The Cauchy-Stieltjes integrals in the theory of analytic functions
We study various Stieltjes integrals as Poisson–Stieltjes, conjugate Poisson–Stieltjes, Schwartz–Stieltjes and Cauchy–Stieltjes and prove theorems on the existence of their finite angular limits a.e. in terms of the Hilbert–Stieltjes integral. These results hold for arbitrary bounded integrands that...
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irk-123456789-1693772020-06-12T01:26:15Z The Cauchy-Stieltjes integrals in the theory of analytic functions Ryazanov, V.I. We study various Stieltjes integrals as Poisson–Stieltjes, conjugate Poisson–Stieltjes, Schwartz–Stieltjes and Cauchy–Stieltjes and prove theorems on the existence of their finite angular limits a.e. in terms of the Hilbert–Stieltjes integral. These results hold for arbitrary bounded integrands that are differentiable a.e. and, in particular, for integrands of the class CBV (countably bounded variation). 2017 Article The Cauchy-Stieltjes integrals in the theory of analytic functions / V.I. Ryazanov // Український математичний вісник. — 2017. — Т. 14, № 4. — С. 548-563. — Бібліогр.: 29 назв. — англ. 1810-3200 2010 MSC. 30С62, 31A05, 31A20, 31A25, 31B25, 35Q15; Secondary 30E25, 31C05, 34M50, 35F45. http://dspace.nbuv.gov.ua/handle/123456789/169377 en Український математичний вісник Інститут прикладної математики і механіки НАН України |
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We study various Stieltjes integrals as Poisson–Stieltjes, conjugate Poisson–Stieltjes, Schwartz–Stieltjes and Cauchy–Stieltjes and prove theorems on the existence of their finite angular limits a.e. in terms of the Hilbert–Stieltjes integral. These results hold for arbitrary bounded integrands that are differentiable a.e. and, in particular, for integrands of the class CBV (countably bounded variation). |
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Ryazanov, V.I. The Cauchy-Stieltjes integrals in the theory of analytic functions Український математичний вісник |
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Ryazanov, V.I. |
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Ryazanov, V.I. |
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The Cauchy-Stieltjes integrals in the theory of analytic functions |
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The Cauchy-Stieltjes integrals in the theory of analytic functions |
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The Cauchy-Stieltjes integrals in the theory of analytic functions |
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The Cauchy-Stieltjes integrals in the theory of analytic functions |
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The Cauchy-Stieltjes integrals in the theory of analytic functions |
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cauchy-stieltjes integrals in the theory of analytic functions |
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Інститут прикладної математики і механіки НАН України |
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The Cauchy-Stieltjes integrals in the theory of analytic functions / V.I. Ryazanov // Український математичний вісник. — 2017. — Т. 14, № 4. — С. 548-563. — Бібліогр.: 29 назв. — англ. |
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AT ryazanovvi thecauchystieltjesintegralsinthetheoryofanalyticfunctions AT ryazanovvi cauchystieltjesintegralsinthetheoryofanalyticfunctions |
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2025-07-15T04:07:14Z |
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Український математичний вiсник
Том 14 (2017), № 4, 548 – 563
The Cauchy–Stieltjes integrals
in the theory of analytic functions
Vladimir I. Ryazanov
(Presented by V. O. Derkach)
Abstract. We study various Stieltjes integrals as Poisson–Stieltjes,
conjugate Poisson–Stieltjes, Schwartz–Stieltjes and Cauchy–Stieltjes and
prove theorems on the existence of their finite angular limits a.e. in terms
of the Hilbert–Stieltjes integral. These results hold for arbitrary bounded
integrands that are differentiable a.e. and, in particular, for integrands
of the class CBV (countably bounded variation).
2010 MSC. 30С62, 31A05, 31A20, 31A25, 31B25, 35Q15; Secondary
30E25, 31C05, 34M50, 35F45.
Key words and phrases. Stieltjes, Poisson–Stieltjes, Schwartz–Stieltjes,
Cauchy–Stieltjes and Hilbert–Stieltjes integrals, harmonic and analytic
functions, angular limits.
1. Introduction
Recall that a path in D := {z ∈ C : |z| < 1} terminating at ζ ∈ ∂D is
called nontangential if its part in a neighborhood of ζ lies inside of an
angle in D with the vertex at ζ. Hence the limit along all nontangential
paths at ζ ∈ ∂D also named angular at the point. The latter is a
traditional tool of the geometric function theory, see e.g. monographs
[3, 11,14,18] and [21].
It was proved in the previous paper [24] that a harmonic function
u given in the unit disk D of the complex plane C has angular limits
at a.e. point ζ ∈ ∂D if and only if its conjugate harmonic function
v in D is so. This is the key fact together with Lemma 2 in Section 5
further to establish the existence of the so–called Hilbert–Stieltjes integral
for a.e. ζ ∈ ∂D and the corresponding result on the angular limits of
Cauchy–Stieltjes integral under fairly general assumptions on integrands,
cf. e.g. [8, 19] and [27], see also [4, 10,11] and [21].
Received 08.08.2017
ISSN 1810 – 3200. c⃝ Iнститут прикладної математики i механiки НАН України
V. I. Ryazanov 549
2. Expansion of the Riemann–Stieltjes integral
First of all, recall a classical definition of the Riemann–Stieltjes inte-
gral. Namely, let I = [a,b] be a compact interval in R. A partition P of
I is a collection of points t0, t1, . . . , tp ∈ I such that a = t0 ≤ t1 ≤ . . . ≤
tp = b. Now, let g : I → R and f : I → R be bounded functions and let
f be in addition nondecreasing. The Riemann–Stieltjes integral of g
with respect to f is a real number A, written A =
∫
I
g df , if, for every
ε > 0, there is δ > 0 such that
|
p∑
k=1
g(τk)[f(tk)− f(tk−1)] − A | < ε (2.1)
for every partition P = {t0, t1, . . . , tp} of I with |tk − tk−1| ≤ δ, k =
1, . . . , p, and every collection τk ∈ [tk−1, tk], k = 1, . . . , p. In other words
and notations,
b∫
a
g df := lim
δ→0
p∑
k=1
g(τk)·∆kf as δ := max
k=1,...,p
|tk−tk−1| → 0 (2.2)
where ∆kf := f(tk) − f(tk−1), k = 1, . . . , p, if a finite limit in (2.2)
exists and it is uniform with respect to partitions {tk} and intermediate
points {τk}. We extend the definition of the Riemann–Stieltjes integral
to arbitrary functions g and f for which the limit (2.2) exists. Let us
start from the following general fact, cf. e.g. [1, 9, 17] and [26].
Lemma 1. Let I = [a,b], g : I → R and f : I → R be arbitrary
functions. If one of the integrals
∫
g df and
∫
f dg exists, then the second
one is so and∫
I
g df +
∫
I
f dg = g(b) · f(b) − g(a) · f(a). (2.3)
Proof. For definiteness, let us assume that there exists the integral
∫
f dg.
Then the identity for the arbitrary integral sum of the integral
∫
g df :
p∑
k=1
g(τk) · (f(tk)− f(tk−1)) = −g(t0)f(t0)
−
p+1∑
k=1
f(tk) · (g(τk)− g(τk−1)) + g(tp)f(tp),
where τ0 := t0 and τp+1 = tp, implies the desired conclusions.
550 Cauchy–Stieltjes integrals and analytic functions
By Theorem 13.1.b in [9] we have the significant consequence of
Lemma 1.
Proposition 1. Let I = [a,b], a function f : I → R be absolutely
continuous and let g : I → R be a bounded function whose set of points
of discontinuity is of measure zero. Then both integrals
∫
I
g df and
∫
I
f dg
exist and the relation (2.3) holds.
Remark 2. It is clear that this formula (2.3) is also valid for complex
valued functions on rectifiable curves because their real and imaginary
parts as functions of the natural parameter can be considered as real val-
ued functions on segments of R. Moreover, the corresponding statements
hold on closed rectfiable Jordan curves J with the relation∫
J
g df = −
∫
J
f dg (2.4)
where we should apply cyclic partitions P of J by collections of cyclic
ordered points ζ0, ζ1, . . . , ζp on J with ζ0 = ζp.
3. On the Poisson–Stieltjes integrals
Recall that the Poisson kernel is the 2π−periodic function
Pr(Θ) =
1− r2
1− 2r cosΘ + r2
, r < 1 , Θ ∈ R. (3.1)
By Proposition 1 and Remark 1, the Poisson-Stieltjes integral
U(z) = UΦ(z) :=
1
2π
π∫
−π
Pr(ϑ− t) dΦ(t) , z = reiϑ, r < 1 , ϑ ∈ R
(3.2)
is well-defined for 2π−periodic continuous functions, furthermore, for
bounded functions Φ : R → R whose set of points of discontinuity is
of measure zero because the function Pr(Θ) is continuously differentiable
and hence it is absolutely continuous.
Moreover, directly by the definition of the Riemann-Stieltjes integral
and the Weierstrass type theorem for harmonic functions, see e.g. Theo-
rem I.3.1 in [7], U is a harmonic function in the unit disk D := {z ∈ C :
|z| < 1} because the functions Pr(ϑ − t) is the real part of the analytic
function
Aζ(z) :=
ζ + z
ζ − z
, ζ = eit, z = re1ϑ, r < 1, ϑ and t ∈ R. (3.3)
V. I. Ryazanov 551
Theorem 1. Let Φ : R → R be a 2π−periodic bounded function
whose set of points of discontinuity has measure zero. Suppose that Φ is
differentiable at a point t0 ∈ R. Then
lim
z→ζ0
UΦ(z) = Φ′(t0) (3.4)
along all nontangential paths in D to the point ζ0 := eit0 ∈ ∂D.
Proof. Indeed, by Proposition 1 with g(t) := Φ(t) and f(t) := Pr(ϑ− t),
t ∈ R, for every fixed z = reiϑ, r < 1, ϑ ∈ R, we obtain that
π∫
−π
Pr(ϑ−t) dΦ(t) =
π∫
−π
Φ(t)· ∂
∂ϑ
Pr(ϑ−t) dt ∀ r ∈ (0, 1) , ϑ ∈ R, (3.5)
because of the 2π−periodicity of the given functions g and f the right
hand side in (2.3) is equal to zero, f ∈ C1 and
∂
∂ϑ
Pr(ϑ− t) = − ∂
∂t
Pr(ϑ− t) ∀ r ∈ (0, 1) , ϑ and t ∈ R.
Now, considering the Poisson integral
u(reiϑ) :=
1
2π
π∫
−π
Pr(ϑ− t) Φ(t) dt,
we see by the Fatou result, see e.g. 3.441 in in [28], p. 53, or Theorem
IX.1.2 in [7], that ∂
∂ϑ u(z) → Φ′(t0) as z → ζ0 along any nontangential
path in D ending at ζ0. Thus, the conclusion follows because just UΦ(z) =
∂
∂ϑ u(z) by (3.5).
Corollary 1. If Φ : R → R is a 2π−periodic continuous differentiable
a.e. function, then (3.4) holds for a.e. ζ ∈ ∂D along all nontangential
paths in D to the point ζ.
Here we denote by arg ζ the principal branch of the argument of
ζ ∈ C with |ζ| = 1, i.e., the unique number τ ∈ (−π, π] such that ζ = eiτ .
Remark 2. Note that the function of interval Φ∗([a, b]) := Φ(b) −
Φ(a) generally speaking generates no finite signed measure (charge) if Φ
is not of bounded variation. Hence we cannot apply the known Fatou
result on the angular boundary limits directly to the Poisson–Stieltjes
integrals, see e.g. Theorem I.D.3 in [11].
552 Cauchy–Stieltjes integrals and analytic functions
Corollary 2. If Φ : R → R is a 2π−periodic bounded function that
is differentiable a.e., then UΦ(z) → Φ′(arg ζ) as z → ζ for a.e. ζ ∈ ∂D
along all nontangential paths in D to the point ζ.
We call Φ : R → C a function of countably bounded variation and
write Φ ∈ CBV(R) if there is a countable collection of mutually disjoint
intervals (an, bn), n = 1, 2, . . . on each of which the restriction of Φ is of
bounded variation and the set R \
∞∪
1
(an, bn) is countable.
Corollary 3. If Φ : R → R is a 2π−periodic bounded function of the
class CBV(R), then UΦ(z) → Φ′(arg ζ) as z → ζ for a.e. ζ ∈ ∂D along
all nontangential paths in D to the point ζ.
4. On the conjugate Poisson–Stieltjes integrals
Recall that the conjugate Poisson kernel is the 2π−periodic func-
tion
Qr(Θ) =
2r sinΘ
1− 2r cosΘ + r2
, r < 1 , Θ ∈ R. (4.1)
By Proposition 1 the conjugate Poisson–Stieltjes integral
V(z) = VΦ(z) :=
1
2π
π∫
−π
Qr(ϑ− t) dΦ(t) , z = reiϑ, r < 1 , ϑ ∈ R,
(4.2)
is well-defined for 2π−periodic bounded functions Φ : R → R whose
set of points of discontinuity is of measure zero because the function
Qr(Θ) is continuously differentiable and hence it is absolutely continuous.
Again, directly by the definition of the Riemann–Stieltjes integral and the
Weierstrass type theorem VΦ is a conjugate harmonic function for UΦ in
the unit disk D because the function Qr(ϑ − t) is the imaginary part of
the same analytic function (3.3).
By Theorem 1 in [24] we have the following significant consequences
from Theorem 1 and Corollaries 1–3.
Corollary 4. Let Φ : R → R be a 2π−periodic continuous function
that is differentiable a.e. Then VΦ(z) has a finite limit φ(ζ) as z → ζ
along nontangential paths for a.e. ζ ∈ ∂D.
Corollary 5. Let Φ : R → R be a 2π−periodic bounded function that
is differentiable a.e. Then VΦ(z) has a finite limit φ(ζ) as z → ζ along
nontangential paths for a.e. ζ ∈ ∂D.
V. I. Ryazanov 553
Corollary 6. Let Φ : R → R be a 2π−periodic bounded function of
the class CBV(R). Then VΦ(z) has a finite angular limit φ(ζ) as z → ζ
for a.e. ζ ∈ ∂D.
The function φ(ζ) will be calculated in the explicit form through Φ(ζ)
in terms of the so–called Hilbert–Stieltjes integral. To prove this fact we
need first to establish one auxiliary result in the next section.
5. On the Hilbert–Stieltjes integral
Lemma 2. Let Φ : R → R be a 2π−periodic bounded function
whose set of points of discontinuity has measure zero. Suppose that Φ is
differentiable at a point t0 ∈ R. Then the difference
VΦ(z) − 1
π
π∫
1−|z|
d { Φ(t0 − t)− Φ(t0 + t)}
2 tan t
2
(5.1)
converges to zero as z → ζ0 := eit0 ∈ ∂D along the radius in D to the
point ζ0.
Proof. First of all, in the case of need applying simultaneous rotations ζ0
to ζ = eit ∈ ∂D and z ∈ D in (3.3), we may assume that t0 = 0. Moreover,
with no loss of generality we may assume that Φ(0) = 0 and Φ′(0) = 0
because, for the linear function Φ∗(t) := Φ(0) + Φ′(0) · t : (−π, π] → R
extended 2π−periodically to R, dΦ∗(t) ≡ Φ′(0) dt, gives identical zero in
the difference (5.1) in view of the oddness of the kernel Qr and tan t
2 .
Note that by the oddness of Qr we have also that
VΦ(r) =
1
2π
π∫
−π
Qr(−t) dΦ(t) =
1
2π
π∫
−π
Qr(t) dΦ(−t) , ∀ r ∈ (0, 1).
Then the difference (5.1) is split into two parts with ε = ε(r) := 1− r :
I :=
1
2π
ε∫
−ε
Qr(t) dΦ(−t),
II :=
1
2π
∫
ε≤|t|≤π
{Qr(t)−Q1(t)} d {Φ(−t)− Φ(t)}.
554 Cauchy–Stieltjes integrals and analytic functions
Integrating I by parts, we have by Proposition 1 and Remark 1 and
the oddness of Qr(t)
I =
1
2π
Qr(ε){Φ(−ε)− Φ(ε)} +
1
π
ε∫
0
{Φ(−t)− Φ(t)} dQr(t). (5.2)
The first summand converges to zero as ε → 0 because |Φ(±ε)| = o(ε)
and
Qr(ε) =
2r sin ε
1− 2r cos ε+ r2
=
2r sin ε
ε2 + 4r sin2 ε
2
≤ 2
sin ε
ε2
≤ 2
ε
. (5.3)
To estimate the second summand in (5.2) note that sin2 t
2 ≤
[
1−r
2
]2
and,
thus,
Q′
r(t) =
2r cos t
1− 2r cos t+ r2
− 4r2 sin2 t
(1− 2r cos t+ r2)2
= 2r
(1 + r2) cos t− 2r
(1− 2r cos t+ r2)2
= 2r
(1− r)2 − 2(1 + r2) sin2 t
2
(1− 2r cos t+ r2)2
≥ 2r
(1− r)2[1− (1 + r2)/2]
(1− 2r cos t+ r2)2
=
r(1 + r)(1− r)3
(1− 2r cos t+ r2)2
,
i.e., Q′
r(t) > 0 for all t ∈ [0, ε]. Since Qr(t) is smooth, it is strictly
increasing on [0, ε]. Hence the modulus of the second summand has the
upper bound
1
π
·Qr(ε) · sup
t∈[0,ε]
{|Φ(−t)|+ |Φ(t)|} ≤ 1
π
· 2
ε
· o(ε) = o(1)
where the inequality follows by (5.3). Thus, the second summand in (5.2)
also converges to zero as ε→ 0.
Now, by oddness of the kernels Qr(t), r ∈ (0, 1) and Q1(t) we obtain
that
II :=
1
π
π∫
ε
{Qr(t)−Q1(t)} d {Φ(−t)− Φ(t)}
where
Q1(t)−Qr(t) =
2 sin t
2(1− cos t)
− 2r sin t
ε2 + 2r(1− cos t)
=
2 sin t
4 sin2 t
2
− 2r sin t
ε2 + 4r sin2 t
2
=
2ε2 sin t
4
(
ε2 + 4r sin2 t
2
)
sin2 t
2
.
V. I. Ryazanov 555
Integrating by parts, we see that the latter integral is equal to
{Q1(ε)−Qr(ε)}·{Φ(−ε)−Φ(ε)} +
π∫
ε
{Φ(−t)−Φ(t)} d {Q1(t)−Qr(t)}.
Here the first summand converges to zero because Φ(−ε) − Φ(ε) = o(ε)
and
Q1(ε)−Qr(ε) =
2ε2 sin ε
4(ε2 + 4r sin2 ε
2) sin
2 ε
2
∼ 1
ε
as ε→ 0. (5.4)
Thus, it remains to estimate the integral
III :=
π∫
ε
{Φ(−t)− Φ(t)} d {Q1(t)−Qr(t)} =
π∫
ε
φ(t) dαr(t)
where φ(t) = Φ(−t) − Φ(t) and αr(t) = Q1(t) − Qr(t). To make it first
of all note that α′
r(t) < 0 for t ∈ (ε, π) because of
α′
r(t) =
2ε2 cos t
4(ε2 + 4r sin2 t
2) sin
2 t
2
−
2ε2 sin2 t (ε2 + 8r sin2 t
2)
4[(ε2 + 4r sin2 t
2) sin
2 t
2 ]
2
= 2· ε
2
δ2
·
[
cos t · sin2 t
2
·
(
ε2 + 4r sin2
t
2
)
− sin2 t ·
(
ε2 + 8r sin2
t
2
)]
= 2 · ε
2
δ2
·
[
ε2 ·
(
cos t · sin2 t
2
− sin2 t
)
− 4r sin2
t
2
·
(
2 sin2 t− cos t · sin2 t
2
)]
= 2 · ε
2
δ2
·
[
ε2 · sin2 t
2
·
(
cos t− 4 cos2
t
2
)
− 4r sin4
t
2
·
(
8 cos2
t
2
− cos t
)]
= −2 · ε
2
δ2
· sin2 t
2
·
[
ε2 ·
(
1 + 2 cos2
t
2
)
+ 4r sin2
t
2
·
(
1 + 6 cos2
t
2
)]
where we many times applied the trigonometric identities
sin t = 2 sin t
2 cos
t
2 and 1 − cos t = 2 sin2 t
2 , 1 + cos t = 2 cos2 t
2 , and
the notation
δ := 2 sin2
t
2
(
ε2 + 4r sin2
t
2
)
.
The above expression for α′
r(t) also implies that |α′
r(t)| ≤ c · ε2
t4
. Thus,
|III| ≤ c · ε2
π∫
ε
|φ(t)| d t
t4
.
556 Cauchy–Stieltjes integrals and analytic functions
Let us fix an arbitrary ϵ > 0 and choose a small enough η > 0 such
that |φ(t)|/t < ϵ for all t ∈ (0, η). Note that we may assume here
that ε < η2
√
ϵ for small enough ε. Consequently, we have the following
estimates
|III| ≤ c · ε2 · ϵ
η∫
ε
d t
t3
+ c · ε2
π∫
η
|φ(t)| d t
t4
≤ c
2
· ϵ + cπ · ϵ ·M
where M = sup
t∈[0,π]
|φ(t)|. In view of arbitrariness of ε and ϵ, we conclude
that the integral III converges to zero as ε→ 0.
Theorem 2. Let Φ : R → R be a 2π−periodic bounded function.
Suppose that Φ is differentiable a.e. Then
lim
z→ξ
VΦ(z) =
1
π
π∫
0
d {Φ(τ − t)− Φ(τ + t)}
2 tan t
2
, ξ := eiτ ∈ ∂D, (5.5)
for a.e. τ ∈ R along all nontangential paths in D to the point ξ.
Here the singular integral from the right hand side in (5.5) is under-
stood as a limit of the corresponding proper integrals (principal value
by Cauchy):
HΦ(τ) :=
1
π
lim
ε→+0
π∫
ε
d {Φ(τ − t)− Φ(τ + t)}
2 tan t
2
=
1
2π
lim
ε→0
∫
|τ−t|≥ε
dΦ(t)
tan τ−t
2
.
(5.6)
We call it as the Hilbert–Stieltjes integral of the function Φ at the
point τ .
Proof. The conclusion of Theorem 2 follows immediately from Lemma 2
and Corollary 5.
Corollary 7. The Hilbert–Stieltjes integral converges a.e. for every
2π−periodic bounded function Φ : R → R that is differentiable a.e.
Remark 3. In particular, the conclusions of Theorem 2 as well as
Corollary 7 hold for every 2π−periodic bounded function of the class
CBV(R).
Of course, Lemmas 1–2, Theorems 1–2, Corollaries 1–7 and the def-
inition of the Hilbert–Stieltjes integral are extended in the natural way
to complex valued functions Φ.
V. I. Ryazanov 557
6. On Schwartz–Stieltjes and Cauchy–Stieltjes integrals
Given a 2π−periodic bounded function Φ : R → R whose set of points
of discontinuity has measure zero, the Schwartz–Stieltjes integral
SΦ(z) :=
1
2π
π∫
−π
eit + z
eit − z
dΦ(t) , z ∈ D, (6.1)
is well–defined by the previous sections and the function SΦ(z) is analytic
by the definition of the Riemann–Stieltjes integral and the Weierstrass
theorem, see e.g. Theorem I.1.1 in [7]. By Theorem 2 and Corollary 2 we
have also the following.
Corollary 8. Let Φ : R → R be a 2π−periodic bounded function that
is differentiable a.e. Then SΦ(z) has finite angular limit Φ′(arg ζ) + i ·
HΦ(arg ζ) as z → ζ for a.e. ζ ∈ ∂D.
It is clear that the definition (6.1), as well as Corollary 8, is extended
in the natural way to the case of the complex valued functions Φ.
Given a 2π−periodic bounded function Φ : R → C whose set of points
of discontinuity has measure zero, we see that the integral
CΦ(z) :=
1
2π
π∫
−π
eitdΦ(t)
eit − z
, z ∈ D, (6.2)
is also well–defined and we call it by the Cauchy–Stieltjes integral. It
is easy to see that CΦ(z) =
1
2 SΦ(z).
Corollary 9. Let Φ : R → C be a 2π−periodic bounded function that
is differentiable a.e. Then
lim
z→ζ
CΦ(z) =
1
2
{Φ′(arg ζ) + i ·HΦ(arg ζ)} (6.3)
for a.e. ζ ∈ ∂D along all nontangential paths in D to the point ζ.
In this connection, note that the Hilbert–Stieltjes integral can be
described in another way for functions Φ of bounded variation.
Namely, let us denote by C(ζ0, ε), ε ∈ (0, 1), ζ0 ∈ ∂D, the rest of the
unit circle ∂D after removing its arc A(ζ0, ε) := {ζ ∈ ∂D : |ζ − ζ0| < ε}
and, setting
IΦ(ζ0, ε) =
1
2π
∫
C(ζ0,ε)
ζdΦ∗(ζ)
ζ − ζ0
, ζ0 ∈ ∂D , where Φ∗(ζ) := Φ(arg ζ),
558 Cauchy–Stieltjes integrals and analytic functions
define the singular integral of the Cauchy–Stieltjes type
IΦ(ζ0) =
1
2π
∫
∂D
ζ dΦ
ζ − ζ0
, ζ0 ∈ ∂D,
as a limit of the integrals I(ζ0, ε) as ε→ 0. By paper [19] we have that
lim
z→ζ
CΦ(z) =
1
2
· Φ′(arg ζ) + i · IΦ(ζ) (6.4)
for a.e. ζ ∈ ∂D along all nontangential paths in D to the point ζ. Com-
paring the relations (6.3) and (6.4), we come to the following conclusion.
Corollary 10. Let Φ : R → C be a 2π−periodic function with
bounded variation on [−π, π]. Then for a.e. τ ∈ [−π, π]
HΦ(τ) = 2 · IΦ(eiτ ). (6.5)
7. Representation of the Luzin construction
The following deep (non–trivial) result of Luzin was one of the main
theorems of his dissertation, see e.g. his paper [12], dissertation [13], p.
35, and its reprint [14, p. 78], where one may assume that Φ(0) = Φ(1) =
0, cf. also [25].
Theorem A. For any measurable function φ : [0, 1] → R, there is
a continuous function Φ : [0, 1] → R such that Φ′ = φ a.e.
Just on the basis of Theorem A, Luzin proved the next significant
result of his dissertation, see e.g. [14, p. 80], that was key to establish the
corresponding result on the boundary value Hilbert problem for analytic
functions in [22].
Theorem B. Let φ(ϑ) be real, measurable, almost everywhere finite
and have the period 2π. Then there exists a harmonic function U in the
unit disk D such that U(z) → φ(ϑ) for a.e. ϑ as z → eiϑ along any
nontangential path.
Note that the Luzin dissertation was published in Russian as the
book [14] with comments of his pupils Bari and Men’shov only after his
death. A part of its results was also printed in Italian [15]. However,
Theorem A was published with a complete proof in English in the book
V. I. Ryazanov 559
[26] as Theorem VII (2.3). Hence Frederick Gehring in [6] has rediscovered
Theorem B and his proof on the basis of Theorem A has in fact coincided
with the original proof of Luzin. Since the proof is very short and nice
and has a common interest, we give it for completeness here.
Proof. By Theorem A we can find a continuous function Φ(ϑ) such that
Φ′(ϑ) = φ(ϑ) for a.e. ϑ. Considering the Poisson integral
u(reiϑ) =
1
2π
2π∫
0
1− r2
1− 2r cos(ϑ− t) + r2
Φ(t) dt
for 0 < r < 1, u(0) := 0, we see by the Fatou result, see e.g. 3.441 in [28],
p. 53, or Theorem IX.1.2 in [7], that ∂
∂ϑ u(z) → Φ′(ϑ) as z → eiϑ along
any nontangential path whenever Φ′(ϑ) exists. Thus, the conclusion
follows for the function U(z) = ∂
∂ϑ u(z).
Remark 4. Note that the given function U is harmonic in the punc-
tured unit disk D \ {0} because the function u is harmonic in D and
the differential operator ∂
∂ϑ is commutative with the Laplace operator ∆.
Setting U(0) = 0, we see that
U(reiϑ) = − r
π
2π∫
0
(1− r2) sin(ϑ− t)
(1− 2r cos(ϑ− t) + r2)2
Φ(t) dt → 0 as r → 0,
i.e. U(z) → U(0) as z → 0, and, moreover, the integral of U over
each circle |z| = r, 0 < r < 1, is equal to zero. Thus, by the criterion
for a harmonic function on the averages over circles we have that U is
harmonic in D. The alternative argument for the latter is the removability
of isolated singularities for harmonic functions, see e.g. [16].
Corollary 5.1 in [22] has strengthened Theorem B as the next, see
also [23].
Theorem C. For each (Lebesgue) measurable function φ : ∂D → R,
the space of all harmonic functions u : D → R with the angular limits
φ(ζ) for a.e. ζ ∈ ∂D has the infinite dimension.
Remark 5. One can find in [25] more refined results which are
counterparts of Theorems A, B and C in terms of logarithmic capacity
that makes possible to extend the theory of boundary value problems
560 Cauchy–Stieltjes integrals and analytic functions
to the so-called A-harmonic functions corresponding to generalizations
of the Laplace equation in inhomogeneous and anisotropic media, see
also [29].
By the well-known Lindelöf maximum principle, see e.g. Lemma 1.1 in
[5], it follows the uniqueness theorem for the Dirichlet problem in the class
of bounded harmonic functions u on the unit disk D = {z ∈ C : |z| < 1}.
In general there is no uniqueness theorem in the Dirichlet problem for
the Laplace equation even under zero boundary data.
Many such nontrivial solutions for the Laplace equation can be given
just by the Poisson–Stieltjes integral
UΦ(z) :=
1
2π
2π∫
0
Pr(ϑ− t) dΦ(t) , z = reiϑ, r < 1, (7.1)
with an arbitrary singular function Φ : [0, 2π] → R, i.e., where Φ is
of bounded variation and Φ′ = 0 a.e. Indeed, by the Fatou theorem,
see e.g. Theorem I.D.3.1 in [11], UΦ(z) → Φ′(Θ) as z → eiΘ along any
nontangential path whenever Φ′(Θ) exists, i.e. UΦ(z) → 0 as z → eiΘ for
a.e. Θ ∈ [0, 2π] along any nontangential paths for every singular function
Φ.
Example 1. The first natural example is given by the formula (7.1)
with Φ(t) = φ(t/2π) where φ : [0, 1] → [0, 1] is the well–known Cantor
function, see e.g. [2] and further references therein.
Example 2. However, the simplest example of such a kind is just
u(z) := Pr(ϑ− ϑ0) =
1− r2
1− 2r cos(ϑ− ϑ0) + r2
, z = reiϑ, r < 1.
We see that u(z) → 0 as z → eiΘ for all Θ ∈ (0, 2π) except Θ = ϑ0.
The construction of Luzin can be described as the Poisson–Stieltjes
integral.
Theorem 3. The harmonic function U in Theorem B has the rep-
resentation
UΦ(z) =
1
2π
π∫
−π
Pr(ϑ− t) dΦ(t) ∀ z = reiϑ, r ∈ (0, 1) , ϑ ∈ [−π, π],
(7.2)
V. I. Ryazanov 561
where Φ : [−π, π] → R is the continuous Luzin function with Φ′ = φ a.e.
Corollary 11. The conjugate harmonic function VΦ has finite an-
gular limits
lim
z→ζ
VΦ(z) = HΦ(arg ζ) for a.e. ζ ∈ ∂D.
Proof. Indeed, choosing in Proposition 1 g(t) = Φ(t) and f(t) = Pr(ϑ−t),
t ∈ [−π, π], for every fixed z = reiϑ, r < 1, ϑ ∈ [−π, π], we obtain that
π∫
−π
Φ(t) · ∂
∂ϑ
Pr(ϑ− t) dt =
π∫
−π
Pr(ϑ− t) dΦ(t) ∀ r ∈ (0, 1) , ϑ ∈ [−π, π]
(7.3)
because by the 2π−periodicity of the given function g and f the right
hand side in (2.3) is equal to zero, f ∈ C1, and
∂
∂ϑ
Pr(ϑ− t) = − ∂
∂t
Pr(ϑ− t) ∀ r ∈ (0, 1) , ϑ ∈ [−π, π] , t ∈ [−π, π].
The relation (7.2) follows from (7.3) because it was in the proof of The-
orem B
U(z) =
1
2π
π∫
−π
Φ(t)· ∂
∂ϑ
Pr(ϑ−t) dt ∀ z = reiϑ, r ∈ (0, 1) , ϑ ∈ [−π, π].
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Contact information
Vladimir I.
Ryazanov
Institute of Applied Mathematics
and Mechanics of the NAS of Ukraine,
Slavyansk, Ukraine
E-Mail: vl.ryazanov1@gmail.com
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