Preorder on the Set of Analytic Functions Текст научной статьи по специальности «Математика»

EDN: XWJHUY УДК 517.55

Preorder on the Set of Analytic Functions

Maria A. Stepanova*

Steklov Mathematical Institute RAS Moscow, Russian Federation

Received 10.03.2023, received in revised form 15.05.2023, accepted 04.07.2023 Abstract. In the paper, we study a preorder, which can be naturally defined on the set of all analytic functions of two complex variables with the help of superpositions of analytic functions of one variable.

Keywords: analytical complexity, superpositions of analytic functions.

Citation: M.A. Stepanova, Preorder on the Set of Analytic Functions, J. Sib. Fed. Univ. Math. Phys., 2023, 16(5), 681-689. EDN: XWJHUY.

Introduction

The celebrated Hilbert 13th problem, even though it has been solved by A. N. Kolmogorov and V.I.Arnold, leaves around itself many interesting questions. Thus, the original problem about the representation of the solutions of a polynomial equation of seventh degree as a finite superposition of continuous functions of two variables could be looked from a different angle, if smooth, analytic or algebraic functions are being considered, instead of continuous functions. Such interpretation of the Hilbert question gave rise to plenty of research and deep results about the representation of functions with the help of superpositions of functions of lesser number of variables in different functional classes (see, e.g., [1-4]).

If we replace superpositions of continuous functions by superpositions of germs of analytic functions, then the beautiful and contensive theory, constructed in papers of V. K. Beloshapka, arises ( [5,6]). This theory allows to introduce a natural preorder on the set of analytic functions of several variables (the definition was proposed by V. K. Beloshapka). We shall restrict ourselves to functions of two variables, although it is possible to study analogous structures also in the case of larger number of variables.

Let two analytic functions fi(x, y) and f2(x, y) be given. We say that fi is not more complex then f2 (fi ^ f2), if f2 is representable as a superposition of analytic functions of one variable and the function fi (see Definition 1 in Section 1). The minimal depth of the superposition is referred to as complexity (with respect to fi) and is designated by N(fi). Preorder ^ brings to the set of analytic functions a new structure, which needs further comprehension.

In the paper, we give answers to the main questions arising here, which, as we see it, are essential for the description of the complete picture. Let us note the following three main points of our exposition:

1) Does there exist a function f (x, y), which is simpler than x + y (i.e., such that f ^ x + y)? Here we suppose that the complexity Nx+y (f ) of the function f with respect to addition x + y

*step_masha@mail.ru © Siberian Federal University. All rights reserved

is greater than one, because otherwise the question ceases to be interesting, as we will show in what follows.

2) Given functions f1 and f2. Is it possible that f1 is expressible in terms of f2 with depth n > 1, and f2 is expressible in terms of f1 with depth m > 1?

3) Given functions f1 and f2. Is it possible that f1 is expressible in terms of f2 with depth n > 1, and f2 is not expressible in terms of f1?

In all the cases the answer is affirmative. But if in 2) and 3) the answer is something we should expect, in 1) the answer is counterintuitive: we obtain that the simpler object x + y is a superposition constructed with the help of the more complex object.

1. Definition of the preorder

The classes Cln = Cln(x + y) of functions of finite analytical complexity are constructed inductively with the help of finite superpositions of analytic functions of one variable and addition x + y. In analogous way the classes Cln(p) can be constructed, if we take as a binary operation an arbitrary analytic function p = p(x,y) of two variables instead of addition. I.e., a function belongs to the class Cl0(p), if it is an analytic function of one variable, and belongs to the class Cln+1(p), if in a neighbourhood of some point its germ has the form C(p(An, Bn)), where An,Bn e Cln(p), and C(t) is an analytic function of one variable. Here the expression C(p(An, Bn)) means the following: for any analytic function in the superposition we take such germ that the whole expression is a well-defined germ. With the help of shifts of the form (x —> x + a, y —> y + b, t —> t + c), where a,b,c e C, we can move the germs of all functions to the origin. Therefore, it could be assumed that all germs are defined at the origin.

If one germ of an analytic function G(x, y) is representable with the help of the superposition of the specified form, then almost all germs are also representable in that way (see [6]). Namely, if G admits a holomorphic branch in some polydisc n e C2, then the representation takes place at a generic point in n (i.e., outside some proper analytic subset in n). A function has complexity n = Nv with respect to the function p, if it belongs to the set Cln(p) \ Cln-1(p). Next, if a function does not belong to the set Cln(p) for any finite value n, then we assume Nv = x>. The complexity is also computed at a generic point. We denote the hierarchy of all functions of finite complexity with respect to p by Cl(p), i.e., Cl(p) = W^=0Cln(p).

Let us recall that a preorder 4 on a set is a binary relation, satisfying the two following properties:

1) reflexivity (f 4 f Vf),

2) transitivity (f1 4 f2,f2 4 fs ^ f1 4 fs yf1,f2, fs).

If in addition to this properties we have one more,

3) antisymmetry (f 4 f2, f2 4 f1 ^ f1 = f2 Vf1, f2),

then we say that on the set a partial order is defined.

With the help of the classes Cln(p) we can introduce a preorder on the set A(x, y) of analytic functions of two variables in the following way:

Definition 1. f1(x,y) =4 f2(x,y) if f2 e Clf).

Reflexivity and transitivity of the relation 4 follow from the definition.

2. Properties of the preorder

On the set of germs of analytic functions f (x,y) acts the group of transformations G = {x a(x), y 3(y), f y(f)},

where a, ¡3, y are germs in the origin of holomorphic invertible functions of one variable, such that a(0) = ¡(0) = y(0) = 0. Here, as above, we assume that the germ of the function f is defined at the origin and vanishes there.

We call two germs [fi]0 and [f2]0 at the origin of the functions fi and f2 gauge equivalent, if there exists an element g e G, such that [fi]0 = go[f2]0. Next, two germs [fi]a and [f2]b at points a = (ai,a2) and b = (bi,b2) are gauge equivalent, if germs of the functions (x — ai,y — bi) —

fi(ai,bi and f2(x — a2,y — b2) — f2(a2,b2)j at the origin are gauge equivalent. Finally, two analytic functions fi and f2 are gauge equivalent, if they have gauge equivalent germs (notation: fi ~ f2).

Let us note an important property of the group G: it acts on the set of all analytic functions without changing their complexity (with respect to any function y). I.e., G o ^Cln(y) \

Cln-i(p)^ = (Cln(p) \ Cln-i(p)^ for all n and y.

2.1. The preorder is not a partial order

Let us show that the introduced preorder is not a partial order, i.e., the property of antisymmetry fails.

Consider two arbitrary different functions fi and f2 of complexity one with respect to addition. The action of G on such functions is transitive, i.e., there exists an element g e G, such that fi = g o f2. This means that the inequalities fi ^ f2 and f2 ^ fi hold, although fi = f2. Thus, if we consider the preorder on the set of all functions, then it is easy to construct the counterexample to the antisymmetry condition. But it seems natural to consider the quotient set of all functions with respect to the action of the group G and ask the question about antisymmetry of the introduced preorder for resulting equivalence classes.

On the set A(x, y)/ ~ j of classes of gauge equivalence the preorder < can be defined as follows: [fi] < [f2], if for representatives fi and f2 of classes [fi] and [f2] of gauge equivalence

we have fi ^ f2. This definition does not depend on the choice of representatives of the classes.

y2 2

Let yi = yi(x,y) = (x — l)y + ——, y2 = ¥2(x,y) = x2 + xy.

1 + y

Theorem 2. 1) yi ^ x + y, and NV1 (x + y) = 2. 2) y2 ^ x + y, and NV2 (x + y) = 2.

Proof. We have:

whence

f(x - V)y + —)y + — = xy2, V 1 + yJ 1 + y

ln{(e - l)Ve + TV^) ^ = x + y,

1 V^/ey; 1 V \[ey

i.e., x + y = ln ) + )) G Ch&i).

We also have x + ye Cl1(p1), because otherwise p1 would have complexity one with respect to addition, but it has complexity two.

Next, xsy = —p2(x2, -p2(x,y)) and xsy ~ x + y holds, i.e., x + y e Cl2(p2). And complexity NV2(x + y) also equals two, because the complexity Nx+y(p2) equals two.

The proof of Theorem 2 is complete. □

Thus, we have

y2 2 x + y 4 (x — 1)y + -- 4 x + y, x + y 4 x2 + xy 4 x + y.

1 + y

y2

And also the functions x + y and p1 = (x — 1)y + --have different complexity with respect to

1+ y

the hierarchy Cl, hence, they belong to the different classes [x + y] and [p1] of gauge equivalence. I.e., the relation < is also not a partial order. The same is true also for the pair of functions x + y and p2 : the classes [x + y] and [p2] are different.

2.2. Increase of dimension of the stabilizer of a function

The functions ф\ and ф2 demonstrate one more interesting effect. Namely, every function f have the stabilizer St(f) with respect to the action of the group G, which consists of all elements g £ G, such that g о f = f. For the dimension of the stabilizer there exist only four possibilities

(see [7]):

1) infinity — in this case the function depends on only one variable,

2) three — in this case f ~ x + y,

3) one — in this case f ~ r(x + y) — x for an analytic function r(t),

4) zero in other cases.

For the functions ф and ф2 the dimensions of the stabilizers are the following: dim St(^2) = 1, dimSt(^i) = 0 (it is easy to check by substituting ф in the published at the web page vkb.strogino.ru criterion of one-dimensionality of a stabilizer, see section "Другое", item "Maple-приложение к статье 'Об аналитических функциях двух переменных с одномерным стабилизатором'"). But on the other hand, the function x + y, having the three-dimensional stabilizer, belongs simultaneously to the classes С12(ф1) and С12(ф2). I.e., the more complicated function (with respect to and ф2) has higher dimension of the stabilizer. Such effect is absent in the standard hierarchy Cl with respect to addition: the simplest function x + y has the largest stabilizer.

2.3. Partial order on the equivalence classes

Using the preorder 4, we can divide the set of analytic functions of two variables into equivalence classes: f1(x,y) is equivalent to f2(x,y), if f1(x,y) 4 f2(x,y) and f2(x,y) 4 f1(x,y) (notation: f1 « f2). Note that the orbits of the action of the group G fall into the same equivalence class. In accordance with the general set-theoretic construction the preorder relation 4 on the set (^A(x, y)/ ~ j of equivalence classes become a partial order relation.

2.4. Existence of nonequivalent functions inside one hierarchy

It follows from Theorem 2 that some functions of different complexity are not equivalent with respect to the relation The question arise: can all the functions from the hierarchy Cl(p) fall

into the same equivalence class for all p? We will give an example of a function $ = $(x,y), such that among the functions from the hierarchy Cl($) there exist nonequivalent ones. For this purpose we will write down the function $ of infinite complexity with respect to the standard hierarchy Cl, such that the function x + y has complexity two with respect to the hierarchy Cl($).

We need some auxiliary constructions. Let F(x, y) be a function, satisfying the following three conditions:

1) F(x,y) = F(y,x),

2) F(x,y) = F(—x,y),

3) the function y = x(x, w) given by the implicit equation F(x, y) = w has infinite complexity.

It can be shown that functions F(x, y) with these properties exists, and there are many functions of the same sort (see below), but we will give explicit examples of functions F(x, y) with the specified properties.

To begin with, there exists an entire function 0, such that the implicit equation

0(y) + xy = w (2)

defines the function ^(x,w) of infinite complexity (see [8]). At the same time, there are plenty of such functions 0 (further we will only need that there are more than two of them). By the implicit function theorem, ^(x,w) is defined everywhere outside a proper analytic subset of the space C2. Note that equation (2) for all possible 0 defines the general solution of the Hopf

dy dy equation — = y^—.

dx dw

Then the function F(x, y) = 0(y2) + x2y2 + 0(x2) satisfies all three conditions given above. It is clear that conditions 1) and 2) are satisfied. Let us check condition 3).

If 0(y2) + x2y2 + 0(x2) = w, then y = x(x,w) = ■\J^(x,w — 0(x2)). Since ^(x,w) has infinite complexity, then x(x, w) also has infinite complexity, because the change of coordinates {x —> x, w —> w — 0(x2)} does not change complexity, and also extracting a root does not change complexity.

Let $(x, w) = x(x + w,w) + w. The function $(x, w) also has infinite complexity. Theorem 3. x + we Cl2($).

Proof. Since $(x, w) has infinite complexity with respect to w + x, then w + x e Cl1($). Let us show that

$(—$(x,w),w)= x + 2w. (3)

As before in the definition of the complexity classes, we understand equation (3) as equality of the function x + 2w to the correctly chosen composition of germs of the function $. Here we consider this equality only in those points, in which it is possible to choose germs so that the composition is well-defined (these are generic points). I.e., we will show that equality (3) holds at a generic point.

Let $(—$(x, w), w) = x^ — ^x(x + w,w) + w^ + w, w^ + w = u. Then

X^ — x(x + w,w),w^ = u — w,

which is equivalent to

w = F ^ — x(x + w,w),u — W. Using properties 2) and 1), we obtain

w = F (x(x + w,w),u — W = F ^u — w, x(x + w, w)j,

whence

x(x + w,w) = x(u' — w,w

Therefore we can suppose u — w = x + w, whence u = x + 2w. Following back the chain of equalities obtained in the proof of the theorem, we obtain that the function u = x + 2w satisfies the equality $(—$(x,w), w) = u. After the change |x —> x, w —> 2} we get what we need. The proof of Theorem 3 is complete. □

Corollary 4. Any function f (x,y) of finite complexity with respect to addition belongs to the hierarchy Cl($), but in the same time $ e Cl(f).

2.5. Pairs of functions of different finite relative complexity

Thus, we have constructed the function such that N^(x + y) = 2, but Nx+y($) = to for all natural n. The given above construction can be slightly modified to obtain examples of functions $n, such that (x + y) = 2, but Nx+y ($n) = v(n) for some natural v(n) e [n — 2,n + 2]. For this purpose we need to construct a function Fn(x,y) satisfying the following three conditions:

1) Fn(x,y) = Fn(y,x),

2) Fn(x,y) = Fn( x, y),

3) the function y = xn(x,w), given by the implicit equation Fn(x,y) = w, has complexity not less than n — 1 and not greater than n +1.

To construct the functions Fn(x,y) with the given properties it is necessary to choose, as above, an entire function such that the implicit equation

^n(y) + xy = w

defines the function ^n(x,w) of complexity n (see [8]). Note that there are also plenty of such functions ^>n.

Then the function Fn(x,y) = ^n(y2) + x2y2 + ^n(x2) satisfies all the three conditions. It is clear that conditions 1) and 2) are satisfied. Let us check condition 3).

If ^n(y2) + x2y2 + ^n(x2) = w, then y = Xn(x,w) = y/^n(x,w — ^n(x2)). Since ^n(x,w) has complexity n, then the complexity of the function xn(x,w) differs from n by no more than one, because the coordinate change {x —> x, w —> w — ^n(x2)} can change complexity by not more than one, and extracting a root does not change complexity.

Let $n(x, w) = xn(x + w,w) + w. Complexity v(n) of the function $n(x, w) differs from the complexity of the function xn by no more than two, therefore n — 3 ^ v(n) ^ n + 3.

2.6. Metric and continuous dependence on function

For functions f and g, which are holomorphic in U, the metric pu (f,g) is introduced in a standard way with the help of the enumerable sequence of seminorms in the Frechet space.

In the same time, it is clear that there is a continuous dependence of the classes Cln(p) on the function p. Namely, let a function f = f (B, S,p)(x,y) e Cln(p) be given. Here B = (B1(t),..., BN(t)) for N = (2n+1 — 1) is the set of functions of one variable, which enter into the superposition of the function f (B, S, p)(x, y), and S is a scheme of the composition, i.e., a way of arrangement of the functions from the set B inside the superposition (see [6] for more information on schemes). Consider a function f = f (B, S, p)(x,y) e Cln(p), which differs from f only by replacing p by p. And let the functions p and p have holomorphic elements (p(U ),Uj and (p(U ),Uj in a common domain U. Then V e > 0 3 S > 0, such that for

pu(p(U),p(U)) < S the functions f,f have holomorphic elements fvand fvin a

common domain V = V(e), for which pv(fv, fv) < e holds.

In particular, from the property of continuity follows that the condition that a function does not belong to the class Cln(p) is an open condition. Namely, we have the following

Proposition 5. Let f e Cln(p). Then there exists a number S > 0; such that for pu(p(U),p(U)) < S we have f e Cln(p).

Proof. Indeed, assume the contrary, i.e., let for all S > 0 there exist a function pg(U), such that pu(p(U),pg(U)) < S, and f e Cln(pg) holds. But the sequence of functions pg converges to p in the metric for S —> 0, whence by the property of continuity we obtain f e Cln(p), which contradicts the condition. □

2.6. Almost all hierarchies contain nonequivalent functions

Choose a function $, such that the following condition hold:

1) there exist a domain U, such that elements $(U) and $(U) of the functions $ and $ are defined in it,

2) p($(U), $(U)) <e,

3) $ e ci($).

Condition 3) can be attained, because the set Cl($) is equal to the countable union of nowhere dense sets Cln($). Hence, Cl($) can not include an entire neighbourhood of the function $(U).

For small enough e we obtain that $(—$(x, w), w) e Cl($), because $(—$(x, w), w) e Cl($) and the condition that a function does not belong to the set Cl($) is an open condition by continuous dependence of the differential criteria Jn(p) on p. This means that the set Cl($)/ « also contain more than one equivalence class. It is also clear that almost all (in the sense of the metric pu) functions $ possess this property.

2.7. Intersecting hierarchies, which do not coincide

Suppose that two hierarchies Cl(p1) and Cl(p2), based on different functions p1 and p2, intersect, i.e., there exists a function f, such that f e Cl(p1) and f e Cl(p2), and also f depends on both variables x and y. This takes place, for example, if p2 e Cl(p1) or p1 e Cl(p2) holds. In this case one hierarchy is nested within one another. It is natural to ask: is it the only reason for intersection? We will show in what follows that the answer is negative - intersecting hierarchies are not necessary nested within one another.

Consider the set F of functions F(x,y), satisfying properties 1) and 2) in (1), which are holomorphic in a domain U and for which the nonequality pu($(U),F) < e for some e > 0

holds. The set F defines the set Y of functions y = x(x,w). The set Y is a complete metric space, which can be represented as the countable union Y = of the sets Yn, consisting

of functions of complexity n. Moreover, the sets Yn for n < to are nowhere dense. By the Baire theorem we obtain from here that the set can not be represented as an at most countable union of nowhere dense sets. Note that coincide with the set of all functions y = x(x,w), given by property 3) in (1).

The set Cln($) is nowhere dense for all fixed n. Hence, there exists a function $ eY (and there are plenty of such functions), such that x + y e Cl2($i), but $ e Cln($) Vn. We also can obtain the condition $(U) e Cln($i) Vn in an analogous way.

This means that the hierarchies Cln($) and Cln($i) intersect (in the function x + y and some others), but are not nested within one another. It is clear that the intersection contains the set Cl(x + y).

In conclusion, let us formulate a question, which was left out of our considerations.

Question 6. Fix a function y. Does there exist the minimal element in the hierarchy Cl(y)?

Here we mean the following. As was shown, there exists the function in terms of which we can express the function x + y, such that the hierarchy Cl($) contains the hierarchy Cl(x + y), and moreover, this inclusion is strict. The question arise: is it possible to find an infinite sequence of functions , j = 1,2, 3,..., such that the strict inclusion Cl(^) c Cl(^2) c Cl(^3) C ... holds?

This work was supported by the Russian Science Foundation, project no. 19-11-00316, https:// rscf.ru/en/project/19-11-00316/.

References

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[2] G.M.Henkin, Nonisomorphism of certain spaces of functions of different numbers of variables, Funct. Anal. Appl., 1(1967), no. 4, 306-315. DOI: 10.1007/BF01076010

[3] A.G.Vitushkin, G.M.Henkin, Linear superpositions of functions, Russian Math. Surveys, 22(1967), no. 1, 77-125. DOI: 10.1070/RM1967v022n01ABEH001204

[4] A.G.Khovanskii, The representability of algebroidal functions by superpositions of analytic functions and algebroidal functions of one variable, Funct. Anal. Appl., 4(1970), no. 2, 152-156. DOI: 10.1007/BF01094492

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[7] V.K.Beloshapka, Stabilizer of a function in the gauge group, Russ. J. Math. Phys., 24(2017), 148-152. DOI: 10.1134/S1061920817020029

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Отношение предпорядка на множестве аналитических функций

Мария А. Степанова

Математический институт им. В. А. Стеклова РАН Москва, Российская Федерация

Аннотация. В работе изучено отношение предпорядка, которое естественно определяется на множестве аналитических функций двух комплексных переменных с помощью суперпозиций аналитических функций одной переменной.

Ключевые слова: аналитическая сложность, суперпозиции аналитических функций.