CC BY
6712 Probl. Anal. Issues Anal. Vol. 7 (25), Special Issue, 2018, pp. 12-30
DOI: 10.15393/j3.art.2018.5451
The paper is presented at the conference "Complex analysis and its applications" (COMAN 2018), Gelendzhik - Krasnodar, Russia, June 2-9, 2018.
UDC 517.574
K. P. Isaev
ON ENTIRE FUNCTIONS WITH GIVEN ASYMPTOTIC
BEHAVIOR
Abstract. We study approximation of subharmonic functions on the complex plane by logarithms of moduli of entire functions. In the theory of series of exponentials these entire functions are the main tool. In questions of decomposition of functions into a series of exponentials, the subharmonic function, as a rule, satisfies the Lipschitz condition. We prove the theorem on approximation of such subharmonic functions. Also we prove the theorem on joint approximation of two subharmonic functions.
Key words: subharmonic function, approximation, entire function, Riesz measure
2010 Mathematical Subject Classification: 31A05, 30E10
1. Introduction. Entire functions with prescribed asymptotic behavior serve as the main tool in constructing the exponential series. The problems of existence and constructing entire functions with prescribed asymptotic properties arose as inner problems of the theory of entire functions. This problem was dealt with by many mathematicians. See, for example, [1], [3]— [5], [8]. The next theorem is proved in [8].
Theorem A. [8] For each subharmonic on the plain function u of finite order greater than zero and for each ft < 0, there exists an entire function f satisfying the relation
|u(A) - ln |f (A)|| = O(ln(|A| + 1)), A / E, |A| rc,
where the exceptional set E can be covered by a set of disks B(wk ,rk) so that E|wfc|^Rrfc = OR), R ~^<x>.
By B(z,t) we denote the open disk with center z and radius t.
© Petrozavodsk State University, 2018
In applications, approximated subharmonic functions often have some additional properties that can refine the asymptotics. On the other hand, one usually requires not just to estimate the size of the exceptional set, but, to a greater extent, to get its construction. In questions of decomposition of functions into series of exponentials, subharmonic functions, as a rule, satisfy the Lipschitz condition (see, for example, [6]). Therefore, we consider the approximation of such functions separately.
For a measure m we denote the ^-measure of the disk B(z, t) by ^(z, t) and let t) = m(0, t). The set of zeros of an entire function L is denoted by N(L). The notation A(x) — B(x), x G X, means that for some constant C > 0, for all x G X, the estimate A(x) ^ CB(x) holds true.
In the present work we prove the following theorem.
Theorem 1. Let u be a subharmonic function on the plane, and m be the Riesz measure of u. If for some M > 0, for all points z G C the condition
p(z,t) < Mt,t G (0; 1), (1)
holds true, then there exists an entire function f with simple zeros An such that for some S G (0; 1) the disks Bg(An) = B(Xn, S(|An| + 1)-1) are pairwise disjoint and the function satisfies the relations
| ln |f (A)| — u(A)| < Aln(|A| + 1) + C, A G J Bs(An),
n
| ln |f'(A)| — u(A)| < Aln(|A| + 1)+ C', A G N(f).
Here the constant A > 0 is independent on the constant M and the function u, and constants C, C', S depend on M, but not on u.
Note that if a subharmonic function u satisfies for some constant K > 0 the Lipschitz condition |u(z) — u(w)| ^ K|z — w|, z,w G C, then its Riesz measure satisfies condition (1): ^(z,t) < Ket, z G C, t > 0. This follows from Jensen's formula
2n r
1 J u(z + reiv)d^ = u(z) + J ^^ dt, z G C, t > 0.
0 0 In fact, the Lipschitz condition implies the estimate
r 2 n
r W f M(z,t) ,n ^ 1
v r ^foi) dt ^ — i ^(z + reiv) — uUMdf < Kr. e' J t 2n J
In applications of approximation theorems, one often needs joint approximation of several subharmonic functions. In the next theorem we consider the joint approximation of two subharmonic functions which Riesz measure satisfying condition (1).
Theorem 2. Let Uj, j = 1, 2, be two subharmonic functions on the plane, and ^j, j = 1,2, their Riesz measures satisfy the conditions
^(z,t) < Mt,t / (0; 1), (2)
and the measure satisfies the condition
[ M2(r)dr ^ _ ^
J < (3)
1
Then there exist entire functions fj, j = 1,2, such that all zeros of f = = f1f2 are simple; for some S > 0 the disks Bg(A) = B(A, c>(|A| + 1)-1), A / N(f), are pairwise disjoint; and for some constants B, C, C > 0 the relations
| ln |fj(A)| — uj(A)| < Bln(|A| + 1) + C, A / U Bg(z),
zeN (fj)
| ln |fj (A)| — Uj (A)| < B ln(|A| + 1) + C', A / N (fj),
hold true. Here the constant B> 0 is independent on the constant M and the functions Uj, and the constants C, C', S depend on M, but not on Uj.
2. Proof of Theorem 1.
Lemma 1. Let u be a subharmonic function on C, u(0) = 0, and its Riesz measure ^ satisfies (1). Then there exists a subharmonic and infinitely differentiable function v on C satisfying the conditions u(A) ^ < v(A) < u(A) + M, Av(A) < c, A / C, where the constant c> 0.
Proof. Let a(t) / Cw(R); a(t) = 0, t / (0; 1); a(t) > 0, t / (0; 1); and
1
J ta(t) dt = -1.
If a(A) = a(|A|), A G C, and dm is the planar Lebesgue measure, then
v(A) := j a(A — w)u(w)dm(w) = j a(w)u(A — w)dm(w), A G C,
c c
is a subharmonic and infinitely differentiable function (see [7, p. 51]). By definition v(A) — u(A) = fC(u(w) — u(A))a(A — w)dm(w). Converting to polar coordinates and using Jensen's formula, we obtain
1 t
v(A) — u(A) = 2nJ a(t)(J ds^tdt.
0 0
It is obvious that v(A) > u(A), A G C. By condition (1) this implies v(A) — - u(A) ^ M fc a(A)dm(A) = M. The first statement of lemma is proved. Let us estimate Av. Considering u as a distribution, we obtain Av(A) = = n fc a(A — w)d^(w). If a = maxt a(t), then in view of (1), we have Av(A) < nap(A,1) < naM. □
Thus, we can prove Theorem 1 assuming function u to satisfy condition
Au(A) < M, A G C. (4)
Let us show that it is enough to prove Theorem 1 for M =1. In fact, suppose the following theorem is proved.
Theorem 1'. Let v G be a subharmonic function on the plane, v(0) = 0, and Av(A) ^ 1, A G C. Then there exists an entire function g with simple zeros wn such that for some ¿0 G (0; 1) the disks B(wn,S0(lwnl + 1)-1) are pairwise disjoint and the function satisfies the relations
| ln |g(w)| — v(w)| < A0 ln(|w| + 1) + C0 A G U B(wn, MK| + 1)-1),
n
| ln |g'(A)| — v(A)| < A0 ln(|A| + 1) + C0, A G N(g).
Here the constants A0,C0,C0 > 0 are independent on the function v. Let u satisfy (4). If M > 1, then we consider the function v(w) =
= u{~M). Since Vv(z,t) = , then ^v(z,t) < t for t < 1.
By Theorem 1', there exists a function g that satisfies the appropriate
estimates. Let us take the function f(A) = g(MA) with simple zeros
1w An = — wn. Under the mapping w —> — the pairwise disjoint disks
M
M
B(wn, cS0(|wn| + 1) 1) are mapped to the disjoint disks B(An, (M|An| + + 1)"1), and outside these disks the estimate
| ln |f (A) | - u(A) | < Ao ln(| A| + 1) + Ao ln M + Co holds true. Since
(5)
M2 (|Anl + 1)-1 < rn := M (M |An| + 1)-1 < M (I An | + 1)-1,
the disks B(An, —^(|An| + 1) ^ are also pairwise disjoint. We extend M 2
estimate (6) outside these disks. Let H be the minimal harmonic majorant of u on B(An,rn); then, by Green's formula and by condition (4) for A G B(An,rn), we have
0 < H (A) - u(A)= J G(A, z)du(z) =
B(A„,r„)
M
G(A, z)Au(z)dm(z) <— G(A, z)dm(z)
B(A„,r„)
B(A„,r„)
where G(A, z) is Green's function of the Dirichlet problem for B(An,rn). Taking into account that for the function A(A) = |A — An|2 the relation AA(A) = 4 holds true and the minimal harmonic majorant of A is identically equal to r^, we obtain the estimate
M
1
0 < H (A) - u(A) = — max >n - |An - z|2) < —, A G B(An,rn).
4 z£B(A„,r„)
By the maximum principle and by (6) for A G B(An,rn) we have
(6)
H (A) - ( ln |f (A)| - ln
|A - An|
^ max |u(z) - ln |f (z)|| ^
|z-A„|=r„
< Ao ln(|An| + rn +1) + Ao ln M + Co < Ao ln(|An| +1) + Ao ln(2M) + Co.
(7)
1
n
n
If
^(|An| + 1)-1 < |A — An| < rn < ^(M|An| + 1)"1,
M2W1 V "nl ^'n ^ M
ln |--1 < ln M and therefore
then
|H(A) — ln |f(A)|| < A0ln(|A| + 1) + (A0 + 1) ln(2M) + C0,
where A G B(An,rn) \ B(An,dM(|An| + 1)-1). This together with (6) implies the required estimate for the function f
|u(A) — ln |f (A)|| < A0 ln(|A| + 1) + (A0 + 1) ln(2M) + C0 + 4m,
where A G B^An, -^"2 (|An| + 1) 1 j. Passing to the limit as A ^ An in (7) and applying (6), we obtain the required estimate for f '(An):
|u(An) — ln |f'(An)|| < A0 ln(|An| + 1) + C0 + 4^+lnM,n G N. To prove Theorem 1', one more lemma is needed.
Lemma 2. Let u be a smooth subharmonic function, and Au satisfy (4) with M = 1. Denote the square centered at the origin with sides of length 3n and parallel to the axes by Qn, n G N U {0}. Then
Qn+1 \ Qn = U Qn,j, n G N U {0},
j=1
where Qn,j are the squares obtained by shifting the square Qn by the vectors (±3n,0), (0, ± 3n), (±3n, ±3n) and enumerated counterclockwise, starting from the square which intersects with the positive ray of the real axe. There exists a subharmonic function u with the Riesz measure ju such that
1) inside the squares Qn,j the function u is smooth and estimate (4) holds true;
2) ¡u(Qn,j) is a non-negative integer;
3) the estimate |u(A) — U(A)| < 45ln(|A| + e) + 142, A G C, holds true.
n
Proof. Let M(Q„,j) := m„,j + q„,j, j = 1, 2,..., 8, n G N, where q„,j = = {M(Q„,j)} G [0; 1) is the fractional part of M(Q„,j)• Let q+ = J2j q„,j G G [0;8), q- = ^j(q„,j — 1) G [—8;0). Define the sequence q„ as follows: let qo = {m(Qo)}; if qj are defined for j < k — 1, then, as qj ^ 0,
let qfc := q-; qfc := q+ otherwise. Thus, "=o qfc G (—8; 8), n G N. Then define the sequence of natural numbers No, N„,j, j = 1,..., 8, n G N. Let No = [m(Qo)]; if q„ = q-, then N„,j = M(Q„,j) — (q„,j — 1); and if q„ = q+, then Nn,j = M(Q„,j) — qn,j. Thus, either N„,j = m„,j + 1 or N„,j = m„,j. The restriction of the measure m to the square Q„j is denoted by M„j, N0
Mo = M Iq0 Let ^o = , „ s Mo, if m(Qo) = 0, and M(Qo)
N„
M„,j = (Q' ) M„,j, j = 1,..., 8, n G N,
if M(Q„,j) = 0. If m(Qo) = 0, then Mo = 0. And if M(Q„,j) = 0, then M„,j = 0. Then M„,j (C) = N„,j are non-negative integers. Let vo = = Mo — Mo, v„,j = M„,j — M„,j; then
8
v„,j(C) G ( —1;1), v„j (C) G (—8; 8). (8)
j=i
Let
TO
v = Vo + V„,j, V+ = Vo + ^ ,V = — ^
then v± are non-negative measures and v = v + — v-. At that,
v±
j=i
Let us prove that
A
w
8
(U = q± G (—8; 8).
n(A):= /ln|1 — -|dv(w) < 45ln(|A| + e) + 142, A G C. (9) w
c
Then Lemma 2 holds true for the function M(A) = u(A) — n(A).
Choose A G Qn+1 \ Qn. If w G Qm+1 \ Qm, then 3m < |w| < —3m+1
and |ln(|Z| + 1)| < 2|Z| for |Z| < 1; therefore
/ln 11--|dv (w)
w
<
c\Q„
Similarly, we have
= 2
Qn+ m+1
/ln h--Uv(w)
w
3n / J 3m ^
m=2
ln
1 — w
2 / n_1 on—m ,/2\
-H* m(85— °-_r +£ <1072 <">
Qn-1 Let us prove that
|A|
72 N 1 72 ' 2
|v(t)| = |v(B(0,t))| < 17, t > 0.
Indeed, if t < —, then B(0,t) c Q2 therefore
<
(10)
(12)
|v(t)| < |v(Q1)| + 5 |v1,j(C)| < 9.
j=1
3 3n
For t ^ —= we denote the maximal natural number, for which —= < t
a/2 A/2
by n. Then Qn c B(0,t) and
3n+2 3 3n+1 3 -= —7 > 7= t > t.
2 72 72 72
Hence, Qn+2 D B(0,t). Thus, in view of (8), we obtain
n+1 8
|v(t)| < |v(Qn)| + Vi,j(C)|< 17.
i=n j=1
Let un be the restriction of the measure v to the square Qn. Then
3n
J ln dv(w)|< ln 3||v (Qn_1)| +||
Un_ 1 (t)
t
dt
+2
Q
By condition (4) with M = 1, in view of (12), we have
^n-1
(t)
dt
r u(iK f 17dt _
</ ^-dt + /- < 1 + 17ln(|A| + e).
Jo t J t
Hence,
Qn-i
and using (11), we obtain
[ ln ^ dv(w) < 18 + 17ln(|A| + e),
J |w|
ln
Qn-i
1 - A
w
dv(w) < 17 ln( lA| + e) + 33.
(13)
If w G Qn+2 \ Qn-1, then |w| G [^; ^ 3n+2] and |A| G [3-; ^ 3-+1 ] Therefore, for w G B(A,1) the estimate
V2 3-n < A - w
9 <
< 51,
holds true. Hence, 1 ln |1 — W11 < ln(|A| + e) + 4. Consequently,
i ln 11 — -|dv(w) < 24ln(|A| + e) + 96. (14)
w
(Qn+2\Q„-i)\B(A,1)
It remains to estimate the integral over the disk B(A,1). The measure v is a part of the measure m by construction, so it satisfies (1) with M = 1. Taking this into account and integrating by parts, we obtain
J ln |A — w|dv(w) < 1, J ln |w|dv(w) < nln(|A| + e). B(A,1) Hence,
B(A,1)
ln|1 - A|
w
dv(w) < nln(|A| + e) + 1.
B(A,1)
This, together with (10), (13) and (14), implies (9). □
Lemma 3. There exist measures Mn, Mn (C) = 1, and rectangles Pn, n G N, such that
!) En Mn =
2) interiors of convex hulls of supports of the measures Mn are pairwise disjoint;
3) support of the measure Mn is located in Pn, n G N;
4) sides of the rectangles Pn are parallel to the axes and the ratio of length of the sides for rectangles Pn lies in the interval [3-1; 3];
5) each point of the plane belongs to at most four rectangles Pn;
6) if Fn is convex hull of support of the measure Mn, then
diam Fn < 2V2 min |A| + V2.
Proof. Let us apply theorem 1 from [8] to restrictions of the measure M on the squares Qn,j. After renumbering, we obtain a set of unit measures satisfying properties 1-5 of Lemma 3. Property 6 follows from the corresponding property of the squares Qn,j. □
Let us continue the proof of Theorem 1'. Centres of mass of the unit measures Mn constructed in Lemma 3 are denoted by An:
j w d,Mn{w) = An, n G N.
By Mn we denote the restriction of the measure m to the square Qn, and let nn be the potential of this measure
nn(A) = / In 1 - — dMn(A).
J c w
Then the measure Mn satisfies the conditions of theorem 3 from [8]. In the terminology of this paper, each point A G C for each s = s(A) G (0; 1] is (n, s)-normal with respect to the measure M, by (4). Hence, this theorem shows that for the polynomial
Pn(A) = n l1 - £
the relation MA) - In |Pn(A)|| < A ln(|A| + 1) + B ln(s(A) + 1) + C holds true outside the set of discs Bk(s) = B(Ak,s(Ak)), Ak G Qn. Here constants A, B, C are independent on m and n. Thanks to the latter fact, we
can justify the passage to the limit in the usual way. As a result, we see that there exists an entire function f with simple zeros at the points A„ satisfying the condition
|M(A) — ln |f (A)|| < Aln(|A| + 1) + B ln(s(A) + 1) + C, A G U B„(s). (15)
We need to show that for sufficiently small S > 0, the discs B„ = B„(A„, S(|A„| + 1)-1) are pairwise disjoint. Let us estimate the distance d„ from the point A„ to the boundary of convex hull F„ of support of the measure M„. Let w„ be a point where this distance is attained:
|A„ — w„| = inf{|A„ — w|, w G F„}.
Let w„ — A„ = eiVn |A„ — w„| and z = Tw = (A„ — w). Under such transformation, the image F* of the hull F„ is located in the half-plane {Re z < d„} and the image of the measure dM*(z) = dMn(A„ — eiVnz) satisfies the conditions
j dM*(z) = 1, j zdM*(z) = 0,
dM* (z) = - x„(z)Am(A„ — eiVn z)dm(z), n
where x„(z) is the characteristic function of the set F*. Let S(x) = 1 X„(x + ¿y)AM(A„ — eiVn (x + iy))dy.
-to
Then S(x) is a compactly supported function on the segment [a; d„] and by statements 6 in Lemma 3, 0 < S(x) < 3n(|A„| + 1) := M„. Moreover, it follows from the properties of m* that
/S(x)dx =1, /xS(x)dx = °.
The next lemma is proved in [2] (see [2, Proposition 2]).
Lemma 4. [2] Let S(x) be a non-negative continuous compactly supported function satisfying the conditions
1) conv supp S = [a; d], 2) sup S(x) < M0 < <x>,
x
d d 8)/ S(x) dx = 1' 4)/ xS(x) dx = °
Then d >
6M0
Lemma 4 implies that dn > (1 + |An |)—1, n € N. By the property 2 in Lemma 3, the disks Bn = B(An, ¿(1 + |An|) —1), 5 < ^^, are pairwise disjoint. In particular, each point A outside these disks is (n,(1 +1An |) 1 )-normal with respect to the measures f and v = ^k 5(Ak), here 5(w) is a unit mass concentrated at the point w. By relation (15), the estimate |i(A) - ln If (A) || < A ln(|A| + 1) + C, A € U n Bn, holds true outside the disks Bn. By usual tricks and by Cauchy's formula f) = 2ni fdz), one can obtain the necessary estimates for the derivatives at the points An.
3. Proof of Theorem 2.
By Theorem 1, for each of the functions Uj there exists an entire function fj satisfying the estimates
| ln |fj(A)|-Uj(A)| < Aln(|A| + 1)+C, A € (J B(A, 5(|A| + 1) —1), (16)
Aew (/,)
| ln |fj (A)| — Uj (A)| < A ln(|A| + 1) + C', A € N (fj).
At that, the disks Bs (A) = B(A,5(|A| + 1) —1), A € N (fj), are pairwise disjoint for each of j = 1, 2 separately. Let v2 be the Riesz measure of ln |f2|. Let us formulate the properties of the measures f2 and v2 in the next lemma.
Lemma 5.
1) f2(t) = o(t), v2(t)= o(t),t —(17)
1 2
2) if A € J Bs(z), then ( V2 (A,T)dT < 2A ln(1 + |A|) + C''.
zew(/2) 0 T
1
Proof. The first assertion of the lemma is an immediate consequence of condition (3). To prove the second assertion, we use Jensen's formula for the function v(z) = u2(z) — ln |/2(z)| with the Riesz measure v = — v2:
2n r
1 f <\ , ivw _ , f v(A,r)dr
¿/^(A + re^)dp = v(A) + J
By (17), there exists R > 2 such that ^2(t) < 4 for every t > R. We take |A| > R and we let r(w) = ¿(1 + |w|)-1. The disk B(wk,r(wk)) can intersect with the circles C(A,r), r G [2; 1), only if the center wk lies in B(A, 1 + S). For these disks r(wk) < 2r(A), therefore,
£ r(wfc) < 2r(A)M2(1 + S + |A|) < 1.
wfc £B(A,1+5)
Hence, there exists a number r G [2; 1) such that the circle C(A,r) does not intersect with the exceptional set. By construction, the inequality |v(z)| = |u2(z) — ln |/2(z)|| < Aln(1 + |z|) + C holds true on this circle. The same inequality holds true at the point A. By Jensen's formula, we obtain
r
f M2(t) — V2 (t)
t
dt
< 2Aln(1 + |z|) + C1.
This, together with (2), implies the second assertion of Lemma (5). □
Next we prove a lemma on exceptional sets.
Lemma 6. If estimate (5) holds true for u2 and S, then estimate (5) holds true for u2 and each S' G (0; S) outside the discs B(A, S'(|A| + 1)-1), A G N(/2), possibly with another constant C depending on S'.
Proof. In fact, let us take an arbitrary AGE' := |J f) B(z, S'(|z| + + 1)-1). If, in addition, A G E := UzeN(l) B(z,S(|z| + l)-1), then the estimates hold true by the hypothesis of the lemma. If A G E, then there exists a number n such that r' := S'(1 + |An|)-1 < |A — An| < r := S(1 + |An|)-1. Let G(A, w) be Green's function for the disk B(An, r), then there are representations
U2(A) = h(A) — J G(A,w)dM2(w), ln |/2(A)| = ho(A) — G(A,A„),
B(A„ ,r)
T
where the functions h, h0 are the harmonic majorants of u2 and ln |f21 respectively on the disk B(An,r). The difference |h(A) — ho(A)| is estimated by the maximum principle for harmonic functions. For some constant C1
|ho(A) — h(A)| < Aln(1 + |A|) + Ci.
We estimate the potential of the measure using (2). Let h1 be the harmonic majorant of u2 on the disk B(A, 2r). Then for S < 2
j G(A, w) d^(w) = h(A) — m2(A) ^
B(A„ ,r)
2r
< hi(A) — U2(A) = J
^2 (A, t) dt
< 2Mr < M.
And finally G(A, An) = ln < ln £ = ln (£), A / B(Ari, r'). Thus,
|u2(A) — ln |f2 (A) || < A ln(l + | ) + Ci-+ M + ln(jr ) for A </B(An,r'). □
Let z G N (fi). Then there are no other zeros of the function f1 in the disk B s (z) and there can be only one zero of the function f2 in this disk. If w G N (f2) p| B s (z), then we move this point to the nearest point w' on the boundary of the disk B s (z). Let N (f2) = {wk }£=1. We denote by wk the point wk if wk does not belong to the union of the disks B| (A), A G N(f1), and the point w'k if the point wk belongs to one of
these disks. Also, we denote by N the set of points wk. Obviously, the disks B s (A), A G N (fO U N, are pairwise disjoint.
Lemma 7. Let
n2(A) = £ ln
k=1
1 — —
Wk
r(A) = £ ln
k=1
1 — A
Wk
, A G C. (18)
Then outside the set of pairwise disjoint disks E := |Jf ^ n B| (w) for each e > 0 and for some constant C the estimate
MA) — 5r(A)| < eln(1 + |A|) + C
t
holds true.
Proof. The convergence of the series in (18) follows from (3). Since
ln 11 - I - ln
I Wk I
1 - ^
wik
^ C(n), for each n G N, then
without loss of generality one can consider only with sufficiently large absolute value. Thus, let us assume that |wk| > 25. Since for |z — w| < 1 and |z| > 25
23 1 + |z| 25
+ | | ' (19)
— ^ 24
^ —, 24'
1 + |w
then we can assume that |wk — | < |(1 + |wk|)-1 and, so Then, taking into account the simple inequality | ln |1 — Z|| ^ 21Z| for |Z| ^ 1, we have
Wk — Wk Wk
< 2.
ln |1 -
Wfc - Wfc Wfc
< 2|wfc - wfc | <
5
By (17), we obtain
œ — œ
^^ | ln | —1 = V|ln|l -
k=1 k k=1
|wfc |
Wfc - Wfc Wfc
|wfc|(1 + |wfc|) '
œ
< Y
k=1
5
Iw |(1 + |wfc|)
5 dv2(w)
<
5dv2(t)
= 25
v2(t)dt
:= Ci.
(20)
J |w|(1 + |w|) t2 J t3
1 1 1
Fix the point A. Let /1(A) be the set of indexes k such that |wk | > 2|A|, let /2(A) be the set of indexes k such that |wk | < ^, and let /3(A) be the set of indexes k such that ^ < |wk | < 21A|. We denote by J1(A) the set of indexes k G 13(A) such that |A - wk| > 1. Let for some p > 1 J2(A) be the set of indexes k G 13(A) such that 1 > |A - wk | > p5(1 + |A|) 1, and let J3(A) be all other indexes k G 13(A).
Let k G 11(A), then |A| < |wk|/2 and |A - wk| > |wk|/2; therefore
Wk—Wk A—Wk
< 1. Hence,
and
fce/i
ln
ln
A - Wfc
A - Wfc
A - Wk
A - Wfc
ln
1
Wk - wk
A - wk
<
21A |
25dv2 (w) |w|(1 + |w|
<
25
|wfc |(1 + |wfc |)'
<
2A
25dv2(t) t(1+1)
< C2.
(21)
Let k G I2(A), then |A — wk| ^ |A|/2; therefore Hence,
ln lA — wk 1
A — wk 1
ln |1 -
wk — wfc
A — wk
<
Wfc-Wfc
A—wk
26
< 1 for |A| > 1.
IA | (1 + |wfc |):
and
E m
fce/2
A — wk
A — wk
|A|/2
<
26dv2(w)
| A|/2
<
26dv2(t)
J |A|(1 + |w|) ^ 7 |A|(1 +1)
i i
< C3. (22)
Let k G J1(A). Then
wv — wk
A — wv
<
26
1
^ -,
3(1 + |wv|) " 2'
and by (17), we have A — wk
ln
keJi
A — wk
2A
<
46dv2(t)
2A
<
86dv2(t)
J 3(1+1) ^ J 3(|A| + 2)
|A|/2 |A|/2
< C4. (23)
It remains to estimate the difference of potentials for k, such that |A — wk| < 2. For k G J2(A), due to the choice of p, we still have
wv — wk
1
^ -. 2
A — wk |
Therefore, by (19), for r = p6(1 + |A|) 1 we have
E m
keJ2
Hence,
A — wk
A — wk
E m
keJ2
< 26 E 3 ^
keJ2
1
(A — wk )(1 + |wk|)
<
6 2 dv2(A, t)
1 + |A|
A — wy A — wk
26v2(A,2) 6 J v2(A,t)dt
1 +1 A 1 + TT|A|y t2 .
t
By assertion 2 of Lemma 5, we have v2(A,t) < 2Aln(1 + |A|) + C, hence, (A, 1) ^ C d
T+aT ^ C5 and
1 1 2 2
5 f v2(A, t)dt
<
1 + |A|7 t2
r r
By the last two estimates and by (8), we obtain
A — wfc
5 f v2(A,t)dt 2A, , ^
(1+m) 2( t ) < — ln(1 + |A|)+ C6. (1 + |A|)rJ t p
ln
fceJ2
A — wfc
2 A
< — ln(1 + |A|)+ C7. p
(25)
The number of indexes k G is finite and bounded by some absolute
constant N. Thus,
ln
\ — Wk A-Wfc
ln
fceJa
< Const , AGE, k G J3(A). Hence, A — wfc
A — wfc
^ C8.
(26)
Since
MA) — 5r(A)| < £
, 1 wfc
ln
+E
ln
1
A — wk
A — uk
then by (20)-(7), (5)-(9), the assertion of Lemma 7 follows. □
Let us now complete the proof of Theorem 2. By condition (3), we have the representation f2(A) = eg(A) k (1 — W^), A G C, where g is an
entire function. Let /(A) = eg(A) k e1 — ^^), A G C. Then by Lemma 7 for each e > 0, we have
| ln |f2(A) | — ln |/(A)|| = |n2(A) — 5r(A)| <
< eln(1 + |A|)+ C, A GE = y Bs (w).
ȣN(/2) U N
By Lemma 6, outside the set E, we have an estimate with some constant C and an arbitrary e > 0
|u2(A) — ln |/(A)|| < |u2(A) — ln |/2(A)|| + |n(A) — 5r(A)| <
< (A + e)ln(1 + |A|) + C, A GE = y B|(w). (27)
wew (/2) U N
We extend this estimate to the union of sets |Jk(Bs (wk) \ Bs (wk)). The
>| (wk) \ B
union of B s (wk) I J Bs (wk) is a subset of the disk Bm (wk ) c B 14s (wk) c
8^8 24 ^ 24
C B^ (wk). Thus, the estimate (27) is satisfied outside the pairwise disjoint discs B iis (wk). By Lemma 6, it is also satisfied outside the discs Bs (wk).
Acknowledgment. This work was supported by RFBR (project 18-0100095 A)
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Received May 15, 2018. In revised form, September 16, 2018. Accepted September 17, 2018. Published online September 20, 2018.
Institute of Mathematics, Ufa Federal Research Centre,
Russian Academy of Sciences (Institute of Mathematics UFRC RAS)
112, Chernyshevsky str., Ufa, Russia, 450008
Bashkir State University (BashSU)
32, Zaki Validi str., Ufa, Russia, 450076
E-mail: [email protected]
