EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS
Vol. 17, No. 3, 2024, 1685-1690
ISSN 1307-5543 – ejpam.com
Published by New York Business Global

Some Generalization of Riesz Type Inequalities for
Harmonic Mappings on the Unit Disk

Elver Bajrami

Department of Mathematics, University of Prishtina, Mother Teresa, No. 5, 10000,
Prishtina, Kosovo

Abstract. In this paper a new generalized norm is defined and Riesz type inequalities for harmonic
functions on the unit disk are discussed by applying it. Also sharp constants are obtained for certain
special values considered in the reverse case of the standard Riesz inequality.

2020 Mathematics Subject Classifications: 30H05, 30H10, 47B06

Key Words and Phrases: Harmonic functions, Riesz inequality, Isoperimetric inequality

1. Introduction and statement of main results

Let U = {z ∈ C : |z| < 1} be the unit disk and let T = {z ∈ C : |z| = 1} be the
unit circle in the complex plane. For p > 1 we define the Hardy class hp as the class of
harmonic mappings f = g + h, where h and g are holomorphic mappings defined on unit
disk U ⊂ C. Norm in this space is defined as:

||f ||p = ||f ||hp = sup
0<r<1

Mp(f, r) < ∞,

where

Mp(f, r) =

(∫
T
|f(rζ)|pdσ(ζ)

)1/p

.

Here σ is probability measure on T. With Hp, we denote the subclass of holomorphic
mappings that belong to the class hp. For the theory of Hardy spaces in the unit disk we
refer to [11], [7] and [8].

Based on results of Verbitsky [12], Kalaj in [9] proved these inequalities

|||f |||p ≤ Ap||f ||p, ||f ||p ≤ Bp|||f |||p,

DOI: https://doi.org/10.29020/nybg.ejpam.v17i3.5282

Email address: elver.bajrami@uni-pr.edu (E. Bajrami)

https://www.ejpam.com 1685 © 2024 EJPAM All rights reserved.



E. Bajrami / Eur. J. Pure Appl. Math, 17 (3) (2024), 1685-1690 1686

where Ap =
(
1− | cos p

π |
)−1/2

, Bp =
√
2 cos π

2p and

|||f |||p = sup
0<r<1

(∫
T
(|g(z)|2 + |h(z)|2)p/2dσ

)1/p

.

These results are used to prove and improve some results about isoperimetric inequali-
ties which are given in [2] and generalized in [6] for a special case. Also, this inequality was
generalized in several directions. Let us mention Beckenbach’s results: the same inequality
holds where in place of |f |p we have a positive logarithmically subharmonic function. This
kind of generalizations can be found in [3]. We refer interested readers to [4], [5] and [9].

The following definition for a generalized norm of this kind is next given: for harmonic
mapping f = g + h̄ ∈ hp, (hg)(0) = 0, q > 1 we define new norm ||| · |||p,q = ||| · |||hp as
follows

|||f |||p,q = sup
0<r<1

(∫
T
(|g(rz)|q + |h(rz)|q)p/qdσ(z)

)1/p

.

Or using limr→1 in last expression we have that

|||f |||p,q =
(∫

T
(|g(z)|q + |h(z)|q)p/qdσ(z)

)1/p

.

We see that the norm defined in [9] is a special case where q = 2. In the same paper, this
inequality is proved for the special value (q = 2). In this paper our main aim is to find
best constant Ap,q in the inequality

||f ||p ≤ Ap,q|||f |||p,q. (1.1)

A reverse case of this kind of inequalities is generalized from Melentijević in [10] for a
special value of p, t and q. There, these inequalities are analyzed using Riesz’s projection
operator. Let us say that the Riesz projection operator is not bounded on L1(T).

2. Main results and strategy of proofs

As the authors of paper did in [9] and [1], in order to prove results we use pluri
subharmonic function. First, a lemma is proved, which is applied for the proof of the
main theorem.

Lemma 1. Let p > 2 and q > 1. Then for complex numbers z = |z|eit and w = |w|eis we
have

|z + w| ≤ Cp,q(|z|q + |w|q)p/q −Dp,q|zw|
p
2 cos

(π − |t+ s|)p
2

.

where Cp,q and Dp,q are defined as below

Cp,q = 2
p− p

q sinp
π

2p
,

and
Dp,q = 2

3p
2
− p

q sinp
π

2p
cot

π

2p
.



E. Bajrami / Eur. J. Pure Appl. Math, 17 (3) (2024), 1685-1690 1687

Based on homogeneity of expression (as in proof of Lemma 2 in [1]), we can assume
that |z| = r < 1 and w = 1. So, rather than proving the last lemma, we will present and
prove the next lemma.

Lemma 2. Let p > 2 and q > 1. Then this sharp inequality hold

(1 + r2 + 2r cos t)p/2 ≤ 2p sinp
π

2p
(
1 + rq

2
)p/q

− 2
3p
2
− p

q r
p
2 sinp

π

2p
cot

π

2p
cos

(π − |t+ s|)p
2

for 0 ≤ r ≤ 1, 0 ≤ t ≤ π.

Proof. Define

P (r, t) = (1 + r2 + 2r cos t)p/2 − 2p sinp π
2p

(
1+rq

2

)p/q
+2

3p
2
− p

q sinp π
2p cot

π
2pr

p
2 cos (π−|t+s|)p

2 (2.1)

We must prove that P (r, t) ≤ 0. Calculate partial derivative

2

p

∂P (r, t)

∂r
= 2(r + cos t)(1 + 2r cos t+ r2)

p
2
−1 − 2p sinp

π

2p
rq−1

(
1 + rq

2

) p
q
−1

+ 2
3p
2
− p

q sinp
π

2p
cot

π

2p
r

p
2
−1 cos

(π − |t|)p
2

and

∂P (r, t)

∂t
=

p

2
(1 + 2r cos t+ r2)

p
2
−1(−2r sin t)

+ 2
3p
2
− p

q sinp
π

2p
r

p
2 cot

π

2p
sin

(π − t)p

2
.

Using equality ∂P
∂t = 0, from the last equation we have

(1 + 2r cos t+ r2)
p
2
−1 =

2
3p
2
− p

q sinp π
2pr

p
2 cot π

2p sin
(π−t)p

2

pr sin t
(2.2)

and substitute in equation 2
p
∂P
∂r = 0 to get

2(r + cos t)
2

3p
2
− p

q sinp π
2pr

p
2 cot π

2p sin
(π−t)p

2

pr sin t
− 2p sinp

π

2p
rq−1

(
1 + rq

2

) p
q
−1

+2
3p
2
− p

q sinp
π

2p
cot

π

2p
r

p
2
−1 cos

(π − t)p

2
= 0



E. Bajrami / Eur. J. Pure Appl. Math, 17 (3) (2024), 1685-1690 1688

or (
1 + rq

2

) p
q
−1

= 2
p
2
− p

q cot
π

2p
r

p
2
−q

(
(r + cos t)

sin (π−t)p
2

p sin t
− cos

(π − t)p

2

)
. (2.3)

Now, if we substitute expressions from (2.2) and (2.3) in (2.1) and avoid the factor

2
3p
2
− p

q sinp π
2p > 0, we get

P (r, t) =
r

p
2 cot π

2p

pr sin t(
(1 + r2 + 2r cos t) sin

(π − t)p

2

−1 + rq

2rq−1
(r sin

(π − t)p

2
+ sin(t− tp

2
)) + r cos

(π − t)p

2
sin t

)
.

After some transformation we obtain

P (r, t) =
r

p
2 cot π

2p

pr sin t

(
sin(t− tp

2
)

(
2r +

1 + rq

2rq−1

)
+ sin

tp

2

(
1 + r2 − 1 + rq

2rq

))
.

Finally, for such r, t and p, we have(
3rq + 1

2rq−1

)
sin(t− tp

2
) +

(
rq(1 + 2r2)− 1

2rq

)
sin

tp

2
≤ 0

and
r

p
2 cot π

2p

pr sin t
≥ 0.

which give P (r, t) ≤ 0 and prove the lemma.
At finish, we analyze inequality P (r, t) ≤ 0 at the boundary points.
For r = 0 our inequality is transformed as

P (0, t) = 1− 2p sinp
π

2p

(
1

2

) p
q

≤ 1− 2
p
2 sinp

π

2p
≤ 0

and hold for this case.
For r = 1 we have

P (1, t) = 2p sinp
π

2p

(
cosp t

2

sinp π
2p

− 1 + 2
p
2
− p

q cot
π

2p
cos

p(π − t)

2

)
≤ 0

With substitute of variable t = π − 2y and using derivative on variable y, this inequality
also hold.



E. Bajrami / Eur. J. Pure Appl. Math, 17 (3) (2024), 1685-1690 1689

For t = 0 our inequality has a form

P (r, 0) = (1− r)p − 2p sinp
π

2p

(
1 + r2

2

) p
q

.

As in the case of r = 0, we can transform and prove inequality

P (r, 0) = (1 + r2)
(
1− 2

p
2 sinp

π

2

)
≤ 0.

For t = π we see that

P (r, π) = 2pr
p
2 sinp

π

2p


(
1+r2

r − 2
) p

2

sinp π
2p

+ cot
π

2p
−
(
1 + r2

2r

) p
2

 .

As in the [9], using transformation a = 1+r2

2r ≥ 0 we get R(a) = (2a−2)
p
2

sinp π
2

+ cot π
2p − a

p
2 .

It is easy to see that R(a) is increasing, and finally P (r, π) ≤ 0.

The main outcome of this paper is given in the next theorem.

Theorem 1. Let 1 < p, q < ∞ and assume that f = g + h̄ ∈ hp is a harmonic mapping
on the unit disk with ℜ(g(0)h(0)) ≤ 0. Then we have the following sharp inequality

∥f∥hp ≤ 2
1− 1

q max{sin π

2p
, cos

π

2p
}
(∫

T
(|g|q + |h|q)p/q

)1/p

.

Proof. Applying Lemma 2.1, Lemma 2.2 and integrating over T, 0 < r < 1 and letting
r → 1− we get∫

T
|g(z) + h(z)|p ≤ 2

p− p
q sinp

π

2p

∫
T
(|z|q + |w|q)p/q −Dp,qT (z),

in case of p > 2. For 1 < p < 2∫
T
|g(z) + h(z)|p ≤ 2

p− p
q cosp

π

2p

∫
T
(|z|q + |w|q)p/q −Dp,qT (z).

Since
∫
T T (z) ≥ 0 (because of subharmonicity). We conclude that

∥f∥hp ≤ 2
1− 1

q max{sin π

2p
, cos

π

2p
}
(∫

T
(|g|q + |h|q)p/q

)1/p

.

Finally, we see that constant in the norm (1.1) has a form

Ap,q = 2
1− 1

q max{sin π

2p
, cos

π

2p
}

which coincide with result in [9], for a special case when q = 2.



REFERENCES 1690

Acknowledgements

I would like to express my gratitude to the anonymous referees for their helpful com-
ments that have improved the quality of the paper.

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