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11
Journal of Siberian Federal University. Mathematics & Physics 2020, 13(5), 622—630
DOI: 10.17516/1997-1397-2020-13-5-622-630 УДК 517.53
On the Differentiation in the Privalov Classes
Eugenia G. Rodikova*
Bryansk State University Bryansk, Russian Federation
Faizo A. Shamoyan^
Saratov State University Saratov, Russian Federation
Received 22.09.2019, received in revised form 25.04.2020, accepted 26.07.2020 Abstract. The invariance of the Privalov classes with respect to the differentiation operator is studied. Keywords: Privalov spaces, the Bloch-Nevanlinna conjecture, differentiation operator. Citation: E.G. Rodikova, F.A. Shamoyan, On the Differentiation in the Privalov Classes, J. Sib. Fed. Univ. Math. Phys., 2020, 13(5), 622-630. DOI: 10.17516/1997-1397-2020-13-5-622-630.
Introduction
Let C be the complex plane, D be the unit disk on C, H(D) be the set of all functions, holomorphic in D. For all 0 < q < we define the Privalov class of function nq as follows
(see [11]): ^
ng = (f € H(D) : sup -L T (ln+ \f (reie)\)q dd < +C .
ln+ \a\ = max(ln \a\, 0), Va € C.
The classes nq were first considered by 1.1. Privalov in [11]. If q =1 the Privalov class coincides with the Nevanlinna class N of analytic functions in D with bounded characteristic
1 n
T(r, f) = — f ln+ \f (rel°)\d0, 0 ^ r < 1. This is well-known in scientific literature (see [9]).
2n -n
Using Holder's inequality, it is easy to prove the inclusion chain:
n (q> 1) C N C n (0 <q< 1).
Since for all 0 < q < q'
(ln+ \f\)q < (ln+ \f \ + 1)q < (ln+ \f \ + 1)q < 2q' • ((ln+ \f\)q' + ^ ,
we have
nq' C nq.
In the case of 1 < q < the Privalov spaces were studied by M. Stoll, V. I. Gavrilov, A. V. Subbotin, D.A. Efimov, R. Mestrovic, Z. Pavicevic, etc. The monograph [6] contains a brief overview of their results. Certain results were extended to the case 0 < q < 1 by the first author of this paper (see [13]). Notice that the case 0 < q < 1 was little studied. The questions
* [email protected] [email protected] © Siberian Federal University. All rights reserved
of interpolation in the Privalov classes, as well as properties of root sets of analytic functions from these classes were investigated in recent works by the authors (see [14-16,20]).
In this paper we study a question of the invariance of the classes nq with respect to the differentiation operator. In other words, we verify the validity of the Bloch-Nevanlinna conjecture in the Privalov spaces.
The assumption, known as the Bloch-Nevanlinna conjecture, was clearly formulated by Nevan-linna in 1929 (see [9]) as follows: a derivative of any analytic function in the unit disk with bounded characteristic is a function of bounded characteristic.
The famous result refuting this hypothesis belongs to O.Frostman (see [5]). He proved that there is a Blaschke product whose derivative is not a function with a bounded characteristic.
Subsequently, many counterexamples that refute the Bloch-Nevanlinna conjecture were constructed in the works of others such as H. Fried (1946), W.Rudin (1955), W.Hayman (1964), P.Duren (1969), J.Anderson (1971), L.-Sh. Khan (1972), et. al. D.Campbell and G.Weeks [1] provide a brief overview of these results, as well as a general approach to the construction of such examples.
The invariance with respect to the integro-differential operators of other classes of analytic functions have been studied by many mathematicians. A brief overview of their results is contained in the work of S. V. Shvedenko [22]. In particular, a closure of the classes of analytic functions in a disk with the restrictions on Nevanlinna's characteristic function regarding the operations of differentiation and integration was studied by F. A. Shamoyan, I. S.Kursina, V. A. Bednazh (see [19]).
We state the Bloch-Nevanlinna conjecture in the Privalov spaces: for whatever q > 0, the derivative of a function from the class nq belongs to the class nq.
The paper is organized as follows. In the first part of the article we refute the Bloch-Nevanlinna conjecture in the Privalov spaces for all 0 < q < In the second part of the
article we indicate the class to which the derivative of any function from the Privalov space belongs.
1. The Bloch-Nevanlinna conjecture for the Privalov spaces
The following statement is true. Theorem 1.1. The Bloch-Nevanlinna conjecture fails in the spaces nq, 0 < q <
In other words, the Privalov spaces nq are not invariant under the differentiation operator for all 0 < q < not only for q = 1.
In the sequel, unless otherwise noted, we denote by c, c1,... ,cn(a, ¡3,...) some arbitrary positive constants depending on a, 3,..., whose specific values are immaterial.
Proof of this statement reproduces the arguments from [21], the method goes back to the work of Hayman [8].
Let A be a sufficiently large positive integer, 0 < a < 1, Hbe the class of bounded analytic functions in D. We define a function fx as follows:
fx = £ A-
k=0
It is obvious that fx € H(D), and \fx\ < £ A-k(1-a) = —--, that is fx € H~. Since
k=0 A a - 1
H~ c n, we have fx € nq for all 0 <q <
k(1-a) Xk
z
In the same time we have
hoo
f = £ A^ -1. (1)
k=0
Show that f' € nq. We fix n € N and denote rn = exp(—a/An), rn ^ 1 — 0, n ^ +c. Let un(z) be the n-th term of the series (1):
un(z) = Aanzxn-1.
By Sn(z) we denote the n-th partial sum of the series (1):
n-1
Sn(z) = £ Aakzxk-1, k=0
and by Rn(z) we denote the n-th remainder of the series (1):
z)= > Aakzx -1.
Rn(z)= ]T X
k=n+1
We estimate these sums on the circle \z\ = rn.
г—1 n-1 n-1
\Sn(z)\ ^ Aakrnk-1 = £ Aak exp (— ^ • (Ak — 1)) = exp (Aak exp (—a • A-(n-k>) <
k=0 k=0 k=0
< exp () E = exp (£) • ^^ = ^ exp(—a — 1) • a),
where A(A, a) = exp
a
a + ^r)
1 ) (1 — A-na • e) a") Aa — 1
< 1 for A > A0. 4
Therefore we have \Sn(z)\ < 1 \un(z)\. Now we estimate Rn(z) on the circle \z\ = rn.
■____/ x ■____\am
\Rn(z)\ < E exp(Aak ^^ ■ A
k=n+1 m=1
(z)\ expl An)" exp (aAm)'
Since exp(aAm) ^ exp(maA) for m ^ 1 and sufficient large A,
exp (aAm)
m=1 v '
so we have
+ A<am Aa
^ exp (aAm) ^ eax — Aa ,
m=1 v /
Aam Aa 1
\Rn(z)\ < exp(2a + 1)\un(z)\exp(aAm) < eoX—< 4\un(z)
for A > A1.
As a result, we obtain:
f (z)\ > 1 \u(z)\, \z\ = rn,
for A > max(A0, A1). But
Thus, we have
2
ln \un(z)\ ^ ca ln-, n =1, 2,...
1 rn
/n
(ln+ \f' (rneid )\) q dd > cqa lnq.
-n
1 — rn
this means that f' € nq. Theorem 1.1 is proved.
1
2. On the differentiation in the Privalov spaces
An important place in the theory of analytic functions belongs to the Nevanlinna N-class of analytic functions in D with bounded characteristic T(r, f). It was introduced by A. Ostrovsky and brothers R. Nevanlinna and F. Nevanlinna (see [10]). As noted above, N = ni. Unlike the class N, the area Nevanlinna class is defined as follows (see ibid.):
N={f e H (D,f h+f ^+4
or equivalent to this
1 n
N = ¡f e H(D) : J j ln+ I f (reie) | dddr <
0 -n
The area Nevanlinna classes are a natural generalization of the classes N. As it was established in the works [2,17], these classes are close with respect to the properties of root sets and the factorization of functions. The class N is included in the scale of the Nevanlinna-Djrbashian classes Na (see ibid.):
Na = j f e H(D) : ^ (1 - r)aT(r,f )dr < j , a > -1, and in the scale of Sq-classes of F.A. Shamoyan (see [18]):
sq = j f e H (D) : j0 (1 - r)aTq (r,f )dr < , a > -1, 0 <q<
Similar to the definition of the area Nevanlinna class, for all 0 < q < we introduce the area Privalov class:
1 n
n q = ¡f e H (D) : / i (ln+ I f (reie) | )q dddr < .
0 -n
It is clear that II1 = N. Using Holder's inequality, it is easy to prove that IIq C Sg for q > 1 and n q D Sq for 0 <q< 1.
The main result of the second part of this paper is the following theorem.
Theorem 2.1. If f e nq (0 < q < and function f has no zeros, then f' e IIq.
To prove this statement, we need auxiliary statements. Theorem 2.2 (see [13]). If f e nq, (0 <q< 1), then
ln+ M(r, f) = o((1 - r)-1/q), r ^ 1 - 0, (2)
where M(r, f) = max | f (z) | , and the estimate is exact.
\z\=r
Lemma 2.3 (The Minkowski inequality, see [7], p. 178). Let {fk}+=1 be the sequence of nonnegative functions. For all 0 < p < 1 the following inequality is valid:
1/p { . } 1/p
J {E ffc(x)} dx > ffcp(x)dxj
Lemma 2.4 (see [6], p. 144). Let P(r,0) denote the Poisson kernel in D, i.e.
1- r2
p (r,e) =-1-.
v ' 1 + r2 — 2r cos e For each real number q there exist finite positive constants cq, dq, such that
1 r*
1 r
Cq4q(r) < 2n J Pq(r, e)de < dq4q(r),
where
4q(r) = <
(1 — r)q, (< 2, VT—r ln (1 +
0 + r—r)
1-r)-q 2,
(1 — r)1-q, q> ^.
Proof of Theorem 2.1. Let z = reie, t = Reip, 0 <r <R< 1. Since f € H (D) and function f has no zeros, we have, by the Schwarz formula, that:
1 p2n i |
ln f (z) = — In \f (t)^ -+zdp + iC
2n J0 t — z
where the main branch of the logarithm is chosen. Differentiate (3) by z:
f '(z) = 1 f (z) W0
ln \f (t)\
(t — z)
rd
(3)
f' (z) = M /"ln \f (R
n J 0
2n
whence
\f'(z)\ < ^ /2n ln+ \f (Re*
Reip
(Re* — reie )2 R
dP-,
n J 0
\f'(z)\ <
|f(z)
ln+ \f (Reicp) \ •
R2 — 2Rr cos(p — e) + r2 1
dP,
1 — 2R cos(p — e) + R?
dp.
nR J0
Let us consider 3 cases.
Case 1. We assume that 0 < q < 1. Rewrite the last inequality in the form:
\f'(z)\ < t(ln+ \f(Reip)\)q2 • (ln+ \f(Reip)\)
nR J0 1 — 2R cos(p — e) + R?
Applying Holder's inequality with exponents 1 and —1—, we have:
q 1 — q
1
(4)
dp.
\f'(z)\ <
\f (z) nR
(ln+ \f (Reip)\)q
(ln+ \f (Reip)\)1+q
'0 (1 — 2R cos(p — e) + R?)1/(1 q)
dp
1-q
1
t
ip
q
2
n
0
Since the function f belongs to the class nq, we have by Theorem 2.2:
\f'(z)| <
\f (z
nR (1 - Д)(1-92)/9 (1 - RL)
" f2n ( ( r NNl/(1-q)
J0 (p(R>*> - e))
1-q
where P (r — is the Poisson kernel. We use the Poisson kernel estimate for- > - from
' 1 - q 2
Lemma 2.4:
\f'(z)\ <
\f (z
Dq
nR (1 - R)(1-q2)/q (1 - RL) (1 - Rf
Suppose R -
1 + r
. After elementary transformations we obtain:
\f'(reiti)\ < A, ■\f(re
i9\
(1 - r)(i+q)/q *
We proceed with the logarithm of the last inequality and take into account that ln+ IabI < ln+ |a| + ln+ IbI, a > 0,b> 0:
ln+ If' (reW )I < ln+ If (re-)| + •
Next, raise both sides to the power q, and take into account (a + b)q < aq + bq for all a > 0, b > 0, 0 < q < 1, after integration over d e [-n, n] we have:
I" (ln+ \f'(reie )\)q d0 ^J* (ln+ \f(reie)\)q dß + Bq + (ln
1
q
q +Vln(1 - r)(i+q)/v '
Since f e nq we have:
£ (ln+ If' (re18 )I)q d9 < Bq + 2^ln (1 - r)1(1 + q)/^g -
Integrate over r e [0,1]. In view of the convergence of the integrals on the right-hand side of the inequality, we conclude that f' e IIq. Case 2. Now we suppose that q > 1.
Applying Holder's inequality with exponents q and 1 +--in (4), we obtain
q -1
\f'(z)\ <
\f (z)
nR a - R2)
(ln+ \f (RelLp)\)q dp
1/q
1-1/q
Since the function f belongs to the class nq, we have
\f'(z) \ ^—Щттcq nR (1 - RL) q
Г - ß1+- ^
1-1/q
We use the Poisson kernel estimate for 1 +--> ~ from Lemma 2.4:
q - 1 2
\f'(z)\ < cq
\f (z
1
nR a - R2) a - R)i/q
cq £q
£
С
q
q
2
1
2
2
0
0
1 + r
Suppose that R = —-—, then we have:
\f (z)\ <
q-1
(1 - r)
We proceed with the logarithm of the last inequality and take into account that ln+ \ab\ < ln+ \a\ +ln+ \b\, a > 0, b > 0:
ln+ \f'(z)\ < ln+ \f (z)\ + ln-Cq-
q
(1 - r) q-1
Further, raise both sides to the power q, and take into account (a + b)q < aq + bq for all a > 0, b > 0, 0 < q < 1. After integration in 0 G [—n, n] we obtain:
/n /*n ^f
(ln+ \f'(reie)\)q dd ^ (ln+ \f (reie)\)q dd + ln-q-
-n J—n (1 — r) '
Since f G nq, we see that:
/■k C
(In+ \f'(reie)\)q do < aq + ln-q—.
n (1 — r) —
Integrate over r G [0,1]. In view of the convergence of the integrals on the right-hand side of the inequality, we conclude that f' G nq.
Case 3. We assume q =1. Using the estimate of S. N. Mergelyan for a function of the Nevanlinna class (see [12, c. 84]), we get from (4):
\f '(z)\ < C-\fZ\-^ /2n P (— 0) dp,
U nR(1 — R) (1 — Jo ^-R )
whence by the property of the Poisson integral
\f'(z)\ < C-\M\-^.
nR(1 — R)(1 — Rr)
Further, the proof repeats the argument for Case 2. Theorem 2.1 is completely proved. □
Remark 2.1. Note that W. Hayman indicates the invariance of the class nq, (1 < q < with respect to the integration operator [8].
First author was financially supported by Russian Foundation for Fundamental Research, project number 18-31-00180.
Second author was financially supported by Russian Foundation for Fundamental Research, project number 17-51-15005.
References
[1] D.Campbell, Wickes, The Bloch-Nevanlinna conjecture revisited, Bull. Austral. Math. Soc., 18(1978), 447-453.
[2] M.M.Djrbashian, On the problem of the representation of analytic functions, Soobshch. inst. matem. i mehan. Acad. Nauk Arm. SSR., 2(1948), 3-40 (in Russian).
[3] P.L.Duren, On the Bloch-Nevanlinna conjecture, Colloq. Math., 20(1969), 295-297.
[4] P.L.Duren, Theory of Hp spaces, Pure and Appl. Math., NY: Academic Press., Vov. 38, 1970.
[5] O.Frostman, Sur les produits des Blaschke, Kungl. Fysiografiska Sallskapets i Lund Forhan-dlingar, [Proa. Roy. Physiog. Soa. Lund], 12(1942), no. 15, 169-182.
[6] V.I.Gavrilov, A.V.Subbotin, D.A.Efimov, Boundary properties of analytic functions (further contribution), Moscow, Publishing House of the Moscow University, 2012 (in Russian).
[7] G.Hardy, Inequalities, G.Hardy, J.Littlewood, G.Polia, Translate from English: S. B. Stechkin (eds.); V.I.Levin (transl.), Moscow, GITTL, 1948.
[8] W.K.Hayman, On the characteristics of functions meromorphic in the unit disk and of their integrals, Acta. math., 112(1964), no. 3-4, 181-214.
[9] R.Nevanlinna, Le theoreme de Picard-Borel et la theorie des fonctions meromorphes, Paris, Gauthiers-Villars, 1929.
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11] I.I.Privalov, Boundary properties of single-valued analytic functions, Moscow, Izd. Moscow State University, 1941 (in Russian).
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[22] S.V.Shvedenko, Hardy classes and the spaces of analytic functions associated with them in the unit disc, polydisc, and ball, Itogi Nauki i Tekhniki. Ser. Mat. Anal., 23(1985), 3-124 (in Russian).
О дифференцировании в классах И. И. Привалова
Евгения Г. Родикова
Брянский государственный университет Брянск, Российская Федерация
Файзо А. Шамоян
Саратовский государственный университет Саратов, Российская Федерация
Аннотация. В статье исследуется инвариантность классов И. И. Привалова относительно оператора дифференцирования.
Ключевые слова: класс Привалова, гипотеза Блоха-Неванлинны, оператор дифференцирования.
