1
arXiv:1701.03429v1 [math.CV] 12 Jan 2017
ON SOME RIESZ AND CARLEMAN TYPE INEQUALITIES FOR
HARMONIC FUNCTIONS ON THE UNIT DISK
DAVID KALAJ AND ELVER BAJRAMI
A BSTRACT. We prove some isoperimetric type inequalities for real harmonic
functions in the unit disk belonging to the Hardy space hp , p > 1 and for complex harmonic functions in h4 . The results extend some recent results on the
area. Further we discus some Riesz type results for holomorphic functions.
1. I NTRODUCTION
AND STATEMENT OF MAIN RESULTS
Throughout the paper we let U = {z ∈ C : |z| < 1} be the open unit disk in
the complex plane C, and let T = {z ∈ C : |z| = 1} be the unit circle in C. The
normalized area measure on U will be denoted by dσ. In terms of real (rectangular
and polar) coordinates, we have
1
1
dσ = dxdy = rdrdθ, z = x + iy = reiθ .
π
π
p
For 0 < p < +∞ let L (U, σ) = Lp denote the familiar Lebesgue space on U
with respect to the measure σ. The Bergman space Ap (U) = Ap is the space of all
holomorphic functions in Lp (U, σ). For a fixed 1 ≤ p < +∞, denote by Ap0 the
set of all functions f ∈ Ap for which f (0) = 0. For a function f in Lp (U, σ), we
write
Z
1/p
kf kp =
p
U
|f (z)| dσ
.
Bergman space bp of harmonic functions is defined similarly.
The Hardy space hp is defined as the space of (complex) harmonic functions f
such that
Z 2π
dt
|f (reit )|p )1/p < ∞.
kf khp := sup (
2π
0≤r<1 0
p
If f ∈ h , then by [1, Theorem 6.13], there exists
f (eit ) = lim f (reit ), a.e.
r→1
and f ∈ Lp (T). It can be shown that there hold
Z 2π
Z 2π
dt
p
it p dt
|f (eit )|p .
=
kf khp = lim
|f (re )|
r→1 0
2π
2π
0
p
Similarly we define the Hardy space H of holomorphic functions.
File: kalbaj.tex, printed:
2017-1-13, 1.24
1
2010 Mathematics Subject Classification: Primary 47B35
Key words and phrases. Harmonic functions, Ries inequality, Isoperimetric inequality.
1
ON SOME RIESZ AND CARLEMAN TYPE INEQUALITIES FOR HARMONIC FUNCTIONS
2
The starting point of this paper is the well known isoperimetric inequality for
Jordan domains and isoperimetric inequality for minimal surfaces due to Carleman
[3]. In that paper Carleman, among the other results proved that if u is harmonic
and smooth in U then
Z
Z 2π
1
e2u dxdy ≤
eu dt)2 .
(
4π
U
0
By using a similar approach as Carleman, Strebel ([13]) proved the isoperimetric
inequality for holomorphic functions; that is if f ∈ H 1 (U) then
Z
Z
1
2
|f (z)| dxdy ≤
(1.1)
( |f (eit )|dt)2 .
4π
U
T
This inequality has been proved independently by Mateljević and Pavlović ([11]).
In [6] have been done some generalizations for the space.
In this paper we prove the following results.
Theorem 1.1. Let f be a real harmonic function. For p > 1 we have that
kf kb2p ≤ Cp kf khp
(1.2)
where
Cp = M2p =



π
cos 4p
π
cos 2p
π
cos 4p
π
sin 2p
, if 1 < p ≤ 2;
, if p ≥ 2.
The inequality (1.2) extends the main result in [2], where Bajrami proved the
same result but only for p = 4. But the wrong constant appear in [2], due to a wrong
citation to the Duren approach [5, p. 67-68]. However the same method produces
the same constant namely C4 = 2 sin1 π ≈ 2.56292. We were not able to check
16
if (1.2) is sharp. We are thankful to professor Hedenmalm who drawn attention
to his paper [7], where is treated a problem which suggests that the inequalities
from Theorem 2.1, which we use in order to prove (1.2), are maybe not sharp.
The inequality (1.2) improves similar results for real harmonic functions proved
by Kalaj and Meštrović in [10] and by Chen and Ponnusamy and Wang in [4]. We
expect that the inequality (1.2) is true for complex harmonic mappings for every
p > 1. Kalaj and Meštrović in [10] proved it for p = 2, namely they obtained that
z
1
C2 = 2 sin
π ≈ 1.3. On the same paper the example fa (z) = ℜ
1−az , when a ↑ 1
8
produces the constant C0 = (5/2)1/4 ≈ 1.257, so the constant C2 is not far from
the sharp constant. In this paper we extend it for p = 4.
Namely we have
Theorem 1.2. If f ∈ h4 (U) is a nonzero complex harmonic function then f ∈ b8
and there hold the inequality
kf kb8 ≤
1
π kf kh4 .
2 sin 16
The main ingredient of the proofs are some sharp M. Riesz type inequalities
proved for Hardy space by Verbitsky ([14]):
ON SOME RIESZ AND CARLEMAN TYPE INEQUALITIES FOR HARMONIC FUNCTIONS
3
Proposition 1.3 (Riesz type inequality). Let p > 1. For every holomorphic mapping there hold the sharp inequalities
Lp kℜf khp ≤ kf khp ≤ Rp kℜf khp ,
provided |arg f (0) − π2 | ≥
Rp =
(
1
π ,
cos 2p
1
π ,
sin 2p
π
2p̄ ,
p
or f (0) = 0, where p̄ = max{p, p−1
} and
if 1 < p ≤ 2;
if p ≥ 2.
and Lp =
(
1
π ,
sin 2p
1
π ,
cos 2p
if 1 < p ≤ 2;
if p ≥ 2.
In addition we refer to the references [8] and [12] for related results.
By results of Verbitsky we prove some similar inequalities for Bergman space
(Theorem 2.1).
From Proposition 1.3, we obtain that, if f = u + iv is holomorphic with f (0) =
0, then
kf kpH p ≤
(1.3)
Rpp
(kukphp + kvkphp )
2
and
Lpp
(kukphp + kvkphp ) ≤ kf kpH p .
2
(1.4)
Now we formulate a similar result, where
respectively bigger constants for p close to 4.
Rpp
2
and
Lpp
2
are replaced by smaller,
Theorem 1.4. Let p ≥ 2 and let f = u + iv be a holomorphic function on the unit
disk so that f (0) = 0 and let f ∈ H p . Then for p ∈ [2, 4] we have
1/p
1/p
p
p
(1.5)
kf kH ≤
kukphp + kvkphp
p−1
and for p ≥ 4 we have
(1.6)
kf kH p ≥
p
p−1
1/p
kukphp + kvkphp
1/p
.
1/p
p
, then 2−1/p Lp ≤ Cp if p ≥ 4 and Cp ≤ 2−1/p Rp ,
Remark 1.5. If Cp = p−1
if p1 ≤ p ≤ 4, so inequalities (1.5) and (1.6) improve the inequalities (1.3) and
(1.4), if p1 ≤ p ≤ 4, and if p ≥ 4, respectively. Here p1 ≈ 2.42484 is the only
solution of the equation
1/p
1
p
= 2−1/p
π , p ≥ 2.
p−1
sin 2p
The proofs are presented in sections 2 and 3 and 4.
ON SOME RIESZ AND CARLEMAN TYPE INEQUALITIES FOR HARMONIC FUNCTIONS
2. P ROOF
OF
4
T HEOREM 1.1
From Proposition 1.3 we obtain
Z
Z
Z
|f (reit )|p dt ≤ Rpp
|ℜf (reit )|p dt.
Lpp
|ℜf (reit )|p dt ≤
T
T
T
So
Lpp
Z
0
1
rdr
Z
it p
T
Z
1Z
r|f (reit )|p dt
0
T
Z
Z 1
|ℜf (reit )|p dt.
≤ Rpp
rdr
|ℜf (re )| dt ≤
0
T
Thus we have the following theorem
Theorem 2.1. Let p > 2 and f = ℜf + iℑf ∈ bp (U). If |arg f (0) − π2 | ≥
f (0) = 0, then we have
Lp kℜf kbp ≤ kf kbp ≤ Rp kℜf kbp
and
Lp kℑf kbp ≤ kf kbp ≤ Rp kℑf kbp .
From now on we will use the shorthand notation
Z 2π
Z
1
f (eit )dt
f :=
2π
0
T
and
1
f :=
π
U
Z
Z
f (z)dxdy, z = x + iy.
U
Thus for p > 2 we have
Z
Z
1
( |f |p )1/p
( |ℜf |p )1/p ≤
Lp U
U
Z
1
≤
( |f |p/2 )2/p
Lp T
Z
Rp/2
( |ℜf |p/2 )2/p
≤
Lp
Z T
= Mp ( |ℜf |p/2 )2/p ,
T
where
Mp =



π
cos 2p
cos πp
π
cos 2p
sin πp
This finishes the proof of Theorem 1.1.
, if 2 < p ≤ 4;
, if p ≥ 4.
π
2p ,
or
ON SOME RIESZ AND CARLEMAN TYPE INEQUALITIES FOR HARMONIC FUNCTIONS
3. P ROOF
OF
5
T HEOREM 1.2
We assume that f (z) = g(z) + h(z), where h(0) = 0, and g and h are holomorphic on the unit disk.
Lemma 3.1. The function |a|2 + |b|2 is log-subharmonic, provided that a and b are
analytic.
Proof. We need to show that f (z) = log(|a|2 + |b|2 ) is subharmonic. By calculation we find
a′ ā + b′ b̄
fz = 2
|a| + |b|2
and so
a′ ā + b′ b̄ aā′ + bb̄′
a′ ā′ + b′ b̄′
− 2
fz z̄ = 2
2
|a| + |b|
|a| + |b|2 |a|2 + |b|2
(|a′ |2 + |b′ |2 )(|a|2 + |b|2 ) − |āa′ + b̄b′ |2
(|a|2 + |b|2 )2
which is clearly positive.
=
Now from the isoperimetric inequality for log-subharmonic functions (e.g. [9,
Lemma 2.2]), we have
Lemma 3.2. For every positive number p and analytic functions a and b defined
on the unit disk U we have that
2
Z
Z
2
2 p
2
2 2p
(|a| + |b| )
.
(|a| + |b| ) ≤
T
U
Further let
L=
Then
L=
≤
Z
2 4
(|g + h̄| ) =
k=0
4 X
k=0
U
U
4 Z
X
4
k
4
k
Z
Z
U
(|g|2 + |h|2 )k (2ℜ(gh))4−k
2
U
(|g|2 + |h|2 + 2ℜ(gh))4 .
2 4 k/4
((|g| + |h| ) )
Z
( |2ℜ(gh)|4 )(4−k)/4 .
U
Let E4 = cos π8 . From Lemma 3.2 and Theorem 2.1 we have
Z
Z
4 X
4
4−k
2
2 4 k/4
L≤
E4 ( (|g| + |h| ) ) ( (2|gh|)4 )(4−k)/4
k
U
U
k=0
Z
Z
4 X
4
≤
E44−k ( (|g|2 + |h|2 )4/2 )2k/4 ( (2|gh|)4/2 )2(4−k)/4 .
k
T
T
k=0
ON SOME RIESZ AND CARLEMAN TYPE INEQUALITIES FOR HARMONIC FUNCTIONS
Further we have
X :=
=
=
=
Z
Z
T
|g + h̄|4
T
(|g|2 + |h|2 + 2ℜ(gh))2
Z
ZT
T
(|g|2 + |h|2 )2 + 4(|g|2 + |h|2 )ℜ(gh)2 + 4(ℜ(gh))2
(|g|2 + |h|2 )2 + 4(|g|2 + |h|2 )ℜ(gh) + 2|gh|2 .
Let
A=
and
Z
B=
2
T
(|g| + |h| )
Z
2
T
Then
2 2
2
4|g| |h|
1/2
1/2
.
Z
Z
Z
2
2
(|g| + |h| )ℜ(gh) ≤ |( ((|g|2 + |h|2 ))2 )1/2 ( |gh|2 /2)1/2
T
T
T
√
2
= AB ·
.
4
Further we have
√
2 2
B2 √
2
X≥A +
− 2AB = (A −
B) .
2
2
Furthermore
Z
A2 − B 2 =
and thus
T
(|g|2 − |h|2 )2 ≥ 0
√
2− 2 2 2
) B
X≥(
2
(3.1)
and similarly
X≥
(3.2)
√ !2
2− 2
A2 .
2
Hence
Z
4 X
4
4 Z
2
2
4
√
(|g + h̄|) ≤
|g + h̄|
k
2
−
2
U
T
k=0
Z
2
2(1 + E4 ) 4
√
|g + h̄|4 .
=
2− 2
T
8
E44−k
6
ON SOME RIESZ AND CARLEMAN TYPE INEQUALITIES FOR HARMONIC FUNCTIONS
So
kg + h̄kb8 ≤
where E4 = cos
Here
1
2 sin
π
16
π
8
=
√
√
2+ 2
.
2
s
7
2(1 + E4 )
√ kg + h̄kh4 ,
2− 2
Finally we have that
kg + h̄kb8 ≤
1
π kg + h̄kh4 .
2 sin 16
≈ 2.56292. This finishes the proof of Theorem 1.2.
4. P ROOF
OF
T HEOREM 1.4
We use the following form of Green formula
(4.1)
r
Z
2π
0
∂G(reit )
dt =
∂r
Z
∆G(z)dxdy.
|z|≤r
Let p > 2 and let f = u + iv be an analytic function. Let q =
and define
Fǫ (z) = |qǫ + |f (z)|2 |p/2
p
p−1
and ǫ > 0
Uǫ (z) = |ǫ + u(z)2 |p/2
Vǫ (z) = |ǫ + v(z)2 |p/2
Then by direct calculation we obtain
(4.2)
∆Fǫ = p(qǫ + |f |2 )p/2−2 (2qǫ + p|f |2 )|f ′ |2 ,
(4.3)
∆Uǫ = p(u2 + ǫ)p/2−2 |f ′ |2 (ǫ + (p − 1)u2 ),
and
(4.4)
If p = 4, then
∆Vǫ = p(v 2 + ǫ)p/2−2 |f ′ |2 (ǫ + (p − 1)v 2 ).
3
∆Fǫ .
4
Lemma 4.1. Let f be a holomorphic function defined on the unit disk U. For
p > 4 and z ∈ U and ǫ > 0 we have
p−1
∆(Uǫ + Vǫ ) ≤
∆Fǫ .
p
∆Uǫ + ∆Vǫ =
For 1 ≤ p ≤ 4 and z ∈ U and ǫ > 0 we have
p−1
∆Fǫ .
∆(Uǫ + Vǫ ) ≥
p
ON SOME RIESZ AND CARLEMAN TYPE INEQUALITIES FOR HARMONIC FUNCTIONS
8
Proof. Let f = u + iv = reis and define
∆(Uǫ + Vǫ )
.
Q(s) =
∆Fǫ
Then from (4.2), (4.3) and (4.4) we have
−2+ p
2
ǫ + r 2 cos2 s
ǫ + (p − 1)r 2 cos2 s
Q(s) =
−2+ p
2
ǫp
(2ǫ + (p − 1)r 2 ) −1+p
+ r2
+
ǫ + r 2 sin2 s
We should prove that
−2+ p
ǫ + (p − 1)r 2 sin2 s
.
−2+ p
2
ǫp
(2ǫ + (p − 1)r 2 ) −1+p
+ r2
Q(s)
First of all
Q′ (s) =
2
(−2 + p)r 2
≥ if p < 4;
≤ if p ≥ 4.
ǫp
p−1
+ r2
2− p
2
cos s sin s
2ǫ + (p − 1)r 2
where
T
−3+ p
2
−3ǫ + (1 − p)r 2 cos2 s
T = ǫ + r 2 cos2 s
−3+ p
2
3ǫ + (p − 1)r 2 sin2 s .
+ ǫ + r 2 sin2 s
Then T = 0, if and only if
1
L = S 2 (6−p) = R :=
where
3ǫ + (p − 1)r 2 sin2 s
,
3ǫ + (p − 1)r 2 cos2 s
ǫ + r 2 sin2 s
.
ǫ + r 2 cos2 s
If p = 6, then cos2 s = sin2 s. If p > 6 then we also have cos2 s = sin2 s, because
if for example cos2 s > sin2 s, then L > 1 > R. Similarly cos2 s < sin2 s implies
L < 1 < R. If 4 < p < 6 and cos2 s > sin2 s , then
S=
R−S =
Thus
ǫ(4 − p)r 2 cos(2s)
< 0.
(ǫ + r 2 cos2 s) (3ǫ + (p − 1)r 2 cos2 s)
S < R < 1.
Since 0 <
6−p
2
< 1, it follows that
6−p
S < R 2 < 1.
2
So L 6= R. If 4 < p < 6 and cos s < sin2 s, then
ON SOME RIESZ AND CARLEMAN TYPE INEQUALITIES FOR HARMONIC FUNCTIONS
S > R > 1.
So
S>R
6−p
2
.
Thus L 6= R.
If p < 4, then cos2 s > sin2 s implies that
1>S>R>R
Finally if p < 4 and
cos2 s
6−p
2
.
2
> sin s. Then
1<S<R<R
6−p
2
.
We proved that the only stationary points of Q are
πj
, j = 0, . . . , 7.
sj =
4
So wee need to show that Q(sj ) ≤ 1 for p ≤ 4 and Q(sj ) ≥ 1 for p ≥ 4.
Let s = 0. Show that
−2+ p
p
2
ǫ + (p − 1)r 2 ≥ 1, if p ≤ 4;
ǫ−1+ 2 p + p ǫ + r 2
Q(0) =
p −2
≤ 1, if p ≥ 4.
2
ǫp
p (2ǫ + (p − 1)r 2 ) p−1
+ r2
Let ǫ = tr 2 , t > 0. Then
2− p p
p
2
pt
t−1+ 2 + (1 + t)−2+ 2 (p − 1 + t)
1 + p−1
.
Q(0) =
p − 1 + 2t
By using convexity of the function k(x) = xp/2−2 , for p < 4 we have
p
p
−2+ p2
t−1+ 2
p − 1 + pt + 2t2
(1 + t)−2+ 2 (p − 1 + t)
.
+
>
p − 1 + 2t
p − 1 + 2t
p − 1 + 2t
Further
p
−2+ p2
p − 1 + pt −2+ 2
p − 1 + pt + 2t2
>
.
p − 1 + 2t
p−1
So Q(0) ≥ 1.
For 4 ≤ p we need to show that
p
i.e.
p
(1 + t)−2+ 2 (p − 1 + t)
t−1+ 2
+
<
p − 1 + 2t
p − 1 + 2t
p − 1 + pt
p−1
−2+ p
2
,
p
−2+ p
p
2
(1 + t)−2+ 2 (p − 1 + t)
t
t−1+ 2
+
< 1+t+
.
p − 1 + 2t
p − 1 + 2t
p−1
The last inequality is equivalent with the trivial inequalities
−2+ p
p
p
2
p
t
t(t−2+ 2 − (t + 1)−2+ 2 )
<0< 1+t+
− (1 + t)−2+ 2 .
p − 1 + 2t
p−1
9
ON SOME RIESZ AND CARLEMAN TYPE INEQUALITIES FOR HARMONIC FUNCTIONS
10
For s = π/4, and ǫ = r 2 t,
2− p2
Q(s) = 2
So
p − 1 + pt
(p − 1)(1 + 2t)
2− p
2
.
≥ 1, if p ≤ 4;
≤ 1, if p ≥ 4.
The other cases can be treated in a similar way.
Q(s)
Proof of Theorem 1.4. Assume that p ≤ 4. By using the Green formula (4.1), and
Lemma 4.1 we obtain
Z r Z 2π
Z 2π
∂
ρ∆Fǫ dρdt.
Fǫ dt =
r
∂r
0
0
0
Z r Z 2π
p
≤
ρ(∆Uǫ + ∆Vǫ )dρdt
p−1 0 0
Z 2π
∂
p
(Uǫ + Vǫ )dt.
=r
p − 1 0 ∂r
Dividing by r and integrating in [0, 1] with respect to r we obtain
Z 2π
[Fǫ (eit ) − Fǫ (0)]dt
0
Z 2π
Z 2π
p
it
it
[Vǫ (e ) − Vǫ (0)] dt.
[Uǫ (e ) − Uǫ (0)]dt +
≤
p−1
0
0
Letting ǫ → 0 we obtain
Z 2π
|f (eit )|p dt ≤
0
p
p−1
Z
2π
0
|u(eit )|p dt +
Similarly we prove the related inequality for p ≥ 4.
Z
0
2π
|v(eit )|p dt.
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FACULTY OF NATURAL S CIENCES AND M ATHEMATICS , U NIVERSITY
C ETINJSKI PUT B . B . 81000 P ODGORICA , M ONTENEGRO
E-mail address: [email protected]
OF
M ONTENEGRO ,
D EPARTMENT OF M ATHEMATICS , U NIVERSITY OF P RISHTINA , M OTHER T ERESA , N O . 5,
10000, P RISHTINA , KOSOVO
E-mail address: [email protected]