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Memoirs on Differential Equations and Mathematical Physics Volume 33, 2004, 87–94

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Memoirs on Differential Equations and Mathematical Physics

Volume 33, 2004, 87–94

G. Khuskivadze and V. Paatashvili

ON A PROPERTY OF HARMONIC

FUNCTIONS FROM THE SMIRNOV CLASS

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classe(L1p1), L02q2)) (i.e., for functions satisfying the inequality (2)) in a simply connected domain with the Lyapunov boundaryLalmost everywhere onL there exist the angular boundary values which on the partL2 of the boundary form an absolutely continuous function.

2000 Mathematics Subject Classification. 31A05, 35J05.

Key words and phrases: Harmonic functions, Smirnov classes of har- monic functions, Zaremba’s problem, absolute continuity.

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On a Property of Harmonic Functions from the Smirnov Class 89

The boundary value problems for harmonic functions are, as usual, con- sidered in different functional classes and the character of their solvability depends considerably on the choice of a class of unknown functions.

When considering Zaremba’s mixed boundary value problem, the bound- ary of the domain is divided into two partsL1 andL2and it is required to find a harmonic function from a class A such that on the portion L1 the boundary function of that function and on the portion L2 the boundary function of its normal derivative take preassigned values. In the capacity of the classA, one of the possible sets is the set of harmonic functions such that the integralp-means are bounded “near”L1 and the integralq-means of their partial derivatives are bounded “near”L2. Since in the role of L1

andL2there appear finite unions of arcs, it is natural to consider weighted integral means with singularities at the ends of those arcs.

Proceeding from the above reasoning, in the works [1, 2] the authors, in connection with the study of Zaremba’s problem, have introduced the classese(L1p1), L02q2)).

As far as the boundedness of integral means is taken as the basis in de- termining Smirnov classes of analytic functions, the above-introduced class is naturally called the Smirnov class of harmonic functions.

In [1,2], the solution of the mixed boundary value problem, besides its be- longing to the classe(L1p1), L02q2)), is required to be absolutely continu- ous onL2. However, it turns out that any function from the above-indicated class possesses the latter property. In the present paper we prove this fact.

In Section 10 we present the definition of the classe(L1p1), L02q2)) and cite some properties of functions from that class established in [2] which will be needed in the sequel. In Section 20we prove absolute continuity on L2

of the boundary function of the function frome(L1p1), L02q2)).

10. LetDbe a simply connected domain bounded by a simple rectifiable curveLand letLk = (Ak, Bk), k= 1, mbe arcs lying separately onL. By C1, C2, . . . , Cmwe denote the pointsAk,Bktaken arbitrarily. AssumeL1=

m

k=1Lk, L2 =L\L1. D1, D2, . . . , Dn denote the points onL different from Ck; note that the pointsD1, . . . , Dn1 are located on L1 whileDn1+1. . . Dn

onL2. Assume ρ1(z) =

n1

Y

k=1

|z−Dk|αk, ρ2(z) = Y2m

k=1

|z−Ck|αk Yn

k=n1+1

|z−Dk|βk. (1) Letz=z(w) be the conformal mapping of the unit circleU={w:|w|<1} onto the domainD, and letw=w(z) be the inverse mapping. Suppose Γ1= w(L1), Γ2=w(L2), Γj(r) ={w:w=re, e∈Γj},Lj(r) =z(Γj(r)).

We say that a harmonic in the domainDfunctionu(z),z=x+iy=re belongs to the classe(L1p1), L02q2)) if

sup

r

Z

L1(r)

|u(z)ρ1(z)|p|dz|+ Z

L2(r)

∂u

∂x(z)

q

+ ∂u

∂y(z)

q

ρq2(z)|dz|

<∞. (2)

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In the case where D coincides with the unit circle, this class will be denoted byh(Γ1p1),Γ02q2)). For Γ1=γ={t:|t|= 1}and ρ1≡1, we obtain the well-known classhp([3], p. 373).

Statement 1(see [2]). Ifp >1, q >1and for the weightsρ1 andρ2 we have −1p < αk < p10, −1q < γk < q10, −1q < βk < q10 (p0 = p−1p , q0 = q−1q ) andu∈h(Γ1p1),Γ02q2)), then:

(i)there existsσ >1such thatu∈hσ;

(ii)ifvis the function harmonically conjugate tou, thenv∈h(Γ1p11), Γ02q2)), wherep1= p+σ ;

(iii) if, however, u ∈ e(L1p1), L02q2)), then the function U(w) = u(z(w))belongs to the class h(Γ1p1),Γ02q2)), whereω1(w) =ρ1(z(w))×

×pp

|z0(w)|, ω2(w) =ρ2(z(w))pq

|z0(w)|.

Due to this fact, ifu∈h(Γ1p1),Γ02q2)), then:

(a) almost everywhere on γ there exist angular boundary values u+(t), andu(re) can be represented by the Poisson integral of the functionu+;

(b) ifφ(z) =u(z) +iv(z), thenφ∈Hσ and sup

|z|=r

Z

Θ(Γ2)

0(z)|qωq2(z)|dz|<∞,Θ(Γ2) ={θ: 0≤θ≤2π, e∈Γ2} (3) (for the definition of Hardy classesH6see [3], p. 388).

20. Theorem. Let p >1, q > 1, the weight functions ρ1, ρ2 be given by the equalities (1), where αk ∈(−1p,p10), γk, βk ∈(−1q,q10), and let u ∈ h(Γ1p1),Γ02q2)). Then the function u can be continuously extended to every closed arc lying onΓ2. Moreover, the boundary functionu+(t)is such that there exist the limits

u(Ak−) = lim

t→Aku+(t), u(Bk−1+) = lim

t→Bk−1+u+(t), k= 2, m, and the obtained in such a way function is absolutely continuous on Γ2. Moreover, ∂u∂θ+ ∈Lq22).

Proof. It suffices to consider the case wherem= 1, i.e., we assume thatγab

is the arc of the circumferenceγwith the endsaandb, and sup

r

Z

Θ(γ\γab)

(u(re1(re))pdθ+

+ Z

Θ(γab)

∂u

∂x(re)

q

+∂u

∂y(re)

q

ρ2(re)dθ

<∞, (4) where Θ(E) ={θ:e∈E, 0≤θ≤2π}.

Let the function v be harmonically conjugate to the function u, and φ = u+iv. According to Statement 1, φ ∈ Hσ ⊂ H1, and therefore φ(z) possesses angular boundary values almost everywhere on γ. Thus in arbitrarily small neighbourhoods of the points aandb there are the points

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On a Property of Harmonic Functions from the Smirnov Class 91

e

a= eiαe,eb= eieb, ea,eb ∈γab at which there exist angular boundary values φ+(ea),φ+(eb). Moreover,

φ(z) = 1 2π

Z

0

u+(θ)e+z

e−zdθ. (5)

Consider now the domainG⊂U which is bounded by the radii passing through the pointsea,eband by the arc of the circumferenceγ

eaeb ⊂γab. Let us show that φ0 ⊂E1(G) (for the definition of the classes Ep(G), p > 0, see [3], p. 422). Towards this end, it is sufficient to construct a sequence of rectifiable curvesγn ⊂Gconverging to the boundary for which

sup

n

Z

γn

0(z)| |dz|<∞ (6)

(see, e.g., [3], p. 422–423).

Let{ean}and{ebn}be sequences of points onγab, converging respectively to the pointseaandeb. Consider the curves γ1n ={z:z=ρeiαen, n1 < ρ <

rn = 1− 1n}, γ2n ={z : z = ρeieβn, n1 < ρ < rn}, γ3n = {z : z = n1e, e

αn < α < βen}, γ4n = {z : z = rne,αfn < α < βen}, where we put e

αn= argean,βen= argbn and letγn = ∪4

j=1γjn,n >2. It is obvious thatγn

converges to the boundaryG. Let us prove that the inequality (6) is valid forγn.

Let r < rn < ρ; choose a point e between a and ea and a point e betweeneb and b with the condition that there exist φ+(e) andφ+(e).

We write (−2πiφ0) in the form

−2πiφ0(re) = Zβ

α

ρφ(ρe)de (ρe−re)2 +

Z

2π\[α,β]

ρφ(ρe)de (ρe−re)2 =

1(re) +φ2(re). (7) Since the distance fromγn to the arcγ\(e, e) is positive, we get

sup

n

Z

γn

2(z)| |dz| ≤M1sup

ρ

Z

Θ(γ\γab)

|φ(ρe)|dθ <∞. (71)

Estimate now the integrals ofφ1. Z

γn

1(z)| |dx| ≤ X4

j=1

Z

γjn

1(z)| |dz|= X4

j=1

Ijn. (72)

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We have

I1n=

rn

Z

1/n

Zβ

α

φ(ρe)d dθ

1 ρe−ren

dr=

=

rn

Z

1/n

φ(ρe)

ρe−ren − φ(ρe) ρe−ren

Zβ

α

φ0(ρe)

ρe−rendedr. (8) Since φ+(e) andφ+(e) exist and the distance from the points ρe, ρeto γ1n is positive, it follows from (8) that

I1n≤M+

rn

Z

1/n

φ0(ρe)de ρe−ren dr≤

≤M+

rn

Z

1/n

Zβ

α

0(ρe)|dθ q

(ρ−r)2+ 4ρrsin2θ−α2n

dr=M+Jn. (9)

Next, taking into account thatn≥3 and sinx > π2xfor|x|< π2, we have Jn

Zβ

α

0(ρe)|

rn

Z

1/n

q dr

(ρ−r)2+ 4ρrsin2θ−α2n

≤ π 2√ρ

Zβ

α

0(ρe)|

rn

Z

1/n

dr

√rq (θ−αρ−r

n)2+ 1 dθ≤

≤M1

Zβ

α

0(ρe)| Zr

1/n

q dr (θ−αρ−r

n)2+ 1

dθ. (10)

Assuming (ρ−r)|θ−αn|−1=x, we obtain Zr

0

q dr

(θ−αρ−rn)2+ 1

= 1

|θ−αn|

ρ

|θ−αn|Z

ρ−rn

|θ−αn|

|θ−αn|dx

√x2+ 1 ≤

ρ

|θ−αn|Z

0

√ dx

x2+ 1 +

ρ

|θ−αn|Z

1

√ dx

x2+ 1 ≤1 + ln 1 θ−αn

.

The inequality (9) implies that Jn≤M2

Zβ

α

0(ρe2(ρe)|q

1/qZβ

α

2(ρe) ln|θ−αn| |q0 1/q0

.

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On a Property of Harmonic Functions from the Smirnov Class 93

Taking into account that γk < q10, the last inequality, (3) and (9) allow us to conclude that sup

n

I1n <∞. Just in the same way we can establish that sup

n I2n<∞. The estimate forI3n is obvious.

Further,

I4n =

βen

Z

e αn

0(rne)|dθ≤ Z

Θ(γab)

0(rne)|dθ,

and from (3) it follows that sup

n

I4n <∞. Thus sup

n

Ijn <∞, j = 1,4, and therefore (7), (71) and (72) show that the inequality (6) is valid. In particular, we conclude that angular boundary valuesφ0(t) exist almost everywhere onγeaebfor anyea,eb∈γabat whichφ+(ea), φ+(eb) exist. Since sucheaandeblie arbitrarily close toaandb, lim

r→1φ0(re) exists almost everywhere on Θ(γab). By Fatou’s lemma, the expressions (3) yield

Z

Θ(γab)

0(e2(e)|qdθ <∞. (11)

In view of the inequalities−1q < γk <q10,−1q < βk< q10, it is not difficult to establish the existence ofε,ε >0, such that

Z

Θ(γab)

0(e)|1+εdθ≤M <∞. (12)

Sinceφ0 ∈E1(G), the functionφ(z) is continuous onGandφ(t) =φ+(t) is absolutely continuous on the boundary ofG(see, e.g., [4], p. 208). Thus φ(t) is absolutely continuous on the arcsγeaeb and, consequently, is such on every closed arc lying onγab. Moreover,

φ(e) = Zθ

e α

φ0θ(e)dθ−φ(eiαe), αe≤θ≤β.e (13) From (12) and (13) it follows that the limits

θ→(arglima)+φ(e) =φ(a+), lim

θ→(argb)−φ(e) =φ(b−)

exist. Therefore the representation (13) is valid for any θ, e ∈ γab if we replace φ(eiαe) by φ(a+). Hence φ(t) is absolutely continuous on γab. Moreover, the inequality (11) holds. Sinceu(z) = Reφ(z), this implies that all the assertions of the theorem about the function u(z) are true.

Incidentally, we have proved the following

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Statement 2. Ifφ∈H1 and for someε >0 sup

r

Zβ

α

|Reφ0(re)|1+εdθ <∞, 0≤α < β≤2π,

thenφ(z)is continuously extendable to every closed arc lying on the arcγab

witha=e, b=e, there exist the limits

t→a+lim φ+(t) =φ+(a+), lim

t→b−φ+(t) =φ+(b−) and the functionφ+(t)is absolutely continuous on γab.

Let z = t(s) be the equation of the curve L with respect to the arc coordinate. Taking into account the property of the absolute continuity of the function w(t(s)) with respect to s on [0, l] and of the function z(e) with respect to θ on [0,2π], due to the fact that in the case of Lyapunov curves we have 0 < m ≤ |z0(w)| ≤ M (see, e.g., [3], pp. 405, 407, 411), one can, using the above-proven theorem, establish that the statement of the above theorem is valid for any functions of the classe(L1p1), L02q2)) ifL is Lyapunov curve.

References

1. G. Khuskivadze and V. Paatashvili,Zaremba’s problem in one class of harmonic functions.Proc. A. Razmadze Math. Inst.132(2003), 143–147.

2. G. Khuskivadze and V. Paatashvili,On Zaremba’s boundary value problem for harmonic functions of Smirnov classes.Mem. Differential Equations Math. Phys.

32(2004), 29–58.

3. G. M. Goluzin,Geometric Theory of Functions of a Complex Variable. (Russian) Nauka, Moscow, 1966.

4. I. I. Privalov, Boundary property of one-valued analytic functions. (Russian)Nauka, Moscow, 1950.

(Received 24.06.2004) Authors’ address:

A. Razmadze Mathematical Institute Georgian Academy of Sciences 1, Aleksidze St., Tbilisi 0193 Georgia

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