lim r→1f(reiθ) denote the radial limit of f

ON THE TWO QUESTIONS OF LOHWATER AND PIRANIAN

M. GVARADZE

Abstract. The problem we are dealing with consists in the following:

find the necessary and sufficient conditions for the zero measure subset of the circumference at which points the bounded analytic function has no radial limits.

1. For a function f : D → C analytic and bounded in the unit disk D = {z : z ∈ C,|z| < 1} and any point e^iθ ∈ C = {e^iτ : 0 ≤ τ < π} letf(e^iθ) = lim

r→1f(re^iθ) denote the radial limit of f. Fatou proved in 1906 that for such a function there exist radial limits except maybe for a set of pointse^iθ of linear measure 0. Conversely, as Lusin showed in [1], for any setEof measure zero onC there exists a function analytic and bounded in D having no radial limits at the points ofE.

Lohwater and Piranian noticed in [2] that “the set of nonexistence of radial limit is of second category for some bounded regular functions; it can even be a residual set onC; but we do not know of any case where a setE of second category, prescribed without reference to function theory, has been established as the precise set where the radial limit of some bounded regular function fails to exist.”

This is the first question we have to answer in this note. The second one is connected with the following statement (Theorem 8 in [2]).

Theorem. “Let the set E on C be of typesFσ andGδ and of measure zero. Then there exists a function f(z), regular and bounded in D, which has the following properties: for each pointe^iθ in E,

rlim→1infŒŒf(re^iθ)ŒŒ= 0, lim

r→1

ŒŒf(re^iθ)ŒŒ= 1;

1991Mathematics Subject Classification. 30B30.

Key words and phrases. Bounded analytic function, radial limit, modulus oscillation, Cantor set.

447

1072-947X/96/0900-0447$12.50/0 c1997 Plenum Publishing Corporation

for each point e^iθ in C\E the radial limit f(e^iθ) exists and has modulus1, except for a denumerable set of points where f(e^iθ) = 0.”

In 1956 Lohwater and Piranian did not know “whether the converse of the theorem is true.”

Theorems 1 and 2 proved below give answers to these two questions.

2. To prove the theorems we need several lemmas. Some of them are of independent interest. In what follows

C(a, r) =ˆ

z: |z−a| ≤r‰

, D(a, r) =ˆ

z: |z−a|< r‰

; the length of an intervalI will be denoted by the same letterI.

Lemma 1. Let z0=e^it0 and0< α <1.

If z∈D(0,1)\D(αz0,1−α), thenRez0+z z0−z ≤ α

1−α. If z∈C(αz0,1−α), thenRez0+z

z0−z = α 1−α. If z∈D(αz0,1−α), thenRez0+z

z0−z > α 1−α. If 0< t < π/2, then

sup

0≤r≤1

ŒŒ

Œ1 +re^it 1−re^it

ŒŒ

Œ<ŒŒŒcott 2

ŒŒ

Œ. For0< r <1

ŒŒ

Œz0+re^it

z0−re^it −icott−t0

2

ŒŒ

Œ=

= 1−r²

(1−r)²+ 4rsin^2t−₂^t0

ŒŒ

Œ1−i1−r

1 +rcott−t0

2

ŒŒ

Œ. Proof. The linear-fractional mapping ^z_z^0+z

0−z maps the circleC(αz0,1−α) on the straight linex=α/(1−α) and the diskD(αz0,1−α) on the halfplane {z: Rez > α/(1−α)}which imply the first three assertions of the lemma.

To prove the next inequality note that the straight line containing the radius {re^it: 0≤r≤1} is mapped by the mapping ^1+z_1−z on the circle having the center on the imaginary axis and intersecting this axis at the points−itan₂^t and a = icot₂^t. The radius itself is mapped onto the smaller arc of this circle (lying in the first quadrant) with the endpoints 1 and b = ^1+e_1−e^itit. Simple analysis of the triangle with vertices 0,a, andbshows that|a|>|b|.

The last assertion is proved as follows (τ =t−t0):

z0+re^it

z0−re^it −icotτ

2 =1 +re^iτ

1−re^iτ −icotτ 2 =

= 1−r²

(1−r)²+ 4rsin^2τ₂ +i 2rsinτ

(1−r)²+ 4rsin^2τ₂ −icotτ 2 =

= 1−r²

(1−r)²+ 4rsin^2τ₂



1−i1−r 1 +rcotτ

2

‘ .

Lemma 2. ρ(e^it, C(α,1−α))≥2αsin²₂^t, where ρ(E, F) is a distance between the sets E andF, andα∈(0,¹₂).

Proof. It is evident that ρ(e^it, C(α,1−α)) =|e^it−α| −(1−α) =

= q

(1−α)²+ 4αsin²₂^t −(1−α) ≥ 2αsin²₂^t (equality holds if t = kπ, k∈Z).

Lemma 3. Suppose θn = argzn and P^∞

n=1

In|cot^θ−₂^θn| < ∞. Then the functiong(z) = P^∞

n=1

Inzn+z

zn−z has a radial limit of modulus one at the pointe^iθ. Proof. By Lemma 1,ŒŒ P^∞_n=1Inzn+z

zn−z

ŒŒ≤P_∞

n=1In

ŒŒcot^θ−₂^θnŒŒ. Thereforeg is continuous on the radius [0, e^iθ].

Let d(E, F) denote an arcdistance between the subsets E and F of C and∂G be a boundary ofG.

Lemma 4. Let G ⊂ G1 ⊂ G0 be open subsets of the circle. Suppose G= ^∞∪

n=1In, G1 = ^∞∪

n=1Jn and zn ∈In. If e^iθ6∈G0 and d(Jk, ∂G0)>2^kJk, then there exists lim

r→1

P∞ n=1

Inzn+re^iθ zn−ze^iθ. Proof. IfIn ⊂Jk, then

sup

0≤r≤1

ŒŒ

Œzn+re^iθ zn−re^iθ

ŒŒ

Œ≤ŒŒŒcotθ−θn

2

ŒŒ

Œ≤cotd(e^iθ, Jk)

2 ,

and therefore X∞ k=1

X

In⊂Jk

ŒŒ

ŒIn

zn+re^iθ zn−re^iθ

ŒŒ

Œ≤ X∞ k=1

cotd(e^iθ, Jk) 2

X

In⊂Jk

In≤

≤ X∞ k=1

Jk

ŒŒ

Œcotd(e^iθ, Jk) 2

ŒŒ

Œ<∞. Using Lemma 3, we can prove Lemma 4.

The following lemma is the basic one.

Lemma 5. SupposeG=^∞∪

n=1Into be an open set. Letzn∈In,d(zn, ∂In)>

In/4 and let

X

n /∈{nk}

In

d(e^iθ, In) <∞, lim

k→∞

Ink+1

Ink

= 0.

Then the functiong(z) = P^∞

n=1

Inzn+z

zn−z has radial limit at the pointe^iθ if and only if the series

X∞ n=1

Incotθ−argzn

2 (1)

is convergent.

Proof. By Lemma 3 since the function P

n6∈{nk}

Inzn+z

zn−z has radial limit at the pointe^iθ, then without loss of generality we may assume that

nlim→∞

In+1

In

= 0, In+1

In ≤1 2.

Denoteρn= 1−In+1 and suppose thatρN−1<|z| ≤ρN. We have g(z) =

NX−1 n=1

In

zn+z

zn−z−icotθ−argzn

2

‘

+INzN +z zN −z+ +IN+1

zN+1+z zN+1−z +

X∞ N+2

In

zn+z zn−z +i

NX−1 n=1

Incotθ−argzn

2 . (2)

Ifn≥N+ 2, then In

ŒŒ

Œzn+z zn−z

ŒŒ

β2 In

1− |z| ≤2 In

IN+1 ≤2^N−nIN+2

IN+1

, whence

ŒŒ

Œ X∞ n=N+2

In

zn+z zn−z

ŒŒ

βIN+2

IN+1

. (3)

By Lemma 1 In

ŒŒ

Œzn+z

zn−z −icotθ−argzn

2

ŒŒ

Œ≤4π³In 1−r

(θ−argzn)² ≤16π³IN

In. Therefore

ŒŒ

Œ

NX−1 n=1

In

zn+z

zn−z −icotθ−argzn

2

‘ŒŒŒ≤16π³ IN

IN−1

. (4)

If lim

n→∞Incot^θ−arg₂ ^zn = 0, then by Lemma 1, lim

n→∞Inzn+z

zn−z = 0. Thus, by (3) and (4) the right-hand side of (2) has radial limit if and only if (1) is convergent.

Finally, if lim

n→∞In|cot^θ−arg₂ ^zn| ≥δ >0, we have ReIN+1zN+1+ρNe^iθ

zN+1−ρNe^iθ ≥IN+1 1−ρ²_N

(1−ρN)²+ 4ρNsin^2θ−₂^θN ≥

≥IN+1

1−ρN

(1−ρN)²+ 4^IN+12_δ2

= I_N²₊₁ I_N²₊₁+_δ⁴2I_M+1² , since sin^2θ−θ₂^N+1 ≤ ^I

2 N+1

δ² , where θN = argzN+1; ReINzN+ρNe^iθ

zN−ρNe^iθ ≤IN 2(1−ρN)

(1−ρN)²+ 4ρNsin^2θ−₂^θN ≤

≤ 2ININ+1

4ρN(_π^2θ−₂^θN)² ≤ 2ININ+1

4¹_2π¹2(¹₄IN)² = 16π²IN+1

IN

, since|θ−θN| ≥ ¹₄IN,e^iθN 6∈IN,ρN ≥ ¹₂.

By the notationρ⁰_N = 1−p

ININ+1we get ReIN+1

zN+1+ρ⁰_Ne^iθ

zN+1−ρ⁰_Ne^iθ ≤IN+1

2(1−ρ⁰_N) (1−ρ⁰_N)² ≤

≤2 IN+1

1−ρ⁰_N = 2 IN+1

pININ+1

= 2 rIN+1

IN

and

ReIN

zN +ρ⁰_Ne^iθ zN −ρ⁰_Ne^iθ ≤IN

2(1−ρ⁰_N) (I_N²)/(16π²) =

= 32π²IN

pININ+1

I_N² = 32π² rIN+1

IN .

Taking into acount these inequalities we can conclude that the right-hand side of (2) has no radial limit.

Definition 1. We say thatEis anarrangeableset if there exist a count- able set (zmn) = (e^iθmn), a sequence of intervals (arcs) (Imn), and a sequence of positive numbers (αk) satisfying the following conditions:

1. zmn∈Imn, 4d(zmn, ∂Imn)> Imn; 2. lim

n→∞

I_m(n+1)

Imn = 0, m= 1,2, . . .; 3. I(m+1)n⊂ ^∞∪

i=1(Imi\{zmi}),m= 1,2, . . .;

4. √

Imn < 2⁻ⁿαmαm+1d(Imn, ∂Gm−1), m >1, where Gm= ^∞∪

n=1(Imn\ {zmn}),ak &0, P

αk<∞; 5. E= ( ^∞∩

m=1Gm)∪( ^∞∪

m=1Qm), where Qm=n

e^iθ: e^iθ6∈Gm, X∞ n=1

Imncotθ−θmn

2 is divergento .

Lemma 6. Let E be an arrangeable set. Then the function f(z) = exp{−g(z)}, with g(z) = P^∞

m=1

gm(z), gm(z) = _α¹_m P^∞

n=1

Imnzmn+z

zmn−z, has no radial limit on the setE only.

Proof. Since X∞ m=1

1 αm

X∞ n=1

Imn≤ X∞ m=1

1 αm

X∞ n=1

2⁻²ⁿα²_mα²_m+1<∞, we conclude thatf is analytic in the unit disk.

Supposee^iθ6∈G1. We have X∞

m=2

1 αm

X∞ n=1

Imn

ŒŒ

Œcotθ−θmn

2

ŒŒ

Œ<

X∞ m=2

1 αm

X∞ n=1

Imn

1

d(Imn, ∂Gm−1)<

<

X∞ m=2

1 αm

X∞ n=1

2⁻⁴ⁿα²_mα²_m+1d(Imn, ∂Gm−1)<

X∞ m=2

αmα²_m+1<∞. Hence by Lemma 3,g−g1 has radial limit ate^iθ, and thus by our basic lemma we can conclude thatg has no radial limit only atQ1.

Now ife^iθ∈Gm\Gm+1,m= 1,2, . . ., then Xm

k=1

1 αk

X∞ n=1

Ikn|cotθ−θkn

2 |< 1 d(e^iθ, ∂Gm)

Xm

k=1

1 αk

X∞ n=1

Ikn<∞, and

X∞ k=m+2

1 αk

X∞ n=1

Ikn|cotθ−θkn

2 |<

<

X∞ k=m+2

1 αk

X∞ n=1

Ikn

1

d(Ikn, ∂Gk−1)<

X∞ k=m+2

αkα²_k+1<∞.

Therefore arguing as above we come to the conclusion that g has no radial limit atQm+1. Suppose now that e^iθ ∈_m=1^∞∩ Gm, and denoteOm=

∞∪

n=1D(α²_mzmn,1−α²_m),Cm=∂Om.

Letz∈Cm andk≤m; then sincez6∈D(α²_mzmn,1−α²_m), we have Regk(z)≤ 1

αk

X∞ n=1

Ikn

α²_m

1−α²_m ≤2α²_m αk

µGk ≤αm.

Further, ifz∈Cmandk > m, then according to Lemma 2, Rezkn+z

zkn−z ≤ 2

ρ(zkn, Cm) < 1

α²_msin^2θkn−₂^θmi < π²

α²_md²(Ikn, ∂Gk−1). Hence

Regk(z)≤ 1 αk

X∞ n=1

2⁻²ⁿα_k²α²_k+1d²(Ikn, ∂Gk−1) π²

α_m²d²(Ikn, ∂Gk−1) ≤

≤ αkα²_k+1

α²_m ≤αk+1. Thus, ifz∈Cm, then

Reg(z)≤ Xm

k=1

αm+ X∞ k=m+1

αk =mαm+ X∞ k=m+1

αk,

whence

lim

r→1

Reg(re^iθ) = 0. (5)

Sincee^iθ∈Gm, there exists an integer j such thate^iθ ∈Imj. It is clear that the radius [0, e^iθ] intersects the circleC((1−Imj)zmj, Imj) in which

Reg(re^iθ)≥Regm(re^iθ)≥Re 1 αm

Imj

zmj+z zmj−z ≥ 1

2αm

.

Hence, limr→1Reg(re^iθ) =∞, which together with (5) gives lim

r→1|f(re^iθ)|= 0, lim

r→1|f(re^iθ)|= 1.

3. Let us now formulate and prove our theorems.

Theorem 1. There exists an arrangeable set of second category.

Proof. LetQ ={xn} be a countable subset of a unit circle and G0 be an open subset coveringQandµG0<2π(µdenotes the Lebesgue measure).

Coverx1by the interval J1 such that∂J1∩Q= 0 and√

J1 <2⁻¹α1α2

d(J1, ∂G0). Let xk2 be the first element of the sequence Q not belonging to J1. Cover xk2 by the interval J2 such that∂J2∩Q= 0, J1∩J2 =∅,

√J2<2⁻²α1α2d(J2, ∂G0), ^J_J²

1 < ¹₂.

Given J1, J2, . . . , Jn−1, let xkn be the first element of Q not belonging to ⁿ∪⁻¹

k=1Jk. Cover xkn by the interval Jn such that ∂Jn ∩Q = ∅, Jn ∩ (ⁿ∪⁻¹

k=1Gk) =∅, √

Jn <2⁻ⁿα1α2d(Jn, ∂G0), _J^J_nⁿ

−1 < _n¹. Selectz1n such that z1n ∈ Jn,4d(z1n, ∂Jn) > Jn and z1n 6∈ Q. Denote I1n = Jn\{z1n} and G1= ^∞∪

n=1I1n.

Taking G1 instead of G0 and arguing analogously, we will obtain new sequences of the intervalsJn and pointsz2n with the following properties:

(a)z2n∈Jn, z2n 6∈Q;

(b)Jn∩Jk =∅,n6=k;

(c) 4d(z2n, Jn)> Jn; (d)√

Jn<2⁻ⁿα2α3d(Jn, ∂G1);

(e) lim

n→∞

Jn+1

Jn

= 0.

DenoteI2n=Jn\{z2n} andG2=_n=1^∞∪ I2n.

Repeating the above process forG2,G3, etc., we shall get the sequences zmnandImn which define the arrangeable setE. IfQis a countable every- where dense subset, then the obtained setEwill be of second category.

Theorem 2. There exist a setEand a bounded analytic functionf with the following conditions:

(a)E is ofGδ type;

(b)E is not ofFσ type;

(c) ife^iθ∈E, then lim

r→1|f(re^iθ)|= 0, lim

r→1|f(re^iθ)|= 1;

(d) the function f on C\E has radial limit of modulus one except a countable set where the radial limit is zero.

Proof. Consider the Cantor setE without ends on its adjacent intervals. It is clear thatE is of typeGδ and not of typeFσ.

SupposeE= ^∞∩

k=1Gk and P^∞

k=1

αk <1. Without loss of generality we may assume that the ends of component intervalsI1n of G1 belong to adjacent intervals of Cantor’s set. Cover everyI1n∩Eby an open setH1nsuch that

√µH1n < 2⁻ⁿα1α2d(H1n, ∂I1n), where H1n is a finite union of intervals.

Put R1 =_n=1^∞∪ H1n = ^∞∪

k=1J1k. Suppose z1k is the right end of the interval J1k.

Ife^iθ6∈G1, we have X∞

k=1

J1k|cotθ−θ1k

2 | ≤ X∞ n=1

X

J1k⊂I1n

J1k

π

d(H1n, ∂I1n) ≤

≤π X∞ n=1

µH1n

d(H1n, ∂I1n) ≤πα1α2

and ife^iθ∈I1n, then X

J1k∩I1n=∅

J1k|cotθ−θ1k

2 | ≤ π

d(e^iθ, ∂I1n) X∞ n=1

µH1n = πµR1

d(e^iθ, ∂I1n). Hence, according to Lemma 3 and the fact that intervals contained inI1n

are finite, we conclude thatf1(z) = exp{−g1(z)}, and g1(z) = 1

α1

X∞ k=1

J1kz1k+z z1k−z has everywhere a radial limit. PutR1∩G2= ^∞∪

n=1I2n. We assume again that the ends ofI2n belong to the adjacent intervals of Cantor’s set. Cover every I2n∩E by open setsH2n such that√

µH2n <2⁻ⁿα2α3d(H2n, ∂I2n), where H2n is a finite union of intervals. PutR2= ^∞∪

n=1H2n= ^∞∪

k=1J2k. Supposez2k

is the right end ofJ2k. Ife^iθ6∈R1∩G2, we have

X∞ k=1

J2k

ŒŒ

Œcotθ−θ2k

2

ŒŒ

Œ≤πα2α3

and ife^iθ∈I2n, then X

J2k∩I2n=∅

J2k

ŒŒ

Œcotθ−θ2k

2

ŒŒ

Œ≤ πµR2

d(e^iθ, ∂I2n).

Hence, according to Lemma 3 and the fact that intervals contained inI2n

are finite, we conclude thatf2(z) = exp{−g2(z)}, and g2(z) = 1

α2

X∞ k=1

J2k

z2k+z z2k−z has everywhere a radial limit.

Given R1, R2, . . . , Rm−1, put Rm−1∩Gm = ^∞∪

n=1Imn. Assume that the ends ofImnbelong to adjacent intervals of Cantor’s set. Cover everyImn∩E by open setsHmn such that √

µHmn <2⁻ⁿαmαm+1d(Hmn, ∂Imn), where Hmn is a finite union of intervals. Put Rm= ^∞∪

n=1Hmn = ^∞∪

k=1Jmk. Suppose zmk is the right end ofJmk.

Ife^iθ6∈Rm−1∩Gm, we have X∞

k=1

Jmk

ŒŒcotθ−θmk

2

ŒŒ

Œ≤παmαm+1 (6) and ife^iθ∈Imn, then

X

Jmk∩Imn=∅

Jmk|cotθ−θmk

2 | ≤ πµRm

d(e^iθ, ∂Imn) . The functionfm(z) = exp{−gm(z)}with

gm(z) = 1 αm

X∞ k=1

Jmk

zmk+z zmk−z has everywhere a radial limit.

Put

f(z) = Y∞ m=1

fm(z) = expn

− X∞ m=1

gm(z)o . It is evident that if e^iθ =zmk, then lim

r→1f(re^iθ) = 0. SinceRm⊂Gm and Rm⊃E, then ^∞∩

m=1Rm=E.

Ife^iθ6∈E, then there exists an integernsuch thate^iθ 6∈Rn. By (6) we

have X∞

m=n

1 αm

X∞ k=1

Jmk

ŒŒ

Œcotθ−θmk

2

ŒŒ

Œ≤π X∞ m=n

αm+1<∞. Therefore, by Lemma 3, Re P^∞

m=n

gm(re^iθ) has zero radial limit.

Suppose finally e^iθ ∈ E. Then using the definitions and arguing as in Lemma 6, we may conclude that

lim

r→1

ŒŒf(re^iθ)ŒŒ= 0, lim

r→1

ŒŒf(re^iθ)ŒŒ= 1.

References

1. N. N. Lusin, Sur la representation conforme. Ivanovo, Izv. Politekh.

Instituta2(1919), 77–80.

2. A. J. Lohwater and G. Piranian, The boundary behavior of functions analytic in a disk. Ann. Acad. Sci. Fenn., Ser. A.I239(1957), 1–17.

(Received 28.10.1993; revised version 30.11.1994) Author’s address:

A. Razmadze Mathematical Institute of Georgian Academy of Sciences 1, M. Alexidze St., Tbilisi 380093, Republic of Georgia