Variations on the theme of Marcinkiewicz’ inequality

By

V. MATSAEV,* I. OSTROVSKII ? AND M. SODIN*

1

I n t r o d u c t i o n

Let h be a smooth function on R with a compact support, and let

1 /rt h(t)

dt

g(x) = p . v . ~ t - x

be its Hilbert transform. Set f = g + ih and introduce the function

ul(z) = frt H(zy(t)) dr,

H(z)

= log I1 - zl + R e ( z ) ,

called the

logarithmic determinant

of genus one. It is subharmonic in C; and its Riesz measure is d/~y(~) = duf((-1), where

duf

is the distribution measure o f f

uf (E) = meas ({t : ](t) E E } ) , E is a Borel subset of 12, and meas(. ) stands for the Lebesgue measure on R. Let

#y(r) = t~y({Izl < r})

= meas

({111 >__

r - l } ) be the usual counting function of d/~y, and let

hi(r) = py ({Iz -

ir/21 <_

r/2}) + py ({I z +

ir/21 <_

r/2})

=/~y ({IXm(z-l)[ > r - a } ) = meas

({Ibm > r-X})

be its

Levin-Tsuji counting function;

see [11], [18], and [7, Chapter 1]. Then the classical estimates of the Hilbert transform can be easily rewritten as upper bounds of #y(r) by ny(r).

* Supported in part by the Israel Science Foundation of the Israel Academy of Sciences and Humanities under Grants Nos. 93197- l and 37/00-1.

tSupported in part by the INTAS Project No. 96-0858. 289 JOURNALD'ANALYSEMATH~vlATIQUEn Vol. 86 (2002)

290 V. MATSAEV, I. OSTROVSKII AND M. SODIN

(1.1)

For example, Marcinkiewicz' inequality (see [9, Chapter V])

{ 1 f0x 1 f x ~ 1 7 6 0 < A < o o ,

m r ( A )

<_ C "~

smh(S)ds + -~

where

and

ms(A) = meas ({I.fl > ,~}) =

vz({Iwl

>

~})

= # f ( ) ~ - l )

mh(A) = m e a s ({Ihl > A}) = vz({llmw I > A}) = nZ(A-x), reads

(1.2)

From this, one readily obtains

(1.3) #l(r) _< Cr fo ~176 nl(t)t2 dt,

{

", (" ,, + / ? ~

}

#i(r)<_C r ~ - - ~ - d t , O < r < c ~ .

0 < r < o o ,

which is equivalent to Kolmogorov's weak Ll-type inequality Am! (A) < CIIhI[L 1 , 0 < A < oo, and

fo ~176 #l(t) ,. f0 ~176 n/(t)dt 1 < p < 2

(1.4) -~--~-T-ar < C(p) tp+l ' '

which is equivalent to M. Riesz' inequality

(1.5) IIfIIL, ___ C(p)llhllL. 9

The classical proof of inequality (1.1) is based on the interpolation technique which later served as one of the cornerstones of the abstract theory of interpolation of operators in Banach spaces [2].

A natural question arises: what is special about the subharmonicfunction u I (z)

which makes inequality (1.2) valid? The answer is proposed in [14]: a key fact

ensuring this, is the positivity condition

(1.6) ul(z ) > 0 , z E C,

which can be easily checked with the help of Green's formula (see [3, Lemma 5]) or by applying the Cauchy residue theorem (see [ 15]).

This leads to a heuristic principle which suggests that

9 results about the distribution o f the Hilbert transform can be deduced from inequality (1.6) by using methods of subharmonic function theory.

As will be shown, inequality (1.6) is even too strong; and in many cases it suffices to assume that

ul(x ) > 0

on R or to control the negative part u~- = max(0,

-ul)

on g .

The principle shows a path to new results on the Hilbert transform. In [ 15], its application leads to a complete description of the distribution of the Hilbert trans- form of L 1-functions and measures of finite variation. At the same time, putting known estimates of the Hilbert transform into this setting, we arrive at new inter- esting questions about the argument-distribution of the Riesz measure in certain classes of subharmonic functions. For example, the proofs of the inequalities of Kolmogorov and M. Riesz found in [ 14] give new bounds for zeros of polynomials. Positivity condition (1.6) links our work with the theory of uniform algebras and Jensen measures (see [6]) and some recent works related to sharp inequalities for conjugate functions [3], [4], [5].

In this work, we put forward a new approach to the Marcinkiewicz inequality (1.1) (or (1.2)). The methods applied in [14], [16] are too rigid for this. Here we use a different technique.

Here and later on, we use the following notations:

r < r means that there is a positive numerical constant C such that, for each s > 0, r _< Cr

r ~<a r means the same as above but C may depend on a parameter a;

H(z) =

log I1 - zl + Re(z) is the canonical kernel of genus one;

C+ are the upper and lower half-planes.

A c k n o w l e d g m e n t s

The authors thank Marts Ess6n and the referee for helpful suggestions that improved the presentation.

2

M a i n results

Define a subharmonic canonical integral of genus one:

(2.1)

u(z) = / c H(z/~) d#(Q,

where dp is a non-negative locally finite measure on C such that

Io

292 V. MATSAEV, 1. OSTROVSKII AND M. SODIN

Let

~(r) = #({Izl <__ r}) be the usual counting function of the measure d#, and let

n(r) = # ({llmCUz)l > I/r})

=/~ ({Iz - ir/21 <__ r/2}) +/~ ({Iz + ir/21 _< r/2}) be its Levin-Tsuji counting function [11], [18] (see also [7, Chapter 1]).

Let M(r, u) = maXlzl<r u(z).

Then by the Jensen inequality, #(r) <

M(er, u).

In the opposite direction, a standard estimate of the kernel

H(z)

< 1 + Izl' z e c , yields Borel's estimate

(2.3) In particular, (2.4)

fo r i~(t )

f oo #(t)

M(r, u) < r

y

dt + r z

- ~ - dr.

~o(r),

r -~ O,

M(r,u) = ( ~

r -r oo.

T h e o r e m

1. Let u(z) be a canonical integral (2.1) of genus one. Then

(2.5)

M(r,u) < r for ~ d t

+r2 f ~ 1 7 6

~ d t

+ r2 f ~ 1 7 6

m(t'u)

_t2

_{dt ,}

where

1 f2,~

m ( r , u ) = ~ _ ~ j ~

u_(reiOlsinO)l ) dO

r sin 2 O

is the Tsuji proximity function of u.

If the function u is non-negative in C, then the proximity function re(r, u) vanishes; and, applying Jensen's inequality, we arrive at

C o r o l l a r y

1. Let u be a canonical integral of genus one which is non-negative

in C. Then

fo r n(t) f r ~ n(t) #(r)

< r

~

dt + r 2

- ~ - dr.

As we explained in the introduction, this result immediately yields the classical Marcinkiewicz inequality (1.1). In this case, one can assume apriori that the function f is bounded, so that most of the technicalities needed for the proof

of Theorem 1 (see Lemmas 2 and 4 below) are redundant; and our proof of Marcinkiewicz' inequality, being conceptually new, is comparable in length to the classical one.

There is a curious reformulation of Corollary 1. Let A,~ be a measurable space endowed with a locally finite non-negative measure din, let f be a complex-valued measurable function on .A4, and let mI()~ ) =

m({Ifl ___ A})

be the distribution function of f. If

(2.6) f.~ min(l/(t)l, If(t)l 2) din(t) <

then we define the logarithmic determinant of f of genus one

uj(z) = f M H(zy(t)) din(t),

which is subharmonic in C and, moreover, is represented by a canonical integral of genus one. Applying Corollary 1, we obtain

C o r o l l a r y 2. I f f satisfies condition (2.6), and the logarithmic determinant u I is non-negative in 12, then

I z / :

1 x 1 mzmi(s)ds

I

, 0 < ) ~ < ~ .

mI()~) ~ -'~

smlmI(S)ds + -~

In particular, IlfllL,(m) <p I[Im fllLp(m), 1 < p < 2, and m i ( ~ ) < IlImfllz,(,.) 0 < )~ < oc.

Corollary 1 can also be applied to Jensen measures in C. A compactly supported finite measure a in C is called a Jensen measure (with respect to the origin) if, for an arbitrary subharmonic function h in C,

(2.7) h(0) < / hda.

A simple argument shows that (2.7) then holds true for subharmonic functions in a domain G such that 0 E G and supp(a) c G. For a harmonic function, the equality sign must occur in (2.7). Therefore,

294 V. MATSAEV, I. OSTROVSKII AND M. SODIN that is, a is a probability measure; and

f (kda(~)

= O, (2.8)

Define the potential (2.9)

Then, by (2.7) and (2.8),

k = 1, 2, ....

V~(z) = f

log I1 - zfflda(ff) 9

(2.10) 0 < V~(z) <_ log+(clzl), z ~ c ,

for some c > 0. The opposite is also true: if, for s o m e e > 0, the subharmonic function V satisfies (2.10), then it is a potential o f a Jensen measure [8, w

B y condition (2.8), every potential V~ o f a Jensen measure can be represented by a canonical integral o f genus one:

V,,(z) = / c H(zl~)d#(~),

dp(~) = d a ( l / ~ ) . Thus, T h e o r e m 1 is applicable to the potential V,,; and w e obtain

C o r o l l a r y

3. Let a be a Jensen measure in C,

a(A) = Or([Z[ > A), erx(A) = a(lImz] _> A).

Then

I~OX

1~

~176

or(h) < ~

Sal(s)ds + ~

Crl(S)ds.

The class o f Jensen measures is invariant with respect to holomorphic mappings. More precisely, let G be a domain which contains the origin and supp(a), and let F be an analytic function in G, F(0) = 0. Then the push forward F , a is defined b y

f dr.a

= f

Fa ,

where ~ is an arbitrary continuous function in C, and ~ o F is the composition o f $ and F . B y the monotone convergence theorem, this equation also holds for semicontinuous functions. The measure

F.a

automatically has compact support since F is b o u n d e d on supp(a). I f h is subharmonic in C, then h o F is subharrnonic in G and

] hdF.,, = ] ho

>_, h(F(O))

h(O) .

C o r o l l a r y 4. Let a be a Jensen measure in C, and let f = g + ih be an analytic

function in a domain G which contains the origin and supp(a), and f(O) = O. Let

m f , . ( A ) = a ( l f l ~ A), mh,.(A) = a ( l h l ~ A).

Then

(2.11) mLa()~ ) < -~

1/0

smh,a(s)ds + -~ mh,a(s)ds.

Corollary 4 probably holds under the weaker (and more natural) assumption g(0) = 0 rather than f(0) = 0. In that case, using Theorem 2 (see below) one can get an estimate which is slightly weaker than (2.11).

In the next result, we do not assume that u(z) is non-negative in 12; instead, we introduce the quantity

5(r) = . ( r ) + [,,-(r) + u - ( - r ) ] , which we keep under control.

We assume that the integrals

f0 5(t) / o o 6(t)(1 + logt)dt

(2.12)

--~-dt

and - 7

are convergent and define

(2.13) 5*(r) = r - 7 dt + r 2 - 7 1 + log dr.

The function 5" (r) does not decrease, r-26 * (r) does not increase, and therefore (2.14) 6*(r) <6"(2r) <46"(r), O < r < o o .

T h e o r e m 2. Let u(z) be an arbitrary subharmonic function in C represented

by a canonical integral o f genus one, and let the integrals (2.12) converge. Then

2

Observe, that the RI-ISs of (2.5) and (2.15) do not depend on the bound for the integral (2.2). Estimate (2.15) is slightly weaker than (2.5); however, it suffices for deriving estimates of M. Riesz and Kolmogorov, as well as the weak (p, ~ ) - t y p e estimate (see Corollary 6 below).

296 V. MATSAEV, I. OSTROVSKII AND M. SODIN

Fix an arbitrary e > O. Then by the Cauchy inequality

[ ~ o o ~ ]2 [~oo r + logl+e ~) ~, (t)

dt

dt =

t3/2

tl/2~/1 + log 1+~ t

~, ~~176

5*(t) (l +logl+~ t )

r

~ - d s +

- 7 1 +

r ds.

Thus we get C o r o l l a r y 5.

For each e > O,

(2.16)

M ( r , u ) < ~ r / r ~ d t + r 2 / ~ 1 7 6

(l+log3+'~)dt

We do not know whether the term log 3+~ is really needed on the RHS of (2.16). Apparently, our method does not allow us to omit it. According to a personal communication by A. Ph. Grishin, the exponent 3 + e can be improved. However, his technique also does not allow one to eliminate the logarithmic factor entirely.

Rewriting (2.16) in the form

f01~( r8 ) f l ~176 ~(r8)

log3+'s) ds

M(r, u) <~c

ds +

(1 +

83

we immediately obtain

C o r o l l a r y

6. The following inequalities hold for canonical integrals of genus

one:

M. Riesz-type estimate:

fo

(2.17)

#(r) dr s

rp+l rp+l

weak (p, cr )-type estimate:

(2.18) sup ~<p sup

re(0,oo) r• re(0,oo)

and Kolmogorov-type estimate:

(2.19) sup #(r) < sup re(0,~) r re(0,oo)

6(r)

~ d r ~ p

r--~dr,

l < p < 2 ;

M(r,u) ~

dr.

r

M(r,u)

6(r)

rP ~<p sup l < p < 2 ; rE(0,oo) rp '

Estimates weaker than (2.17)and (2.19)were obtained in [141 and [16] under additional restrictions, which now appear to be redundant. Estimate (2.18) is apparently new.

If we assume that d# is supported by R, that is, u(z) is harmonic in C+, then our technique gives a better result:

Let u(z) be a canonical integral of genus one of a measure d# supported by I t Then, for 0 < r < oo,

f[

,,-(t) +,,-(-t) dr.

(2.20) M(r, u) < r t= dt + t3

Note that one cannot replace u - (z) by u + (x) on the RHS of our estimates. For example, the function

O,

is subharmonic in C, harmonic in the upper and lower half-planes C+, represented by the canonical integral of genus one of the measure d#(x) =

cplzlp-ldx (cp > 0),

and non-positive on R.

There is a corollary to Theorem 2 which is parallel to Corollary 2. Let 24 be a measurable space endowed with a locally finite non-negative measure din, and let f : A4 ---r R n+x, n > 1, be a measurable function such that

(2.21) min(ll/(t)ll,

IlY(t)ll 2) din(t) <

where

II. II

stands for the (n + 1)-dimensional Euclidean norm. We start to enumerate the coordinates in R n+l with j = 0, and denote by e0 the vector in R "+1 whose zeroth coordinate is one and whose other coordinates vanish. Let

f~(t) be the j-th coordinate function of f(t), and ](t) = {Ej"=l .f~(t)} 1/2" We define the logarithmic determinant

vi(x ) = f [log

lie0

- zY(t)ll +

xy0(t)]

din(t)

,/At

= / A t [log~/1-2Xfo(t)+x211fl]2+xf0(t)] din(t), x fi R ,

where the integral converges due to assumption (2.21). Then, if the function

W(z) is non-negative on 1~ we may estimate its distribution function mr(A) =

m({lllll _> ~}) by the distribution function m] = m ( { ] > A}) of ]. For this, observe that

298 V. MATSAEV, I. O S T R O V S K I I A N D M. S O D I N

where

f c

is a "complex-valued surrogate" of f:

f c = fo + i].

That is, vf has a subharmonic continuation from t t to C by a canonical integral of genus one

(z) = f ~ H(zfc(t))dm(t),

z E C.

ufc

Next, observe that mr(A) =

m({]~ + ]2 >

A2}) =

mlc (A), and m](A) = mimic (A)

for 0 < A < oo. Hence, Theorem 2 is applicable in this situation. For simplicity, we restrict ourselves to the case when v! is non-negative on the real axis.

Corollary 7. Let f satisfy condition (2.21), and let the logarithmic determinant

v! be non-negative on the real axis. Then, for 0 < A < cx~ and ~ > O,

l foX (

~ )

l f x ~ 1 7 6

m/(A) <, ~-~ s 1 + log 3+'

ml(s) ds + -~

ml(s) ds.

In particular,

IlfllL,(~) <p II]IIL,(~),

I < p <

2,

and

mr(A) <

II]IILI(,~)

"~ A

This corollary may be of some interest in view o f the results of Aleksandrov and Kargaev [1].

Our third result pertains to the more general class of subharmonic functions rep- resented by a generalized canonical integral of genus one. It gives a Kolmogorov- type estimate which can be applied to a wider class of subharmonic functions than those represented by canonical integrals of genus one.

Theorem 3. Let d# be a non-negative locaUy finite measure on C such that

I 'l 2

and suppose the principal value

is finite. Let

Then (2.22) sup O < r < ~ lim ~ d#(() e-~O <_1r ~"

f

uCz) =

lira

!

H(z/r dD(().

e--*O JKl>e

M(r,u) <

dt + limsup p(r)

r "~ r--~O r

Obviously, this result is of interest only when the RHS of (2.22) is finite. Thus, the assumption on the principal value integral is an assumption on its real part (if (2.22) is finite, the imaginary part of this integral is absolutely convergent).

It is easy to see that if the integral (2.2) converges at the origin, then the upper limit on the RHS of (2.22) vanishes; in this case, (2.22) coincides with (2.19).

In fact, our proof yields a stronger result

f ~ u + ( t ) M ( r , u ) f0~ ,(t) /z(r) (2.23) ~ t2 dt + limr~sup - - r < - ~ - dt + limr~0sup ~ , r

which gives control over the positive harmonic majorants of u in the upper and lower half-planes. Applying a known technique for functions of Cartwright class [12], [10], one can extract from (2.23) information about the asymptotic regularity of u and/~ at infinity and near the origin.

Notice that one can also reformulate Theorem 3 in the spirit of Corollaries 2 and 7. We leave this to the reader.

3

Auxiliary l e m m a s

We require several known facts about harmonic and subharmonic functions.

L e m m a 1. Let v be a subharmonic function in the angle

S = { z : O < a r g z < a } , O < a < 2 ~ r ;

let

(3.1) limsup v+(z) < r r E OS;

z-~, zES and let

(3.2) v+(re iO) sin 0 dO = o(r'~/~), r ~ ~ . Then, f o r z = re i~ E S,

t~/a+l dr. I f the majorant ~(t) does not decrease, then the factor sin(IrO/a) on the LHS o f (3.3) can be omitted.

P r o o f . The general case is easily reduced to the special case when S = C+, so that, without loss of generality, we assume that a = ~r. First we show that v(z) is majorized by the Poisson integral of

~(Itl),

and then we estimate this integral.

300 V. MATSAEV, I. OSTROVSKI1 AND M. SODIN

Denote by hR(z) a harmonic function in the semi-disk {Iraz > 0, Izl < R}

with b o u n d a r y values hR(t) = ~(Itl), - R < t < R, and hn(Re i~ = v+(Rei~

0 < 0 < rr. Applying the Poisson-Nevanlinna representation in this semi-disk (see [7, Chapter 1, Theorem 2.3], [12, Section 24.3]), we obtain for z = re i~ r < R,

;

/0"

v(z) < hn(z) = r (z, t) dt+ v + (Rei*)Kz(z, Re i•) de,

R (3.4) where (3.5) K l ( Z , t ) = rsin_____O0 { 1 R 2 } ~r Iz tl 2 IR 2 - z t l 2 ' 1 4Rr(R 2 - r 2) sin r sin 0

(3.6) K2(z'Reir = 27r (R 2 + r 2 - 2Rr cos(r - 0))(R 2 + r 2 - 2Rr cos(r + 0))" B y condition (3.2), the second integral on the R H S o f (3.4) tends to 0 as R ---r oo. Therefore, letting R ~ oo in (3.4), we obtain

v(z) <_ rsin~ f_ "~

,I'(Itl) (3.7)

oo Iz=YI 2

dr.

Making use o f straightforward estimates o f the Poisson kernel, we get

v(z) <

- - - I r r sin0

2

fo ~

O(t)dt + 4--r r f2r ~O(t) dt, o o

and estimate (3.3) follows.

I f the majorant #(t) does not decrease, then we m o d i f y the previous argument:

v z,

<, c

- r d t + r + 4r dt < -

'f:

r d t + + r

F

dt

- r j0 r - - ~ r -7Y- '

completing the proof. []

The next lemma asserts that, under certain conditions, the Carleman integral formula [12, Lecture 24], [7, Chapter 1] holds without remainder.

L e m m a 2. Let v(z) be a subharmonicfunction on D n = {z E C+ : Izl < R}

(i.e., v(z) is subharmonic on some open set which contains Da) which satisfies

/

~ Iv(re '0 sin 19 dO )1 = o ( r ) , r ~ 0 ,

(3.8)

a n d

where

u+(r) =/~{Iz -

irl < r/2}. Then

(3.10)

~ f_11v(t) ( 1,

~ ) dt + - ~ fo v(Reir

r

= fo~ (1~2 R2) I m ' d # ( 0 '

(3.13) and

where the first integral on the LHS is absolutely convergent.

P r o o f . We start with the Nevanlinna representation

/?

v + (t)K1

(z,

t) dt +

/0

v + (Rei*)K2

(z,

Re i*) de

1l

/?

I"

= v(z) +

v-(t)Kl(Z,t)dt +

v-(ReiC~)K2(z, Rei~)dr

11

+ [_ K3(z, ~) dlt(~),

J D R

where the kernels K1 and Kz were defined by (3.5) and (3.6), and (3.11)

K3(z,~)

= l o g l Z - ~ R2 - z ~

z - ~ R 2

Multiply both the left and right hand sides of the Nevanlinna representation by r -1 sin0, integrate it with respect to 19 from 0 to 7r and change the order of integration in all terms. We shall use the formulas

/0

[ (

_1

Kl(re i~ s i n S d 0 = 1 1 1 _ 1 , (3.12) r ~ min ~ , ~-ff 1 fo ~ 1

r

K2(rei~162

= ~ sin~b,

/0

[ (11)1]

(3.14) 1

K3(rei~

= 7rImff min

ICI2, ~

-

Observe that the RHS of (3.12)-(3.14) are non-decreasing functions of r -1. Therefore, making the limit transition r --+ 0, and using the monotone convergence theorem and condition (3.9) of the lemma, we get

lf]nv+(t)

(1 ~ R 2

1)

dt+-~

lfo'~V+(ReiC~)sinCdr

1 R

1

1

= ~ f_11v-(t) (t~

R 2 ) d t + - ~ f o v-(ReiC~)sinedr

1

1

302 v. MATSAEV, I. OSTROVSKII AND M. SODIN

The first and third integrals on the RHS are finite due to condition (3.9). This completes the proof. []

R e m a r k . Condition (3.8) holds for the canonical integrals of genus one defined

in (2.1).

Indeed, if

u(z)

is such an integral, then by (2.4),

fo ~

u+ (re i~ dO

= o(r),

r

O.

Since u(0) = 0, this yields

lu(rei~ dO = 2 u+(re ie) dO - u(re iO) do

~0 21r

< 2 u+(rei~ = o(r), r - ~ O.

The third lemma was proved in [16] (cf. [12, Lecture 26]). Its proof uses the Nevanlinna representation for the semi-disk.

L e m m a 3. Let v(z) be a function which is harmonic in C+, subharmonic in C+ (i.e., on an open set containing C+), and satisfies conditions (3.8) and (3.9) of

Lemma 2. Then, f o r z = re i~ E C+,

1 v - ( 2 r e i~) sin qa d~ + ~ 2r

(3.15) v(rei~ < r t2 dr.

The next lemma is a version o f the Levin integral formula ([11], [7, Chapter 1]) without a remainder.

L e m m a 4. Let v be a subharmonic function in C such that v(z) and v(5)

satisfy conditions (3.8) and (3.9) o f Lemma 2. Then

(3.16)

v(Re~~ I

sinOI)~

= dt,

where n(t) is the Levin-Tsuji counting function, and the integral on the L H S is

absolutely convergent.

P r o o f . It suffices to prove that

- -

Then (3.16) follows by adding to (3.17) a similar formula for the integral from rr to 21r.

First, we prove that the integral on the LHS of relation (3.16) is absolutely convergent. Making use of the notation introduced in (3.5), (3.6) and (3.11), we note that the Nevanlinna formula implies that

f?

f:

fo

Iv(z)l ___ Iv(t)lK1 (z, t) dt +

Iv(Re~*)lK2(z,

Re i*) de + Kz(z, r dIJ(ff).

R a

Set z = l~e ia sin0, multiply the formula by (Rsin 2 0) -x, integrate it with respect to r from 0 to r , and change the order of integration in all terms. We recall the following relations:

and

fo '~ _{K1 (Re i~ sin 0, t) R sin 20 - t 2} dO 1 _{R 2 '} 1

fo '~ K2 (Re i~ sin 0, Re ir dO 2 sin r R sin 28 - R

fo"KZ(Re,OsinO,r , dO

_{R sin 2 6 -}

2.[min(llm~l

_'

R)_~r

_"

Using these relations, we verify that

f0 '~ [ v(Rei~ Rsin20 -< d~ f [ v(t)l ( 1 1 ) t 2 ~2 dt

R

+ ~

Iv(Rei~)lsin~dr

Im

du(r

R

The first integral on the RHS is finite by Lcmma 2, and the third is finite because of condition (3.9). That is, the integral on the LHS of (3.16) is absolutely convergent.

Now, we write the Nevanlinna formula in the form

f]

f[

fo

v(z) = v(t)K1 (z, t) dt + v(Rei*)K2(z, Re i*) de - K3(z, r d#(r

R a

Again, set z = Re i~ sin 0, multiply by (R sin 2 O) -1, integrate with respect to 8 from 0 to ~r and change the order of integration in all terms. We can do this since we already know that the integrals with Ivl instead of v are finite. As a result, we obtain the equation

v(Re i~ sinO) ~ = (3.18)

I? (

v(t) 1 I

)

2

fo

"

R i~ R2 dt+ -~ v(Rei~)sin~dr

[rain (lIm l,

] d./r

304 V. MATSAEV, I. OSTROVSKII AND M. SODIN

Taking into account (3.10), we get

fo

v(Re i~ sin r R s~nn ~ r

f

= 2~r] [ l I m ( l / ( ) l - l/R] d#C0. J{ I1m(l/r

Then (3.17) follows, and the proof is complete. []

In other words, in the assumptions of Lemma 4, the first fundamental theorem for Tsuji characteristics holds without a remainder term:

fo r n(t)

(3.19) ~(r, u) = re(r, u) + - ~ - d r , 0 < r < cr where 1 ~2~ ~(r, u) = ~ u+(rei~ I sin0)l) do r sin 2 0 and 1 fo 2~ m(r,u) = ~ u-(rei~ do r sin 2 0 "

The last lernma was proved in a slightly different setting in [13, w (see also [7, l_.emma 5.2, Chapter 6]):

L e m m a 5. Let u(z) be a subharmonic function in C, and let

1 fo 2n

T(r, u) = ~ u + (re i~ do

be its Nevanlinna characteristic.function. Then, for 0 < R < oo,

(3.20) r----T--- dr < r2 dr.

R e m a r k . The referee pointed out that Lemrna 2 admits the following generalization, which is of independent interest.

L e m m a 2'. Let v(z) be a subharmonicfunction on DR = {z 6 C+ : Izl ___ R }

which satisfies

and(3.9). Then

and

lira 1 f0 '~

-

v(re i~

sin 19 dO = a,

r-~O r

1 f_:v(t)(1 1)

1 fo'~

2-'~

~

R2 dt + - ~

v( Re i~

sin 0 dO

- - - - ' - - q - 2 71" R

where the first integral on the LItS is absolutely convergent.

The referee has also pointed out that, if a function

v(z)

satisfies the conditions of Lemma 2', then Lemmas 3 and 4 continue to hold in modified form: the former with the additional sununand ar on the RHS, the latter with the additional sumrnand

2a/zr

on the RHS (this implies (3.19) with the additional sumrnand

2a/~r

on the RHS).

4

P r o o f o f T h e o r e m 1

Using the monotonicity of T(r, u), Lemma 5, and then Lemma 4, we obtain

T(R, u)

/~o T(r, u)

R2 < 2 J n r3 dr

(3.2o)_< 2

JR["~ T(r,r

2 u) dr

(3.1~) 2

~-~

~

dt +

re(r,

u)

2 f0nn(t )

fff.(t) fffm(t,u)

= -~

--~-dt+2

- i t - d r + 2

t2

dr.

Then the inequality M(r, u) < 3T(2r, u) completes the proof. I-1

5

P r o o f o f T h e o r e m 2

We split the proof into several parts. We assume convergence of the integrals

f~ ~(t)

fo ~

dt

and

-~- log t dt .

Define a measure/~1, supp(#l) C C - , by reflecting through the real axis the part of the measure ~ which lies in the upper half-plane. Formally,

306 V. MATSAEV, I. OSTROVSKII AND M. SODIN

where E C 12 is a Borel set, and E - = {z : ~, E E}. Then the measure #t also satisfies condition (2.2); we denote by Ul (z) its canonical integral of genus one. Observe that

ui(t)

= u(t), so that

~(t, ut) = 6(t, u), t E It.

5.1 E s t i m a t e o f u~-(iy), y > 0. We have

H(iy/~)

= log I1 + y l m ~ - i y R e ~ l - y l m ~

>

1o. I, +

Since the RI-IS is non-positive for y > 0 and ~ E (~_,

ur(iY) < - f~_ [l~ +ylm~l-ylm~] d#(~)

: y2

fO O0 n(t)

t 2 (t + y) dt

(5.D

_{<- Y Jo t2 dr+ Jv t3 tit.}

5.2 E s t i m a t e s o f

u+(rei~

0 < /9 < 7r. Using the harmonicity o f the function ul in the upper half-plane, we transform the lower bound for ul into the upper bound. We shall show that

(5.2)

u+(re i6)sinO < 6*(r),

O < r < o o , O < O < I r , where 6" (r) is defined by (2.13).

Consider the function - u l (z) and apply Lemma 1 to the angles {0 < argz < 7r/2} and {7r/2 < argz < ~'} with

Condition (3.1) holds due to estimate (5.1), and condition (3.2) holds due to estimate (2.4) combined with Jensen's inequality:

f;

ul (rei~

<__

/;

u+(rei~ < M(r, ul) = o(r2), r ~ oo. Therefore,

1/0

/;

-ul(rei~ [ < -~ O(t)tdt + r 2 ---~-dt

l fo*

f ~ u ? ( t ) + u { ( - t ) d t < r- ~- [u{ (t) + u~- (-t)]t dt + r 2 t3 + r --~- ds + r 2 - 7 l + l o g r ds (5.3) < **(r).

Observe that the factor I sin 201 on the LHS of (5.3) can be replaced by sin 0. This follows from inspection of the proof of L e m m a 1 (since, on the imaginary axis, the function - u ( i y ) has an increasing majorant). Alternatively, one may again apply L e m m a 1 to a small angle around the imaginary axis, say in {10 - Ir/21 < 7r/8}. That is, we have

(5.4) - u l ( r e i~ sin0 < 6*(r).

Using L e m m a 3, we obtain

fo" f 2 r uF(t ) + u ~ ( - t )

u+(re i~ sin0 < uF(2re iv) sin r de + r t2 dt < 5*(r) , d o

proving estimate (5.2).

5.3 E s t i m a t e o f u + (rei~ 0 r 0, ~r. Here we prove that, for an arbitrary 0, 0<~/_< I,

6*(r) or___[_ 2 f ~ M(t,u)

(5.5) u+(re i0) < r/i sin0--- ~ + sin2 0 _. t3 dr.

For this, we need several upper bounds for the difference

308 V. MATSAEV, I. OSTROVSKII AND M. SODIN when z, ff E C+.

First,

(5.6)

l Im lIm l< 'z'

We shall use this estimate when Izlllm(1/ff)l > 1. Next, let t = Izl/lr 0 = arg(z), ~ = arg(~). Then

1 [ 4_t.sin O sin ~b ] D = ~log 1 l1 - t e i ( a + ~ ) [ 2 J +2tsinOsin~b < 2t sin e sin ~b + 2t sin e sin ~b - I1_tei(O+~)12 - 2 t cos(0 + 4~) + t 2 = 2t sin o sin ~ yi ~ V ~ , ~ ) ~ ma~(t, t ~)

(5.7) < t sin 0 sin ~ I1 - t e i(~ 12 " If t < 1/2, then

and w e obtain

I1 - te~(a+~)12 > 1;

(5.8) D < t 2 sin 0 sin ~ < tit 2 + r I - x t 2 sin 2 ~ = O ~ + Im ,

with an arbitrary 0 > 0. If t > 1/2, then

I1 - tei(a+~)12 > t 2 sin 2 O, so that (5.7) gives us

(5.9) D < t s i n ~ < 7/ + - - s i n 2 4 ~ = 7/ + Im

" sin 0 '~ sin 2 0 ~/ sin 2 0 '

again, with an arbitrary positive O. W e shall use the b o u n d s (5.8) and (5.9) w h e n

Now, for z e C+, r =

Izl,

we have +

m~l___~

m~l_<~, 1r

m~l_<5,1r

fo r dt(t) -~r2 fJr

~176 dn(t)t--T-

O f or

f~ ~176

d#(t)

< r

- - +

_

+ - -

d~(t) + or 2

"~ sin 2 0

t 2

< a(r) + r

dt + - -

dt+

r/ - ~ sin 2 0 Jr t3

~'(r) + E_~ jr~f

M(t,u)

rj sin 2 0 t3

dt;

the last inequality uses that 0 < T/< 1.

Then, using estimate (5.2) for

u+(z)

in the upper half-plane, we obtain estimate (5.5) for 0 < 0 < ~r. The same argument applies for the lower half-plane, and the proof of (5.5) is complete.

5 . 4 I n t e g r a l i n e q u a l i t y f o r M ( r , u ) . Here we prove the integral inequality

(5.10)

M(r,u)<i6*(r)r2fr~176 t3 dr.

It suffices to prove it only for those values r that satisfy

< r2f~ ~176

M(t,u)

~i*(r)

_

t3 dt,

as otherwise (5.10) holds trivially:

~ f~176

M(t, u)

l i 6 , ( r ) r 2 fr~176

M(t,u) dt

M(r,u) <_ 2 Jr

t3 dr<_

t3 _ .

First, we improve estimate (5.5) near the real axis. Consider the function

u(z)

in the angles {I argzl < ~r/6} and {[argz - 7r I < 7r/6}. On the boundary of these angles,

310 V. MATSAEV, I. OSTROVSKII AND M. SODIN where for an arbitrary 7, 0 < 0 < 1,

9 (r) = o-l~*(r) + or 2 f oo

M(t,u)

_t3

_dr.

Applying L e m m a 1 to

u(z)

in these angles, we obtain for

101 _< ~r/8

and

I~r-01 _< ~r/8

u(rei~ < r-3 for~(t)t2dt + rz fr~ ~-~dt < .('. ) .

The second inequality follows since the function r does not decrease and the function r-2r does not increase.

Thus, for 0 < r < co,

u) < r

= 0-1~'(r) + 7". 2 [ o o M(t, u)

M('.,

_t3

_dr.

Jr

Choosing

we obtain inequality (5.10).

5.5 Solution of the integral inequality (5.10). We set

f oo M(t,u)

M1 ( r ) =

t3

dr.

Then

M(r, u) = - r3M~(r),

and inequality (5.10) takes the form

- M~ (r)r 2 ~< X/6* (r) M1 (r) o r

d r ~ ".2 Integrating this inequality from ~ to r, we obtain

2

On the other hand, since

M('., u)

does not decrease,

M1 (r) _> M('., ~) f~ d_tt3 = M('.,2,.2u)

Therefore,

M('.,u) ~< ".2Ml('.) <'.2

~ d t

t2

6

Proof of Theorem 3

We divide the proof into 4 parts. Set

fo ~ ,~(t) B := limsup #(r) and C :=

dt.

r--+0 r - ' ~

We assume that both B and C are finite; otherwise there is nothing to prove. First, we prove the theorem under the additional assumption

(6.1) supp(/~) C (~_,

and till Section 6.4 we assume that the function

u(z)

is harmonic in C+.

6.1 T h e f u n c t i o n

u(z)

has n o n n e g a t i v e h a r m o n i c m a j o r a n t s in C+.

Consider the function

y f ~

u- (t)dt

U(z)

~Z~

]_~ ( t - 5 7 y2.

This function is harmonic in C+ and

U(x)

_< 0, x e R. Moreover, for y > 0,

U(iy) < -u(iy) = - lim fl

e-.o

I>~.,e(~_ H ( ~ ) d/~(,)

< - ~im f;l>e,;eC_ Re ~ dP(~)

=Y fc Im~d#(r

9

By the Poisson-Nevanlinna representation of harmonic functions in the semi- disk Dn (cf. Section 3), we have

'f0"

U(z) < ~

U(Rei~)K2 (z, Re i~) ddp

2Rr(R + r) fon

<- -~--r~ff

u-(Re'C~)dr

312 V. MATSAEV, I. OSTROVSKII AND M. SODIN where

T(R, u)

is the Nevanlinna characteristic of u.

Note that. for any 5 > 0, the function u can be represented in the form

u(z) = flr

(~ )

d / ~ ( ( ) + f ( l < , l ~ ~

dl~(f) + Re (Z fl(l<, d~ () )

(6.3) =:

u6(z) + v6(z) +

Re z - . 1<6

The well-known Borel estimates

m a x u 6 ( z ) = o(r2), m a x d ( z ) = o(r), r --~ oo,

Izl<r Izl<r

imply that

T(R, u) = o(R2), R --+ oo.

Therefore, by setting R = 2r in (6.2). we get

U+(z)

= o(Iz12), z ~ oo, Irnz > 0.

Applying the Phragm6n-Lindel6fprinciple in the angles {0 < arg z < 7r/2} and {7r/2 < arg z < Ir}, we conclude that

(6.4)

U(z) < Cy,

z = x + iy E

C+ ;

i.e..

-U(z) + Cy

is a non-negative harmonic function in C+. Since

u(z) < - U ( z ) + Cy,

the function

u(z)

also has a non-negative harmonic majorant in 12+. For z E 12-. we write

u(z,-u(~.,=lime_+o

fl,l>e.,ec-_ {log ll-Z/~-

z/,

+ R e [z ( ~ - ~ ) ] } d/~(,)

(6.5) < f c _ Re [z2ilm~] dry(if) < 2CIYl.

Because u(Z,) has a non-negative harmonic majorant in C_. we get the desired conclusion.

6.2 E s t i m a t e o f u(z) n e a r t h e origin. Set

Let us prove that

(6.6) lim sup I(r) < B.

r--+0

For any given e > 0, choose a positive 6 < e such that #(r) < ( B + ~ ) r f o r 0 < r < 6 . Let us represent u by the formula (6.3) with this 6. Since

u6(z) = O(Izl=), z -~ 0,

we have

(6.7)

where

lim sup I ( r ) < limsup I6(r) + Ill; d#(~)

r~O - r-~O ]<6 ~ '

f0 ~"

i6(r ) = _1

v6(reiO)sinOdO"

r

It suffices to show that

(6.8)

I6(r)~B+e+[

Imld#(r

0 < r < 6 .

Jr <1r q

Indeed, if (6.8) is valid, then substituting it into (6.7), we get

limsupI(r)<~B+e+f~

I m ~ d # ( ( ) + f ~ d#(()[

r-,o I<6 I<~ ~ "

Taking the limit as e ~ 0 (then 6 -r 0 as well), we obtain (6.6). To prove (6.8), note that

fo II~ 1 ~ l l s i n o d 0 < l~ 2,~ dO

(

"

_<41flog I + 6 +21rl~ I~-'["

This estimate allows us to change the order of integration and to write (for 0 < r < 6)

314

V. MATSAEV, I. OSTROVSKII AND M. SODIN

Further,

I~(r) = 1 f~<lr

{ - R e f0 '~

(rei~ inO) dO+ fo~H (re~)sinOdO} d/z(()

r r2

< lf~</r

{rIm~ + ~ 5 } d/z(()

<

<lr

d/z(~) + r

t2

and

f~

Im ~ d/z(() + B + e

<

<1(1<6

I62(r)<lf

K r

I<r 21~

d#(()-2#(r)r

Putting these estimates together, we obtain (6.8).

2 f f /z(t)dt

- - - + - ~ B + e .

r

t(t + r)

6 . 3 E s t i m a t e o f u + ( z ) o n t h e r e a l a n d i m a g i n a r y a x e s .

that

~~ u+(t),

u+(iy)

(6.9)

-~ at +

lim sup

< B + C.

oo y ~ + o o y

Since u has a non-negative harmonic majorant in C+, we have

f

'~ lu(t)ldt

l + t 2 < o o ,

and u admits the Poisson representation

(6.10)

u(rei~ ) _

rsin~o

_71"

f~-~

_{oo r2 +}

_{tu --- ~'rt cos ~ +}

u(t) dt

_kr

_{sin ~,}

where

k = lim sup ~

# oo.

y---~+oo

Note that inequality (6.4) implies

u(iy)

> - y f_~

u-(t)dt

- ~r oo t 2 + Y 2 C y = o ( y ) - C y , y - ~ o o ;

i.e., k >_ - C .

Multiplying (6.10) by sin ~o, integrating with respect to ~o from 0 to ~r, and taking into account that

fo '~ sin2 ~o d~o 7r min ( 1 ' 1 )

r 2 + t 2 - 2 r t c o s ~ o = 2

~ ~ ' we get

fo "r

u(re i~~

sin ~o d~o = ~

r f j ~

oo u ( t ) m i n ( 1 ~ - , ~ 1 )

dt+

kTrr

2

Hence

f__~

_u+(t)

_{min ~-~ ~} ( 1 , 1 )

_{dt + k+rr =}

f ~

_u-(t)

_{min ~'7 ~5} ( 1 , 1 )

_{dt + k-Tr +}

_2I(r).

o o

Letting r ~ 0, we obtain by the monotone convergence theorem

ff

oo t2 dt + k+r

=

ffu(,,

t2

dt

+ k-Tr + 2 lim

I(r)

o o r--~O

(it turns out that the last limit exists). Taking into account that k- < C and using (6.6), we get (6.9).

6.4 C o n c l u d i n g steps. From (6.5) and (6.9), we obtain

u + (iy)

u + (iy) + 2Cy

lim sup < lim sup ~< B + C.

~ - , - o o lyl - ~ - - , + o o y

Since u has non-negative harmonic majorants in both upper and lower half-planes, the following inequality holds in the whole plane:

(6.11)

u(z) <~ [Y[ / ?

u+ (t) dt

--

o o ( t - - ~ T y 2 +(B+C)IyI' z = z + i y .

The assertion of Theorem 3 can be obtained from this inequality and (6.9) by applying a known argument (cf. [14], [16]). First, one applies (6.11) and (6.9) to get the upper bound for

u(z)

in the angles {I argz :t: rr/21 < rr/4}; then, using the Phragm6n-Lindelff principle, one obtains the upper bound for

u(z)

in the complementary angles. This gives

u(z) <~ (B + C)[z[

and completes the proof of estimate (2.22) for the special case (6.1).

Now, let # be an arbitrary measure in C satisfying the conditions of Theorem 3 and having finite value C. As in Section 5, we define the measure #1, supp#x C C - , by reflecting with respect to the real axis the part of # which charges C+. Since

316 V. MATSAEV, I. OSTROVSKII AND M. SODIN

the measure #1 also satisfies the conditions of Theorem 3; and we can define the corresponding generalized canonical integral ul (z) of this measure. Then a straightforward estimate (cf. 6.5) shows that for z E (3+

u(z) <_ ul(z)

+ 2 [ Y [ / c IIm~{

dlz(~) ~ (B + C)[z[.

The same estimate holds in the lower half-plane, and the general case of Theorem 3

follows. I7

REFERENCES

[1] A. Aleksandrov and P. Kargaev, Hardy classes o f functions that are harmonic in a half-space,

Algebra & Analysis 5 (1993), no. 2, 1-73 (Russian); English transl.: St. Petersburg Math. J. $

(1994), 229-286.

[2] C. Bennett and R. Sharpley, Interpolation o f Operators, Academic Press, London, 1988. [3] M. Ess~n, Some best constants inequalities for conjugate functions, Internat. Ser. Numer. Math.

103, Birkhiiuser, Basel, 1992.

[4] M. Ess~n, D. E Shea and Ch. S. Stanton, Best constants inequalities for conjugate functions,

J. Comput. Appl. Math. 105 (1999), 257-264.

[5] M. Ess~n, D. E Shea and Ch. S. Stanton, Sharp L log a L inequalities for conjugate functions,

Institute Mittag-Leffier Report No. 37, 1999/2000.

[6] T.W. Gamelin, Uniform Algebras and Jensen Measures, London Math. Soc. Lecture Note Series, Cambridge University Press, 1978.

[7] A.A. Goldberg and I. V. Ostrovskii, Value Distribution o f Meromorphic Functions, Nanka, Moscow, 1970 (Russian).

[8] B. Khabibullin, Sets o f uniqueness in spaces o f entire functions o f one variable, Math, USSR Izv. 39 (1992), 1063-1083.

[9] P. Koosis, Introduction to Hp Spaces, 2nd edn., Cambridge University Press, 1998.

[10] P. Koosis, Lefons sur le thdordme de Beurling et Malliavin, Los Publications CRM, M o n t r ~ , 1996.

[11] B. Ya. Levin, On functions holomorphic in a half-plane, Travaux de l'Universit~ d'Odessa (Math)

3 (1941), 5-14 (Russian).

[12] B. Ya. Levin, Lectures on Entire Functions, Transl. Math. Monographs, Vol. 150, Amer. Math. Soc., Providence, RI, 1996.

[13] B. Ya. Levin and I. V. Ostrovskii, The dependence o f the growth o f an entire function on the distribution o f the zeros of its derivatives, Sibirsk. Mat. Zh. 1 (1960), 427-455 (in Russian); English transl.: Amer. Math. Soc. Transl. (2) 32 (1963), 323-357.

[14] V. Matsaev and M. Sodin, Variations on the theme o f M. Riesz and Kolmogorov, Intern. Math. Res. Notices, no. 6 (1999), 287-297.

[15] V. Matsaev and M. Sodin, Distribution o f the Hilbert transforms o f measures, Geom. Funct. Anal.

10 (2000), 160--18d.

[ 16] V. Matsaev and M. Sodin, Compact operators with Sp-imaginary component and entire functions,

in Entire Functions in Modern Analysis, Boris Levin Memorial Conference, Israel Math. Conf. Proc., Vol. 15, to appear.

[17] R. Nevanlinna, Ober die Eigenschaflen meromorpher Funktionen in einem Winkeiraum, Acta Soc. Sci. Fenn. 50, no. 12 (1925).

[18] M. Tsuji, On Borel's directions o f meromorphic functions of finite order, T6hoku Math. J. 2 (1950), 97-112.

Vladimir Matsaev

SCHOOL OF MATHEMATICAL SCIENCES TEL AVIV UNIVERSITY

RAMAT AVIV, 69978, ISRAEL email: matsaev@math.tau.ac.il lossif Ostrovskii DEPARTMENT OF MATHEMATICS B1LKENT UNIVERSITY 06533 BILKENT ANKARA, TURKEY email: iossif@fen.bilkent.edu.tr

VERKIN INsTrrUTE FOR LOW TEMPERATURE PHYSICS AND ENGUqEERING 310164 KHARKOV, UKRAINE

email: ostrovskii@ilt.kharkov.ua Mikhail Sodin

SCHOOL OF MATI4EMATICAL SCIENCES TEL AVlV UNIVERSITY

RAMAT AVIV, 69978, ISRAEL

email: sodin@post.tau.ac.il