Description of some weighted exponential classes of subharmonic functions

Let C be the complex plane. SH _ao ( C ) ( о , a are positive numbers) be the class of subharmonic

+^

functions u in C such that

j T ( r , u ) e ^{-o r} dr <+^ , where T ( r , u ) is the Nevanlinna’s characteristic of

1 П                          + subharmonic function u, that is T (r, u ) = — I u+(r cosф, r sin^) dф, u += max (u ,0) (see [1-4]).

-П

In complex and real analysis, potential theory and mathematical physics the value of subharmonic functions is very significant (see [1–3, 5]). In the works R. Nevanlinna and W. Hayman (e.g., see [1]) the obtained definition of a class of subharmonic in the plane of the functions, the characteristics of which have exponential growth at infinity. The question of whether faithful same parametric representation for the weight classes, allowing for stronger growth at infinity, say exponential growth, occurs in the theory of entire and meromorphic functions (see [6]).

This paper studied the representing measures of the functions of class SHao (C), as well as the necessary and sufficient condition for such measures.

Statement of the main result

Let о , a >  0 , z, Z e C , Z * 0, P (| Z ) = max 0\£ \ ^a ,1 ] , where [ a ] is the integer part of a real number a .

is a factor of the modified product by K. Weierstrass (see

V

[7, 8]).

Математика

Theorem 1. Let u e SH _ao ( C ) , Ц is representing measure of a function u. Then ц satisfies the condition:

+^

J

n ( t ) e

—

t ^a

m“

— dt <  +^ ,

where n ( t ) = ц ( D_t ) , | z | <  t.

The opposite is also true: let ц is some non-negative Borel measure in C, satisfying the condition (1) , then it is possible to build explicitly subharmonic function of class SH _ao '( C ) , ^O : О > — + o e “ , for which ц will be a representing measure.

ea

Proof of auxiliary assertions

We require some auxiliary assertions for the proof of the theorem.

Lemma 1. Let y( R) is a non-negative monotonically increasing function, for which where a> 0. Then

+^

a

J ^ ( x ) e ^o dx <+^ ,

lim

R ^^

^ ( R^e —^ R a R^

= 0 .

+^

—I

°x' dx = 0.

Proof. The convergence of the integral (2) implies that the lim [ ^ ( x ) e

R ^+^ J

R

+^                                         +^

Let ^(R)= J y(x)e—ox dx. It is clear that ^(R)> ^(R) J e—ox dx . We find the asymptotics of the RR last integral by applying the L'Hospital's rule.

Therefore, lim

R ^^

+^

J e-° x ^a dx lim ^R r ^^ e ~oR^a

^ ( R ) e ^{—o R a}

^ ''

= lim

R ^^

_e —o ^{R a}

e ^a R ^a o R ^a 1 a e

R a — 1

—

= 0 . That is lim ^ ( R ) e ^—o R ^a R ^{a — 1} = 0.

R -^^

oa

.

The lemma is proved.

Lemma 2. Let p ( x ) = e

—(

a

, x e R ₊ , R ₊ = { x e R : x >  0 } , £ _x

~a . Then p ( x + £ _x ) = e ^ ^x ) • e ^{e ( x} ) ,

where f( x ) = o [ — | when x ^+^.

V x 7

Proof. It follows easily from the following simple arguments. It is clear that f.\“       ao 1

V x 7           x V x 7

Then, e ^ ^{( x +£ x} ) = e ^ ^{( x} ) • e ^{e ( x} ), where в ( x ) = o V— 7

when x ^ +^ .

The lemma is proved.

Lemma 3. Let u is an arbitrary subharmonic function in C, while it admits a representation in the form:

u ( z ) = J ln| A p ( z, K ) d ^ ( K ) + h ( z ) ,

C

+^

z,Ke C, ^(K) is an arbitrary non-negative Borel measure in C, such that J n (r )= ^( Dr), c > 0, a > 0, h (z) is harmonic function in C, for which

+^       П

1        — П

n ( t ) e

—I

t ^a

c t ^a

—dt <+^ ,

Then u e SH _a ^- ( C ) when VC : c'> — + c e ^a .

Proof.

Let V p ( ^z ) = J ^ln| A p ( ^{z, K} ) d^ ^K ) . Then ^u ( ^z ) = ^h ( ^z ) ⁺ V p ( ^z ) .

C

It is obvious that h ( z ) belongs to the class under consideration. We will show that V_p ( z ) also included in the class SH _ac ‘ ( C ) for V С = C ( a ) >  c .

We apply the estimate (see [1], p. 94):

К A p ( z, K )|< ;!

, K * 0, z, K e C .

Let's get that: u ( z ) <  J K

C

To continue the evaluation of the function, we can partition the complex plane into sets:

p ( K I)

d ^ ( f ) .

+^

a k = { z e C :2 ^k < | z | <  2 ^{k + 1} } , A o = { z e C : | z | <  1 } . Then u ( z ) < ^ J ^ k =0 A _k ^K

Each of the rings A _k is divided into small rings and we use Lemma 2.

Let 8 k j = { z e C :2 ^k + j 2~ ^{a k} < | z | <  2 ^k + ( j + 1 ) 2 ^{— k a} } , 0 <  j <  |^ 2 ^{k ( a + 1 ) + 1} ^ , where [ a ] is the integer

Na part of the number a . Then Ak c Q 8k j , where Na = I 2k( ) I.

j =0 ’                       ^{L J}

Therefore, ^z

A k

N

a

d*« ) < У J _K

a

j = ⁰ 8 k,j

By Lemma 2, we obtain

z

8 k , j

d ^ ( K ) < K *

p (I K * )

d ^ ( ⁸ k , j ) < 4*

K

It is obvious that

Therefore ^z

a k

p (I K 1 )

n ( 2 ^k + ( j + 1 ) 2 ^{a k} ) , where K is a point of 8 k j .

z

K

p (I K i)

a

d^(K)< Cу J

N a ⁸ , j ^{+ 1} z p ⁽ t )

^j =0 V ,

t

2 ^k + ( j + 2 ) 2 ^{— a k}

J

2 ^k + ( j + 1 ) 2 ^{— a k}

2 ^{k + 1}

n ( t ) dt <  C J ^z k t

k

z t

n ( t ) dt , k = 0,1,...

p ⁽ t )

n ( t ) dt .

Summing over k , we get:

Математика

+^

u

(-")

1t

p (t)

n (t) dt.

Let us estimate the last integral. According to (5) we have that

+^

j e -

п

+^

'°^r j u⁺(re^1<p) d^dr< C j e

¹ °^r j I I     n(t)dtdr.

-п

Having established the convergence of the last integral, we prove the Lemma. To do this, imagine the inner integral as a sum:

V/ V⁽t)

erCAp⁽t)

+~ / \p(t)

I r j I — I    n (t) dt = 11 +12 .

er

Let us prove the boundedness of the second integral by a constant, independently of r . Indeed, \ p (t) r I                        r

— I = exp p (t) In —.

r

However, when — < e t

rr

In — < -1. Then exp p (t) In — < exp (-p (t)).

p(t) = °'t^a J, so p(t)>G'^a-1. Therefore, exp(-p(t))< Ce °t .

-G„^at“/2а ^e< ~ , ° > °, ⁿ (t) < ^e t ,

+^ / pplt\           +“                 +“ (A -ot“ r I r I                   f   ( a -at “ ,   лГ” (t) e dt

We estimate the integral.

er

>1=jf

Consider the estimate for the integral I₂ we get that

+~         п                    +~        er/ pip (t)

j e^{-a r} j u⁺(re¹^) dф< j e ^°^r jf“| ” (t) dtdr .

1          -п                      1          1 ^t¹

To calculate the largest value of the function expp(t)(Inr - Int) on an interval [1;er] we put x = ln r и /(x, t ) = p (t)(x - ln t) .

x-ln te« = 0.

I 11      -1

That is lnr = ln te^a , t = re ^a

t^a = r^ae^-1

This point is the maximum point, therefore

Y( x, t ) = Y x, re “

= p re ^a

r \ p (t) _r

Because max

ln r - ln re “

e

then

er

j n (t) If

1 ^кt

p(t)       er

r^a

e

er

r^a

r^aer

dt = j n(t)e^aedt =e^ae j n (t)dt 11

+^

j e—

a ‘ r^a

r^aer

+^

e

?^ea j n (t)dtdr = j e

—

Ir^{a e}r^r Г A

j n (t)did

= C1jn (t)dt

—I

• e

I r^a

+^

+^

+- J a 1

^—I e

I г^а

—-----n (er) dr.

_ | _r^a—1

er

In view of Lemma 1, we write that j n (t)dt< Ce^a(er)

\a       1

¹ r^a—1.

Using the last estimate, we clarify the values a , that ensure the convergence of the integral:

⁺F —I

J e

I raer                            a          ( f n(t)dtdr : ara — a(er)“-->0, raI a' — aea —

ас

ea

The lemma is proved.

Let us prove the auxiliary theorem 2.

Theorem 2. Let u is an arbitrary subharmonic function in the class SH_aa (C), Ц^) is an arbitrary measure of the function u, in this case ц( D_t) = n (t), 0 < t< +^. Then

+^ j 1

n (t) е

—(

t^a

ata

---dt < +^.

Proof. We integrate the integral by parts. Using the equality of Jensen for subharmonic functions (see [1]), it is enough to estimate the integral:

+^ Г t

I = j j

1 к 1

ⁿ (^r)^dr"

e^—atdt.

r

We integrate I by parts:

a

_ а t at t

j

ⁿ (^r)^dr"

r

+^

+^

+j

e

—at

^{1 — a} t^a

j к 1

ⁿ (^r)^dr"

r

n (t)) ,

+--dt t^a у

According to Lemma 1, we have:

+^

—(

at

r

+^

t^a_У

—(1^—a) J a 1

e^a t^a—1

J--------+ n (t) dt

By the hypothesis of the theorem I< C.

It is clear that in 0 < a< 1 the theorem is proved,

+” e—at^a

1t

Suppose a > 1. Then by Lemma 1:

since from the condition (5) implies that

+^

je

—

t₁r₁

„.a ea ta

dt e "atata—1

r 7 dt

1t

+- e—at^a

Because a>1, then J-----n(t)dt<+^ .

1t

The theorem is proved.

Математика

The proof of the main result

The proof of the direct assertion of theorem 1 follows from theorem 2.

The validity of the converse follows from Lemma 3.

That proves theorem 1 completely.

From Theorem 1 directly follows

Theorem 3. Let a> 0, SHaro = U SHao. Then class SHa„ coincides with the class of subhar-,      о>0        ,                                , monic functions in C, representable in the form u (z ) = J ln| Ap (z ,Z)\ d^(f) + h (z), where Ц is non-C negative measure in C, such that n(t) = ^( Dt), Dt ={ z: |z| < t}, and also satisfies the condition

+^ ∫ 1

n(t)e

t^a

„ a

Ot

dt < +^ for some

a> 0, h an arbitrary harmonic function in C, for which there exists

о > 0: J e~°r^a J |h(re¹^)d^dr< +~. 1        -П