the class Lip(1). A. G. Vitushkin and G. M. Henkin [12] showed that \varphi pq
cannot be taken to be continuously differentiable. Y. Sternfeld [11] showed that the number 2d + 1
in (1.2) cannot be reduced.
Kolmogorov’s result shows that continuous functions admit representation by linear superposi-
tions of form (1.2). Y. Sternfeld [10] proved that bounded functions also admit such representation
* This research was supported by the Science Development Foundation under the President of the Republic of Azerbaijan
(Grant EIF-2013-9(15)-46/11/1).
c\bigcirc V. E. ISMAILOV, 2016
1620 ISSN 1027-3190. Укр. мат. журн., 2016, т. 68, № 12
ON THE UNIQUENESS OF REPRESENTATION BY LINEAR SUPERPOSITIONS 1621
with the natural proviso that the functions gq are bounded. In [2], we start to study properties of
linear superpositions on topology-free spaces and showed that every multivariate function f can be
represented in form (1.2), where gq are univariate functions depending on f. In the current paper, we
continue our research on the representation capabilities of linear superpositions.
Let T be the set of all real functions on X. Note that the above set \scrL is a linear subspace of
T. For a set Q \subset X, let T (Q) and \scrL (Q) denote the restrictions of T and \scrL to Q respectively. We
are interested in sets Q with the property that \scrL (Q) = T (Q). Such sets will be called representation
sets. For a representation set Q, we will also use the notation Q \in RS. Here, RS stands for the set
of all representation sets in X.
Let Q \in RS. Clearly for a function f defined on Q the representation
f(x) =
r\sum
i=1
gi(hi(x)), x \in Q, (1.3)
is not unique. We are interested in the uniqueness of such representation under some reasonable
restrictions on the functions gi \circ hi. These restrictions may be various, but in the current paper, we
require that the values of the representing functions in (1.3) are prescribed at some point x0 \in Q.
That is, we require that
gi(hi(x0)) = ai, i = 1, . . . , r - 1, (1.4)
where ai are arbitrarily fixed real numbers. Is representation (1.3) subject to initial conditions (1.4)
always unique? Obviously, not. We are going to identify those representation sets Q for which
representation (1.3) subject to conditions (1.4) is unique for all functions f : Q \rightarrow \BbbR . In the sequel,
such sets Q will be called unicity sets.
2. Main results. In our earlier paper [2], we characterized representation sets in terms of rather
practical objects called closed paths. A closed path (with respect to the functions h1, . . . , hr ) is a
set of points \{ x1, . . . , xn\} in X such that there exists a vector \lambda = (\lambda 1, . . . \lambda n) with \lambda i \in \BbbR \setminus \{ 0\} ,
i = 1, . . . , n, satisfying the equations
n\sum
j=1
\lambda j\delta hi(xj)(t) = 0 for all t \in Xi, i = 1, . . . , r.
Here \delta a is the characteristic function of a single point set \{ a\} .
For example, the set l = \{ (0, 0, 0), (0, 0, 1), (0, 1, 0), (1, 0, 0), (1, 1, 1)\} is a closed path in \BbbR 3
with respect to the functions hi(z1, z2, z3) = zi, i = 1, 2, 3. The vector \lambda above can be taken as
( - 2, 1, 1, 1, - 1).
In the case r = 2, the picture of closed path becomes more clear. Let, for example, h1 and h2
be the coordinate functions on \BbbR 2. In this case, a closed path is the union of some sets Ak with
the property: each Ak consists of vertices of a closed broken line with the sides parallel to the
coordinate axis. These objects (sets Ak ) have been exploited in practically all works devoted to the
approximation of bivariate functions by univariate functions, although under the different names (see,
for example, [3], Chapter 2). If X and the functions h1 and h2 are arbitrary, the sets Ak can be
described as a trace of some point traveling alternatively in the level sets of h1 and h2, and then
returning to its primary position.
A result of [2] states that Q \in RS if and only if there is no closed path in Q. From this result it
is easy to obtain the following set-theoretic properties of representation sets:
ISSN 1027-3190. Укр. мат. журн., 2016, т. 68, № 12
1622 V. E. ISMAILOV
(1) Q \in RS \Leftarrow \Rightarrow A \in RS for every finite set A \subset Q.
(2) The union of any linearly ordered (under inclusion) system of representation sets is also a
representation set.
(3) For any representation set Q there is a maximal representation set, that is, a set M \in RS
such that Q \subset M and for any P \supset M, P \in RS we have P = M.
(4) If M \subset X is a maximal representation set, then hi(M) = hi(X), i = 1, . . . , r.
Properties (1) and (2) are obvious, since any closed path is a finite set. The property (3) follows
from (2) and Zorn’s lemma. To prove (4) note that if x0 \in X and hi(x0) /\in hi(M) for some i, one
can construct the representation set M \cup \{ x0\} , which is bigger than M. But this is impossible, since
M is maximal.
Definition 2.1. A set Q \subset X is called a complete representation set if Q itself is a representation
set and there is no other representation set P such that Q \subset P and hi(P ) = hi(Q), i = 1, . . . , r.
The set of all complete representation sets of X will be denoted by CRS. Obviously, every
representation set is contained in a complete representation set. That is, if A \in RS, then there exists
B \in CRS such that hi(B) = hi(A), i = 1, . . . , r. It turns out that for the functions h1, . . . , hr,
complete representation sets entirely characterize unicity sets. To prove this fact we need some
auxiliary lemmas.
Lemma 2.1. Let Q \subset X be a representation set and for some point x0 \in Q the zero function
representation
0 =
r\sum
i=1
gi(hi(x)), x \in Q,
is unique, provided that gi(hi(x0)) = 0, i = 1, . . . , r - 1. That is, all the functions gi \equiv 0 on the
sets hi(Q), i = 1, . . . , r. Then Q \in CRS.
Proof. Assume that Q /\in CRS. Then there exists a point p \in X such that p /\in Q, hi(p) \in hi(Q)
for all i = 1, . . . , r and Q\prime = Q \cup \{ p\} is also a representation set. Consider a function f0 :
Q\prime \rightarrow \BbbR such that f0(q) = 0 for any q \in Q and f0(p) = 1. Since Q\prime \in RS,
f0(x) =
r\sum
i=1
si(hi(x)), x \in Q\prime .
Then
f0(x) =
r\sum
i=1
gi(hi(x)), x \in Q\prime , (2.1)
where
gi(hi(x)) = si(hi(x)) - si(hi(x0)), i = 1, . . . , r - 1,
and
gr(hr(x)) = sr(hr(x)) +
r - 1\sum
i=1
si(hi(x0)).
A restriction of representation (2.1) to the set Q gives the equality
r\sum
i=1
gi(hi(x)) = 0 for all x \in Q. (2.2)
ISSN 1027-3190. Укр. мат. журн., 2016, т. 68, № 12
ON THE UNIQUENESS OF REPRESENTATION BY LINEAR SUPERPOSITIONS 1623
Note that gi(hi(x0)) = 0, i = 1, . . . , r - 1. It follows from the hypothesis of the lemma that
representation (2.2) is unique. Hence, gi(hi(x)) = 0 for all x \in Q and i = 1, . . . , r. But from (2.1)
it follows that
r\sum
i=1
gi(hi(p)) = f0(p) = 1.
Since hi(p) \in hi(Q) for all i = 1, . . . , r, the above relation contradicts that the functions gi are
identically zero on the sets hi(Q), i = 1, . . . , r. This means that our assumption is not true and
Q \in CRS.
The following lemma is a strengthened general version of Lemma 2.1.
Lemma 2.1B. Let Q \in RS and for some point x0 \in Q, numbers c1, c2, . . . , cr - 1 \in \BbbR and a
function v \in T (Q) the representation
v(x) =
r\sum
i=1
vi(hi(x))
is unique under the initial conditions vi(hi(x0)) = ci, i = 1, . . . , r - 1. Then for any numbers
b1, b2, . . . , br - 1 \in \BbbR and an arbitrary function f \in T (Q) the representation
f(x) =
r\sum
i=1
fi(hi(x))
is also unique, provided that fi(hi(x0)) = bi, i = 1, . . . , r - 1. Besides, Q \in CRS.
Proof. Assume the contrary. Assume that there exists a function f \in T (Q) having two different
representations subject to the same initial conditions. That is,
f(x) =
r\sum
i=1
fi(hi(x)) =
r\sum
i=1
f \prime
i(hi(x))
with fi(hi(x0)) = f \prime
i(hi(x0)) = bi, i = 1, . . . , r - 1, and fi \not = f \prime
i for some indice i \in \{ 1, . . . , r\} . In
this case, the function v(x) will possess the following two different representations:
v(x) =
r\sum
i=1
vi(hi(x)) =
r\sum
i=1
\bigl[
vi(hi(x)) + fi(hi(x)) - f \prime
i(hi(x))
\bigr]
both satisfying the initial conditions. The obtained contradiction and above Lemma 2.1 complete the
proof.
In the sequel, we will assume that for any points ti \in hi(X), i = 1, . . . , r, the system of equations
hi(x) = ti, i = 1, . . . , r, has at least one solution.
Lemma 2.2. Let Q \in CRS. Then for any point x0 \in Q the representation
0 =
r\sum
i=1
gi(hi(x)), x \in Q, (2.3)
subject to the conditions
gi(hi(x0)) = 0, i = 1, . . . , r - 1, (2.4)
is unique. That is, gi \equiv 0 on the sets hi(Q), i = 1, . . . , r.
ISSN 1027-3190. Укр. мат. журн., 2016, т. 68, № 12
1624 V. E. ISMAILOV
Proof. Assume the contrary. Assume that representation (2.3) subject to (2.4) is not unique, or
in other words, not all of gi are identically zero. Without loss of generality, we may suppose that
gr(hr(y)) \not = 0 for some y \in Q. Let \xi \in X be a solution of the system of equations hi(x) = hi(x0),
i = 1, . . . , r - 1, and hr(x) = hr(y). Therefore, gi(hi(\xi )) = 0, i = 1, . . . , r - 1, and gr(hr(\xi )) \not = 0.
Obviously, \xi /\in Q. Otherwise, we may have gr(hr(\xi )) = 0.
We are going to prove that Q\prime = Q \cup \{ \xi \} is a representation set. For this purpose, consider an
arbitrary function f : Q\prime \rightarrow \BbbR . The restriction of f to the set Q admits a decomposition
f(x) =
r\sum
i=1
ti(hi(x)), x \in Q.
One is allowed to fix the values ti(hi(x0)) = 0, i = 1, . . . , r - 1. Note that then ti(hi(\xi )) = 0,
i = 1, . . . , r - 1. Consider now the functions
vi(hi(x)) = ti(hi(x)) +
f(\xi ) - tr(hr(\xi ))
gr(hr(\xi ))
gi(hi(x)), x \in Q\prime , i = 1, . . . , r.
It can be easily verified that
f(x) =
r\sum
i=1
vi(hi(x)), x \in Q\prime .
Since f is arbitrary, we obtain that Q\prime \in RS, where Q\prime \supset Q and hi(Q
\prime ) = hi(Q), i = 1, . . . , r. But
this contradicts the hypothesis of the lemma that Q \in CRS.
Theorem 2.1. Q \in CRS if and only if for any x0 \in Q, any f \in T (Q) and any a1, . . . , ar - 1 \in \BbbR
the representation
f(x) =
r\sum
i=1
gi(hi(x)), x \in Q,
subject to the conditions gi(hi(x0)) = ai, i = 1, . . . , r - 1, is unique. Equivalently, a set Q \in CRS
if and only if it is a unicity set.
Theorem 2.1 is an obvious consequence of Lemmas 2.1B and 2.2.
Remark 2.1. In Theorem 2.1, all the words "any" can be replaced with the word "some".
Remark 2.2. For the case X = X1\times . . .\times Xn, the possibility and uniqueness of the representation
by sums
\sum n
i=1
ui(xi), ui : Xi \rightarrow \BbbR , i = 1, . . . , n, were investigated in [4] and [5].
Example. Let r = 2, X = \BbbR 2, h1(x1, x2) = x1 + x2, h2(x1, x2) = x1 - x2, Q be the graph
of the function x2 = \mathrm{a}\mathrm{r}\mathrm{c}\mathrm{s}\mathrm{i}\mathrm{n}(\mathrm{s}\mathrm{i}\mathrm{n}x1). The set Q has no closed paths with respect to the functions h1
and h2. Therefore, Q \in RS. By adding a point p /\in Q, we obtain the set Q \cup \{ p\} , which contain a
closed path and hence is not a representation set. Thus, the set Q \in CRS and representation on Q
is unique.
Let now r = 2, X = \BbbR 2, h1(x1, x2) = x1, h2(x1, x2) = x2, and Q be the graph of the function
x2 = x1. Clearly, Q \in RS and Q /\in CRS. By the definition of complete representation sets, there is
a set P \supset Q such that P \in RS and any set T \supset P is not a representation set. There are many sets
P with this property. One of them can be obtained by adding to Q any straight line l parallel to one
of the coordinate axes. Indeed, if y /\in Q \cup l, then the set Q1 = Q \cup l \cup \{ y\} contains a four-point
closed path (with one vertex y, two vertices lying on l and one vertex lying on Q). This means that
Q1 /\in RS and hence Q \cup l \in CRS.
The following corollary can be easily obtained from Theorem 2.1 and Lemma 2.1B.
ISSN 1027-3190. Укр. мат. журн., 2016, т. 68, № 12
ON THE UNIQUENESS OF REPRESENTATION BY LINEAR SUPERPOSITIONS 1625
Corollary 2.1. Q \in CRS if and only if Q \in RS and in the representation
0 =
r\sum
i=1
gi(hi(x)), x \in Q,
all the functions gi, i = 1, . . . , r, are constants.
We have seen that complete representation sets enjoy the unicity property. Let us study some
other properties of the following sets:
(a) If Q1, Q2 \in CRS, Q1 \cap Q2 \not = \varnothing and Q1 \cup Q2 \in RS, then Q1 \cup Q2 \in CRS.
(b) Let \{ Q\alpha \} , \alpha \in \Phi , be a family of complete representation sets such that \cap \alpha \in \Phi Q\alpha \not = \varnothing and
\cup \alpha \in \Phi Q\alpha \in RS. Then \cup \alpha \in \Phi Q\alpha \in CRS.
The above two properties follow from Corollary 2.1. Note that (b) is a generalization of (a). The
following property is a consequence of (b) and property (2) of representation sets.
(c) Let \{ Q\alpha \} , \alpha \in \Phi , be a totally ordered (under inclusion) family of complete representation
sets. Then \cup \alpha \in \Phi Q\alpha \in CRS.
We know that every representation set A is contained in a complete representation set Q such
that hi(A) = hi(Q), i = 1, . . . , r. What can we say about the set Q\setminus A? Clearly, Q\setminus A \in RS. But
can we chose Q so that Q\setminus A \in CRS? The following theorem answers this question.
Theorem 2.2. Let A \in RS and A /\in CRS. Then there exists a set B \in CRS such that A \subset B,
hi(A) = hi(B), i = 1, . . . , r, and B\setminus A \in CRS.
Proof. Since the representation set A is not complete, there exists a point p /\in A such that
hi(p) \in hi(A), i = 1, . . . , r, and A\prime = A \cup \{ p\} \in RS. By \scrM denote the collection of sets M such
that
(1) A \subset M and M \in RS;
(2) hi(M) = hi(A) for all i = 1, . . . , r;
(3) M\setminus A \in CRS.
Obviously, \scrM is not empty. It contains the above set A\prime . Consider the partial order on \scrM
defined by inclusion. Let \{ M\beta \} , \beta \in \Gamma , be any chain in \scrM . The set \cup \beta \in \Gamma M\beta is an upper bound for
this chain. To see this, let us check that \cup \beta \in \Gamma M\beta belongs to \scrM . That is, all the above conditions
(1) – (3) are satisfied. Indeed,
(1) A \subset \cup \beta \in \Gamma M\beta and \cup \beta \in \Gamma M\beta \in RS. This follows from property (2) of representation sets;
(2) hi(\cup \beta \in \Gamma M\beta ) = \cup \beta \in \Gamma hi(M\beta ) = \cup \beta \in \Gamma hi(A) = hi(A), i = 1, . . . , r;
(3) \cup \beta \in \Gamma M\beta \setminus A \in CRS. This follows from property (c) of complete representation sets and the
facts that M\beta \setminus A \in CRS for any \beta \in \Gamma and the system \{ M\beta \setminus A\} , \beta \in \Gamma , is totally ordered under
inclusion.
Thus we see that any chain in \scrM has an upper bound. By Zorn’s lemma, there are maximal sets
in \scrM . Let B be one of such sets. Let us now prove that B \in CRS.
Assume on the contrary that B /\in CRS. Then by Lemma 2.1B, for any point x0 \in B the
representation
0 =
r\sum
i=1
gi(hi(x)), x \in B, (2.5)
subject to the conditions gi(hi(x0)) = 0, i = 1, . . . , r - 1, is not unique. That is, there is a point
y \in B such that for some index i, gi(hi(y)) \not = 0. Without loss of generality we may assume
that gr(hr(y)) \not = 0. Clearly, y cannot belong to B\setminus A, since B\setminus A \in CRS and over complete
ISSN 1027-3190. Укр. мат. журн., 2016, т. 68, № 12
1626 V. E. ISMAILOV
representation sets, the zero function has a trivial representation provided that conditions (2.4) hold.
Thus, y \in A. Let \xi \in X be a point such that hi(\xi ) = hi(x0), i = 1, . . . , r - 1, and hr(\xi ) = hr(y).
The point \xi /\in B, otherwise from (2.5) we would obtain that gr(hr(y)) = gr(hr(\xi )) = 0. Following
the techniques in the proof of Lemma 2.2, it can be shown that B1 = B \cup \{ \xi \} \in RS. Now prove
that B1\setminus A \in CRS. For this purpose, consider the representation
0 =
r\sum
i=1
g\prime i(hi(x)), x \in B1\setminus A, (2.6)
subject to the conditions g\prime i(hi(x0)) = 0, i = 1, . . . , r - 1, where x0 is some point of B\setminus A.
Such representation holds uniquely on B\setminus A, since B\setminus A \in CRS. That is, all the functions g\prime i are
identically zero on hi(B\setminus A), i = 1, . . . , r. On the other hand, since g\prime i(hi(\xi )) = g\prime i(hi(x0)) = 0 for
all i = 1, . . . , r - 1 we obtain that g\prime r(hr(\xi )) = 0. This means that representation (2.6) subject to
the conditions g\prime i(hi(x0)) = 0, i = 1, . . . , r - 1, is unique on B1\setminus A. That is, all the functions g\prime i
in (2.6) are zero functions on hi(B1\setminus A), i = 1, . . . , r. Hence by Lemma 2.1, B1\setminus A \in CRS. Thus,
B1 \in \scrM . But the set B was chosen as a maximal set in \scrM . We see that the above assumption
B /\in CRS leads us to the contradiction that there is a set B1 \in \scrM bigger than the maximal set B.
Thus, in fact, B \in CRS.
Let A be a representation set. The relation on A defined by setting x \sim y if there is a finite
complete representation subset of A containing both x and y, is an equivalence relation. Indeed, it
is reflexive and symmetric. It is transitive on the basis of property (a) of complete representation
sets. The equivalence classes we call C -orbits. In the case r = 2, C -orbits turn into classical orbits
considered by D. E. Marshall and A. G. O’Farrell [8], which have a very nice geometric interpretation
in terms of bolts (for this terminology see [3, 8]). A classical orbit consists of all possible traces of
an arbitrary point in it traveling alternatively in the level sets of h1 and h2. In the general setting, one
partial case of C -orbits are introduced by A. Klopotowski, M. G. Nadkarni, K. P. S. Rao [5] under
the name of related components. The case considered in [5] requires that A \subset X = X1 \times . . .\times Xn
and hi be the canonical projections of X onto Xi, i = 1, . . . , r, respectively. Finite complete
representation sets containing x and y will be called C -trips connecting x and y. A C -trip of the
smallest cardinality connecting x and y will be called a minimal C -trip.
Theorem 2.3. Let A be a representation set and x and y be any two points of some C -orbit in
A. Then there is only one minimal C -trip connecting them.
Proof. Assume that L1 and L2 are two minimal C -trips connecting x and y. By definition, L1
and L2 are complete representation sets. Note that L1 \cup L2 is also complete. Let us prove that the
set L1 \cap L2 is complete. Clearly, L1 \cap L2 \in RS. Let x0 \in L1 \cap L2. In particular, x0 can be one of
the points x and y. Consider the representation
0 =
r\sum
i=1
gi(hi(x)), x \in L1 \cap L2, (2.7)
subject to gi(hi(x0)) = 0, i = 1, . . . , r - 1. On the strength of Lemma 2.1, it is enough to prove that
this representation is unique. For i = 1, . . . , r, let g\prime i be any extension of gi from the set hi(L1 \cap L2)
to the set hi(L1). Construct the function
f \prime (x) =
r\sum
i=1
g\prime i(hi(x)), x \in L1. (2.8)
ISSN 1027-3190. Укр. мат. журн., 2016, т. 68, № 12
ON THE UNIQUENESS OF REPRESENTATION BY LINEAR SUPERPOSITIONS 1627
Since f \prime (x) = 0 on L1 \cap L2, the following function is well defined:
f(x) =
\biggl\{
f \prime (x), x \in L1,
0, x \in L2.
Since L1 \cup L2 \in CRS, the representation
f(x) =
r\sum
i=1
wi(hi(x)), x \in L1 \cup L2, (2.9)
subject to
wi(hi(x0)) = 0, i = 1, . . . , r - 1, (2.10)
is unique. Besides, since L1 \in CRS and g\prime i(hi(x0)) = gi(hi(x0)) = 0, i = 1, . . . , r - 1, repre-
sentation (2.8) is unique. This means that for each function gi, there is only one extension g\prime . Note
that
f(x) = f \prime (x) =
r\sum
i=1
wi(hi(x)), x \in L1.
Now from uniqueness of representation (2.8) we obtain
wi(hi(x)) = g\prime i(hi(x)), i = 1, . . . , r, x \in L1. (2.11)
A restriction of formula (2.9) to the set L2 gives
0 =
r\sum
i=1
wi(hi(x)), x \in L2. (2.12)
Since L2 \in CRS, representation (2.12) subject to conditions (2.10) is unique, whence
wi(hi(x)) = 0, i = 1, . . . , r, x \in L2. (2.13)
From (2.11) and (2.13) it follows that
gi(hi(x)) = g\prime i(hi(x)) = 0, i = 1, . . . , r, x \in L1 \cap L2.
Thus, we see that representation (2.7) subject to the conditions gi(hi(x0)) = 0, i = 1, . . . , r - 1, is
unique on the intersection L1 \cap L2. Therefore by Lemma 2.1, L1 \cap L2 \in CRS.
Let the cardinalities of L1 and L2 be equal to n. Since x, y \in L1 \cap L2 and L1 \cap L2 \in CRS,
we obtain from the definition of minimal C -trips that the cardinality of L1 \cap L2 is also n. Hence,
L1 \cap L2 = L1 = L2.
Let Q be a representation set. That is, each function f : Q \rightarrow \BbbR enjoys representation (1.3).
Can we find gi, i = 1, . . . , r, for a given f ? There is a procedure for finding one certain collection
of gi, provided that Q consists of a single C -orbit. That is, any two points of Q can be connected
by a C -trip. To show this procedure, take some point x0 \in Q and fix it. We are going to find gi
from (1.3) and conditions (1.4). Let y be any point Q. To find the values of gi at the points hi(y),
i = 1, . . . , r, connect x0 and y by a minimal C -trip S = \{ x1, . . . , xn\} , where x0 = x1 and xn = y.
ISSN 1027-3190. Укр. мат. журн., 2016, т. 68, № 12
1628 V. E. ISMAILOV
Since S is a complete representation set, equation (1.3) subject to (1.4) has a unique solution on S.
That is, we can find gi(hi(y)), i = 1, . . . , r, by solving the system of linear equations
r\sum
i=1
gi(hi(xj)) = f(xj), j = 1, . . . , n.
We see that each minimal C -trip containing x0 generates a system of linear equations, which is
uniquely solvable. Since any point of Q can be connected with x0 by such a trip, we can find gi(t)
at each point t \in hi(Q), i = 1, . . . , r.
The above procedure can still be effective for some particular representation sets Q consisting of
many C -orbits. Let \{ C\alpha \} , \alpha \in \Lambda , denote the set of all C -orbits of Q. Fix some points x\alpha \in C\alpha ,
\alpha \in \Lambda , one in each orbit. Let y\alpha be any points of C\alpha , \alpha \in \Lambda , respectively. We can apply the
above procedure of finding the values of gi at each y\alpha if hi(y\alpha ) \not = hi(y\beta ) for all i and \alpha \not = \beta .
For hi(y\alpha ) = hi(y\beta ), one cannot guarantee that after solving the corresponding systems of linear
equations (associated with y\alpha and y\beta ), the solutions gi(hi(y\alpha ) and gi(hi(y\beta )) will be equal. That is,
for the case hi(y\alpha ) = hi(y\beta ), the constructed functions gi may not be well defined.
References
1. Fridman B. L. An improvement in the smoothness of the functions in A. N. Kolmogorov’s theorem on superpositions
(in Russian) // Dokl. Akad. Nauk SSSR. – 1967. – 177. – P. 1019 – 1022.
2. Ismailov V. E. On the representation by linear superpositions // J. Approxim. Theory. – 2008. – 151. – P. 113 – 125.
3. Khavinson S. Ya. Best approximation by linear superpositions (approximate nomography) // Transl. Math. Mono-
graphs. – Providence, RI: Amer. Math. Soc., 1997. – 159. – 175 p.
4. Klopotowski A., Nadkarni M. G., Bhaskara Rao K. P. S. When is f(x1, x2, . . . , xn) = u1(x1) + u2(x2) + . . .
. . .+ un(xn) ? // Proc. Indian Acad. Sci. (Math. Sci.) – 2003. – 113. – P. 77 – 86.
5. Klopotowski A., Nadkarni M. G., Bhaskara Rao K. P. S. Geometry of good sets in n-fold Cartesian product // Proc.
Indian Acad. Sci. (Math. Sci.) – 2004. – 114. – P. 181 – 197.
6. Kolmogorov A. N. On the representation of continuous functions of many variables by superposition of continuous
functions of one variable and addition (in Russian) // Dokl. Akad. Nauk SSSR. – 1957. – 114. – P. 953 – 956.
7. Lorentz G. G. Metric entropy, widths, and superpositions of functions // Amer. Math. Mon. – 1962. – 69. – P. 469 – 485.
8. Marshall D. E., O’Farrell A. G. Approximation by a sum of two algebras. The lightning bolt principle // J. Funct.
Anal. – 1983. – 52. – P. 353 – 368.
9. Sprecher D. A. An improvement in the superposition theorem of Kolmogorov // J. Math. Anal. and Appl. – 1972. –
38. – P. 208 – 213.
10. Sternfeld Y. Uniformly separating families of functions // Isr. J. Math. – 1978. – 29. – P. 61 – 91.
11. Sternfeld Y. Dimension, superposition of functions and separation of points, in compact metric spaces // Isr. J. Math. –
1985. – 50. – P. 13 – 53.
12. Vitushkin A. G., Henkin G.M. Linear superpositions of functions (in Russian) // Uspehi Mat. Nauk. – 1967. – 22,
№ 1. – P. 77 – 124.
Received 02.03.15,
after revision — 14.01.16
ISSN 1027-3190. Укр. мат. журн., 2016, т. 68, № 12
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| id | umjimathkievua-article-1948 |
| institution | Ukrains’kyi Matematychnyi Zhurnal |
| keywords_txt_mv | keywords |
| language | English |
| last_indexed | 2026-03-24T02:15:48Z |
| publishDate | 2016 |
| publisher | Institute of Mathematics, NAS of Ukraine |
| record_format | ojs |
| resource_txt_mv | umjimathkievua/65/af129882538a9a3beb430f735bda4665.pdf |
| spelling | umjimathkievua-article-19482019-12-05T09:32:42Z On the uniqueness of representation by linear superpositions Про єдинiсть зображення через лiнiйнi суперпозицiї Ismailov, V. E. Ісмаілов, В. Є. Let $Q$ be a set such that every function on $Q$ can be represented by linear superpositions. This representation is, in general, not unique. However, for some sets, it may be unique provided that the initial values of the representing functions are prescribed at some point of $Q$. We study the properties of these sets. Нехай $Q$ — така множина, що кожну функцiю на $Q$ можна зобразити в термiнах лiнiйних суперпозицiй. У загальному випадку таке зображення не є єдиним. Проте для деяких множин воно може бути єдиним, якщо початковi значення функцiй з цього зображення задано в деякiй точцi $Q$. Вивчаються деякi властивостi таких множин. Institute of Mathematics, NAS of Ukraine 2016-12-25 Article Article application/pdf https://umj.imath.kiev.ua/index.php/umj/article/view/1948 Ukrains’kyi Matematychnyi Zhurnal; Vol. 68 No. 12 (2016); 1620-1628 Український математичний журнал; Том 68 № 12 (2016); 1620-1628 1027-3190 en https://umj.imath.kiev.ua/index.php/umj/article/view/1948/930 Copyright (c) 2016 Ismailov V. E. |
| spellingShingle | Ismailov, V. E. Ісмаілов, В. Є. On the uniqueness of representation by linear superpositions |
| title | On the uniqueness of representation by linear superpositions |
| title_alt | Про єдинiсть зображення через лiнiйнi суперпозицiї |
| title_full | On the uniqueness of representation by linear superpositions |
| title_fullStr | On the uniqueness of representation by linear superpositions |
| title_full_unstemmed | On the uniqueness of representation by linear superpositions |
| title_short | On the uniqueness of representation by linear superpositions |
| title_sort | on the uniqueness of representation by linear superpositions |
| url | https://umj.imath.kiev.ua/index.php/umj/article/view/1948 |
| work_keys_str_mv | AT ismailovve ontheuniquenessofrepresentationbylinearsuperpositions AT ísmaílovvê ontheuniquenessofrepresentationbylinearsuperpositions AT ismailovve proêdinistʹzobražennâčerezlinijnisuperpozicií AT ísmaílovvê proêdinistʹzobražennâčerezlinijnisuperpozicií |