Int. J. Anal. Appl. (2023), 21:2

Remarks on Some Higher Dimensional Hardy Inequalities

Zraiqat Amjad1, Jebril Iqbal1,∗, Hawawsheh Laith2, Abudayah Mohmammad2

1Alzaytoonah University of Jordan, Jordan
2School of Basic Sciences and Humanities, German Jordanian University, P.O. Box 35247, Amman

11180, Jordan

∗Corresponding author: i.jebril@zuj.edu.jo

Abstract. In this note, we give an elementary proof of Hardy inequality in higher dimensions introduced

by Christ and Grafakos. The advantage of our approach is that it uses the one-dimensional Hardy

inequality to obtain higher dimensional versions. We go further and get some well-known weighted

estimates using the same approach.

1. Introduction

Let Rn be the n-dimensional Euclidean space. Let H(f )(x) be the average of |f | ∈ Lp(Rn) over
the Euclidean ball B(0, |x|), that is,

H(f )(x)=
1

|B(0, |x|)|

∫
B(0,|x|)

|f (x)|dx.

Christ and Grafakos [1] introduced the operator H in order to get a higher dimensional version of the
classical Hardy inequality in one dimension [2]. In fact, they proved the following sharp estimate:

||H(f )||Lp(Rn) ≤
p

p−1
||f ||Lp(Rn), (1.1)

for f ∈ Lp(Rn) and 1 < p < ∞. The estimate 1.1 was obtained by using Minkowski’s convolution
inequality over the space Lp(R+, dt

t
), where R+ is the multiplicative topological group (0,∞). It

Received: Oct. 12, 2022.

2020 Mathematics Subject Classification. 26D15, 47A63.

Key words and phrases. Hardy inequality; weight functions; weighted estimates.

https://doi.org/10.28924/2291-8639-21-2023-2
ISSN: 2291-8639

© 2023 the author(s).

https://doi.org/10.28924/2291-8639-21-2023-2


2 Int. J. Anal. Appl. (2023), 21:2

should be mentioned that the idea of their proof was very elegent, although they did not use the

classical one dimensional estimate(∫ ∞
0

(
1

x

∫ x
0

|F(t)|dy
)p

dx

)1
p

≤
p

p−1

(∫ ∞
0

|F(t)|p dt
)1

p

(1.2)

or any of its variants at any stage of the proof. So it is natural to ask whether we can attain the sharp

estimate 1.1 by invoking the old bottles of Hardy inequalities in [2].

A humble attempt towards answering this question can be done by re-examining the left-hand side

of 1.1. More precisely, using polar coordinates and applying Hölder’s inequality give

||H(f )||Lp(Rn) =
(∫
Rn

(
1

|B(0, |x|)|

∫
B(0,|x|)

|f (y)|dy
)p

dx

)1
p

≤
ωn−1
νn

(∫ ∞
0

(
1

r

∫ r
0

(∫
Sn−1
|f (tθ)|ptn−1dθ

)1
p

dt

)p
dr

)1
p

, (1.3)

where ωn−1 is the surface area of the unit sphere Sn−1 and νn is the volume of the unit ball B(0,1).

Now, let F(t) :=
(∫
Sn−1
|f (tθ)|ptn−1dθ

)1
p

and use 1.2 with the fact that ωn−1 = nνn to obtain

||H(f )||Lp(Rn) ≤
np

p−1
||f ||Lp(Rn). (1.4)

The dependence on the dimension (n) in estimate 1.4 is natural and indicates that our inquiry makes

sense. Based upon this observation, we give a simple and direct proof of 1.1 which reveals that this

inequality is one dimensional in spirit.

Let W and Z be two weight functions on Rn, that is, nonnegative and locally integrable on Rn, and
denote the conjugate exponent of p > 1 by p′ =

p

p−1
. In [3], the authors introduced a weighted

version of Hardy inequality in Higher dimensions which generalizes 1.1. More precisely, they obtained

the following result.

Theorem 1.1. Let W and Z be weight functions on Rn. For 1 < p ≤ q < ∞, the inequality

(∫
Rn
W(x)

(∫
B(0,|x|)

|f (y)|dy
)q

dx

)1
q

≤ C
(∫
Rn
Z(x)|f (x)|p dx

)1
p

(1.5)

holds if and only if

A := sup
α>0

(∫
|x|≥α

W(x)dx

)1
q
(∫
|x|≤α

Z1−p
′
(x)dx

) 1
p′

< ∞. (1.6)

Moreover, if C is the smallest constant for which 1.5 holds, then

A ≤ C ≤ Ap′
1
p′ p

1
q .



Int. J. Anal. Appl. (2023), 21:2 3

We remark here that the condition 1.6 extends the one dimensional condition which can be found

in [4]. As a special and important case of Theorem 1.1, we have, for 1 < p = q < ∞, s > 0,
W(x)= |B(0,x)|−s−1 and U(x)= |B(0,x)|p−s−1, that

(∫
Rn
|B(0, |x|)|−s−1

(∫
B(0,|x|)

|f (y)|dy
)p

dx

)1
p

≤
p

s

(∫
Rn
|B(0, |x|)|p−s−1|f (x)|p dx

)1
p

, (1.7)

and the constant
p

s
being best possible. We notice here that by taking s = p − 1 in 1.7 one can

obtain 1.1. Similarly, it is natural to ask whether we can deduce the estimates 1.5 and 1.7 using

their one dimensional versions. In the following, we introduce a new and simple proofs of the higher

dimensional Hardy inequalities 1.1, 1.5 and 1.7. We show also in Theorem 2.4 that our technique can

be used to produce nice estimates by using the classical Hardy inequalities. It should also mention

that many authors worked on the equivalence between the higher dimensional and one-dimensional

Hardy’s inequalities. For instance, Gord [7] studies Hardy inequalities in higher dimensions where the

averages are taken over appropriate dilates of a given star-shaped regions. We refer the readers to

( [5], [6], [8]) for more background information and relevant work.

2. Proofs and further results

In this section we introduce new and simple proofs of the estimates 1.1, 1.5 and 1.7 using some

elementary tools and based on Hardy’s inequality in one dimension. We start by recalling the following

result.

Theorem 2.1. [2, Theorem 330] Let f be a measurable function. Then

(∫ ∞
0

(
1

x

∫ x
0

|F(t)|dt
)p

xη dx

)1
p

≤
p

p−1−η

(∫ ∞
0

|F(t)|p tη dt
)1

p

(2.1)

holds for 1 < p < ∞ and η < p−1.

Now, we present our proofs of 1.1, 1.7.

Proof of estimate 1.1. Applying polar coordinates and Höldr’s inequality yield

||H(f )||Lp(Rn) =
(∫
Rn

(
1

|B(0, |x|)|

∫
B(0,|x|)

|f (y)|dy
)p

dx

)1
p

≤ n

(∫ ∞
0

(
1

r

∫ r
0

(∫
Sn−1
|f (tθ)|ptn−1dθ

)1
p

t
n−1
p′ dt

)p
r
(1−n)p
p′ dr

)1
p

. (2.2)



4 Int. J. Anal. Appl. (2023), 21:2

Using 2.1 with η =
(1−n)p

p′
and F(t) :=

(∫
Sn−1
|f (tθ)|ptn−1dθ

)1
p

t
n−1
p′ , we obtain

||H(f )||Lp(Rn) ≤ n

(
p

p−1+ (n−1)p
p′

)(∫ ∞
0

(∫
Sn−1
|f (tθ)|ptn−1dθ

)
t
(n−1)p
p′ t

(1−n)p
p′ dt

)1
p

=
p

p−1
||f ||Lp(Rn).

Proof of estimate 1.7. Let F(t) :=
(∫
Sn−1
|f (tθ)|ptn−1dθ

)1
p

t
n−1
p′ dt, η = p−sn−1 and proceeding

as above, we get

(∫
Rn
|B(0, |x|)|−s−1

(∫
B(0,|x|)

f (y)dy

)p
dx

)1
p

≤ ν
−s−1
p

n ωn−1

(∫ ∞
0

(∫ r
0

F(t)dt

)p
r−ns−1dr

)1
p

≤ ν
−s+p−1

p
n

(p
s

)(∫ ∞
0

∫
Sn−1
|f (tθ)|ptnp−ns−1dθdt

)1
p

= ν
−s+p−1

p
n

(p
s

)(∫
Rn
|f (x)|p |x|n(p−s−1)dx

)1
p

=
(p
s

)(∫
Rn
|f (x)|p |B(0, |x|)|p−s−1dx

)1
p

.

Before introducing the proof of Theorem 1.1, we need the following result.

Theorem 2.2. Let u and v be nonnegative measurable functions on (0,∞). If f is a measurable
function, then

(∫ ∞
0

(∫ x
0

f (t)dt

)q
u(x)dx

)1
q

≤ C
(∫ ∞
0

f p(x)v(x)dx

)1
p

(2.3)

holds for 1 < p ≤ q < ∞ if and only if

A := sup
x>0

(∫ ∞
x

u(t)dt

)1
q
(∫ x
0

v1−p
′
(t)dt

) 1
p′

< ∞. (2.4)

Now, we are ready to prove Theorem 1.1. In fact, our proof is simpler than the proof in [4] and it

depends primarily on the appropriate choice of the one dimensional weights.

Theorem 2.3. In order to apply Theorem 2.2 we carefully define the weight functions u, v and the

function F. Let



Int. J. Anal. Appl. (2023), 21:2 5

u(t) :=

(∫
Sn−1

W(rϕ)dϕ

)
rn−1,

v(t) := t(1−n)(p−1)
(∫
Sn−1

Z
−p
′
p (tθ)dθ

)− p
p′

,

F(t) :=

(∫
Sn−1
|f (tθ)|pZ(tθ)dθ

)1
p
(∫
Sn−1

Z
−p
′
p (tθ)dθ

) 1
p′

tn−1.

Now, consider

(∫
Rn
W(x)

(∫
B(0,|x|)

|f (y)|dy
)q

dx

)1
q

=

(∫ ∞
0

(∫ r
0

∫
Sn−1
|f (tθ)|tn−1dθdt

)q
u(r)dr

)1
q

=

(∫ ∞
0

(∫ r
0

∫
Sn−1
|f (tθ)|Z

1
p(tθ)Z

−1
p(tθ)tn−1dθdt

)q
u(r)dr

)1
q

≤
(∫ ∞
0

(∫ r
0

F(t)dt

)q
u(r)dr

)1
q

. (2.5)

Then invoking Theorem 2.2 we have that

(∫
Rn
W(x)

(∫
B(0,|x|)

|f (y)|dy
)q

dx

)1
q

≤
(∫ ∞
0

(∫ r
0

F(t)dt

)q
u(r)dr

)1
q

≤ C
(∫ ∞
0

Fp(t)v(t)dt

)1
p

= C

(∫ ∞
0

(∫
Sn−1
|f (tθ)|pZ(tθ)dθ

)(∫
Sn−1

Z
−p
′
p (tθ)dθ

) p
p′

t(n−1)pv(t)dt

)1
p

= C

(∫ ∞
0

(∫
Sn−1
|f (tθ)|pZ(tθ)dθ

)
tn−1dt

)1
p

= C

(∫
Rn
|f (y)|pZ(y)dy

)1
p

(2.6)

if and only if

A := sup
α>0

(∫ ∞
α

u(t)dt

)1
q
(∫ α
0

v1−p
′
(t)dt

) 1
p′

= sup
α>0

(∫ ∞
α

(∫
Sn−1

W(tϕ)dϕ

)
tn−1dt

)1
q
(∫ α
0

(∫
Sn−1

Z1−p
′
(tθ)

)
tn−1dt

) 1
p′



6 Int. J. Anal. Appl. (2023), 21:2

= sup
α>0

(∫
|x|≥α

W(x)dx

)1
q
(∫
|x|≤α

Z1−p
′
(x)dx

) 1
p′

< ∞. (2.7)

Next, we use the main scheme of the previous proofs to introduce the following result.

Theorem 2.4. Let f be a measurable function and − n
p′ < α <

1
p′ . Then

(∫
Rn

(
1

|B(0, |x|)|

∫
B(0,|x|)

|f (y)||x ·y|−α dy
)p

dx

)1
p

≤ C
(∫
Rn
|f (y) |y|−2α|p

)1
p

for 1 < p < ∞, where C =


Cα,p,n ω1pn−1

νn


( p

np+αn−n

)
.

Proof. Let

G(ϕ,t)=

∫
Sn−1
|f (tθ)|t

n−1
p |ϕ ·θ|−α dθ.

Then applying Holder’s inequality and using the fact that∫
Sn−1
|ϕ ·θ|−αp

′
dθ = ωn−2B

(
1−αp′

2
,
n−1
2

)
:= Cp

′
α,p,n

for α < 1
p′ , we get

G(ϕ,t)≤
(∫
Sn−1
|f (tθ)|tn−1dθ

)1
p

Cα,p,n (2.8)

Now set Dα,p,n = n−1−(α+n−1)p, F(t)=
(∫
Sn−1 |f (tθ)|t

n−1dθ
)1
p t

n−1
p′ −α and use 2.8 to obtain

(∫
Rn

(
1

|B(0, |x|)|

∫
B(0,|x|)

|f (y)||x ·y|−α dy
)p

dx

)1
p

=
1

νn

(∫ ∞
0

∫
Sn−1

(
1

r

∫ r
0

G(ϕ,t)t
n−1
p′ −α dt

)p
rDα,p,n dϕdr

)1
p

≤
Cα,p,n ω

1
p

n−1
νn

(∫ ∞
0

(
1

r

∫ r
0

F(t)dt

)p
rDα,p,n dr

)1
p

. (2.9)

Finally, Applying Theorem 2.1 with η = Dα,p,n and α > − np′ we get the desired result. �

Conflicts of Interest: The authors declare that there are no conflicts of interest regarding the publi-

cation of this paper.



Int. J. Anal. Appl. (2023), 21:2 7

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978-3-0348-8942-1_1.

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[8] W.G. Alshanti, Inequality of Ostrowski Type for Mappings with Bounded Fourth Order Partial Derivatives, Abstr.

Appl. Anal. 2019 (2019), 5648095. https://doi.org/10.1155/2019/5648095.

https://doi.org/10.1090/s0002-9939-1995-1239796-6
https://doi.org/10.1007/978-3-0348-8942-1_1
https://doi.org/10.1007/978-3-0348-8942-1_1
https://doi.org/10.1017/s0308210500021818
https://doi.org/10.1155/2019/5648095

	1. Introduction
	2. Proofs and further results
	References