A STUDY OF THE REAL HARDY INEQUALITY
MOHAMMAD SABABHEH
Department of Science and Arts
Princess Sumaya University for Technology
Amman 11941 Jordan
Real Hardy Inequality
Mohammad Sababheh
vol. 10, iss. 4, art. 104, 2009
EMail: sababheh@psut.edu.jo
Title Page
Received:
11 November, 2008
Accepted:
20 October, 2009
Communicated by:
L. Pick
2000 AMS Sub. Class.:
Contents
JJ
II
42A16, 42A05, 42B30.
J
I
Key words:
Hardy’s inequality, inequalities involving Fourier transforms.
Page 1 of 18
Abstract:
We show that some Hardy-type inequalities on the circle can be proved to be true
on the real line. Namely, we discuss the idea of getting Hardy inequalities on the
real line by the use of corresponding inequalities on the circle. In the last section,
we prove the truth of a certain open problem under some restrictions.
Go Back
Full Screen
Close
Contents
1
Introduction
3
2
Setup
5
3
From The Circle To The Line
7
Real Hardy Inequality
4
Another Hardy Inequality
10
Mohammad Sababheh
vol. 10, iss. 4, art. 104, 2009
Title Page
Contents
JJ
II
J
I
Page 2 of 18
Go Back
Full Screen
Close
1.
Introduction
When McGehee, Pigno and Smith [4] proved the Littlewood conjecture, many questions regarding the best possible generalization of Hardy’s inequality were asked.
The longstanding question is: Does there exist a constant c > 0 such that
∞
X
|fˆ(n)|
(1.1)
n=1
n
≤ ckf k1 + c
∞
X
|fˆ(−n)|
n=1
n
1
for all f ∈ L (T)? The truth of this inequality is an open problem. Many attempts
were made to answer this question and many partial results were obtained. We refer
the reader to [2], [3], [6] and [7] for some partial results.
Almost all articles in the literature treat Hardy-type inequalities on the circle and
a very few articles treat them on the real line.
In [8] it was proved that a constant c > 0 exists such that for all
Z x
1
1
f ∈ g ∈ L (R) :
g(t)dt ∈ L (R)
−∞
we have
Z ∞ ˆ
|f (−ξ)|2
|fˆ(ξ)|2
2
dξ ≤ ckf k1 + c
dξ.
ξ
ξ
0
0
Although this is not the first proved Hardy inequality on the real line, its proof is the
first proof which uses the construction of a bounded function on R whose Fourier
coefficients have some desired decay properties.
We have two main goals in this article. The first is to prove Hardy inequalities on
the real line using well known inequalities on the circle and the second is to prove a
real Hardy inequality on the real line which is related to the open problem (1.1).
Z
∞
Real Hardy Inequality
Mohammad Sababheh
vol. 10, iss. 4, art. 104, 2009
Title Page
Contents
JJ
II
J
I
Page 3 of 18
Go Back
Full Screen
Close
Proving a Hardy inequality usually involves quite a difficult construction and this
is because of the way we prove such inequalities. Again we refer the reader to [2],
[4], [6], [7] and [8] for more information on how and why we construct bounded
functions with desired Fourier coefficients.
Real Hardy Inequality
Mohammad Sababheh
vol. 10, iss. 4, art. 104, 2009
Title Page
Contents
JJ
II
J
I
Page 4 of 18
Go Back
Full Screen
Close
2.
Setup
Let L1 denote the space of all integrable functions (equivalent classes) defined on R.
For f ∈ L1 , we define the Fourier transform of f to be
Z
ˆ
f (ξ) =
f (x)e−ixξ dξ.
R
If fˆ ∈ L1 then the inversion formula for f holds:
Z
1
f (x) =
fˆ(ξ)eiξx dξ.
2π R
If f ∈ L2 then the Fourier transform of f is defined to be the L2 − limit:
Z n
ˆ
f (ξ) = lim
f (x)e−iξx dx.
n→∞
−n
In this case fˆ ∈ L2 and
1
f (x) = lim
n→∞ 2π
Z
n
fˆ(ξ)eixξ dξ,
in the L2 sense.
−n
The Plancheral theorem then says:
1
kf k2 = √ kfˆk2 .
2π
Lemma 2.1. For f, g ∈ L2 we have
Z
Z
1
ˆ
f (x)g(x)dx =
fˆ(ξ)g(ξ)dξ.
2π R
R
Real Hardy Inequality
Mohammad Sababheh
vol. 10, iss. 4, art. 104, 2009
Title Page
Contents
JJ
II
J
I
Page 5 of 18
Go Back
Full Screen
Close
We refer the reader to any standard book in analysis for more on these concepts,
see for example [5].
Observe that if f ∈ L1 is such that fˆ is compactly supported, then fˆ ∈ L2 and
hence f ∈ L2 .
Now, given f ∈ L1 (R), define
(2.1)
ϕN (t) = 2π
∞
X
fN (t + 2πj),
where fN (x) = N f (N x).
j=−∞
Mohammad Sababheh
vol. 10, iss. 4, art. 104, 2009
Then we have
(2.2)
n
ϕN ∈ L1 (T), ϕ̂N (n) = fˆ
N
Real Hardy Inequality
and
lim kϕN kL1 (T) = kf kL1 (R) .
N →∞
For a discussion of this idea we refer the reader to [1, p. 160-162].
Thus, the equations in (2.1) and (2.2) enable us to move from a function integrable
on the line to a function integrable on the circle. This idea will be used efficiently in
the next section to prove some Hardy-type inequalities on the real line using only a
corresponding inequality on the circle.
Title Page
Contents
JJ
II
J
I
Page 6 of 18
Go Back
Full Screen
Close
3.
From The Circle To The Line
In this section we discuss how one can use a Hardy inequality on the circle to obtain
an inequality on the real line.
Recall that Hardy’s inequality on the circle states that a constant C > 0 exists
such that for all f ∈ L1 (T) with fˆ(n) = 0, n < 0 we have
∞
X
|fˆ(n)|
n=1
n
Real Hardy Inequality
≤ Ckf k1 .
Mohammad Sababheh
vol. 10, iss. 4, art. 104, 2009
As a matter of notation, let H 1 (R) be defined by
n
o
H 1 (R) = f ∈ L1 (R) : fˆ(ξ) = 0, ∀ξ < 0 .
Title Page
Contents
In the following theorem we use the above stated Hardy’s inequality to get the
well known Hardy inequality on the line. We should remark that proving such an
inequality (on the line) is a difficult task, but transforming the problem from the
circle to the line greatly simplifies the proof.
Theorem 3.1. There exists an absolute constant C > 0 such that
Z ∞ ˆ
|f (ξ)|
≤ Ckf k1
ξ
0
II
J
I
Page 7 of 18
Go Back
Full Screen
Close
for all f ∈ H 1 (R).
Proof. Let f ∈ H 1 (R) be arbitrary and let, for N ∈ N, ϕN (as in (2.1) and (2.2)) be
such that:
ϕN ∈ L1 (T),
JJ
lim kϕN kL1 (T) = kf kL1 (R)
N →∞
and ϕ̂N (n) = fˆ(n/N ).
Clearly ϕ̂N (n) = 0, ∀n < 0, hence Hardy’s inequality applies and we have
∞
X
|ϕ̂N (n)|
n
n=1
≤ CkϕN kL1 (T) .
However, this implies that
N
X
1 |fˆ(n/N )|
≤ CkϕN kL1 (T) .
N n/N
n=1
We take the limit as N → ∞ to get
Z 1 ˆ
|f (ξ)|
(3.1)
≤ Ckf k1 .
ξ
0
Thus, we have shown (3.1) for any function in H 1 (R). Now we proceed to prove the
inequality stated in the theorem. That is, we would like to replace the upper limit of
the above integral by ∞. For this, let M ∈ N be arbitrary and put h(x) = f (x/M ).
Then
khk1 = M kf k1 and ĥ(ξ) = M fˆ(M ξ).
Now apply (3.1) on h to obtain
Z 1 ˆ
Z 1
|ĥ(ξ)|
|f (M ξ)|
≤ Ckhk1 ⇒
dξ ≤ Ckf k1 .
ξ
ξ
0
0
Put M ξ = t to get
Z
(3.2)
0
M
|fˆ(ξ)|
dξ ≤ Ckf k1 .
ξ
Now, letting M tend to ∞, we obtain the result.
Real Hardy Inequality
Mohammad Sababheh
vol. 10, iss. 4, art. 104, 2009
Title Page
Contents
JJ
II
J
I
Page 8 of 18
Go Back
Full Screen
Close
In fact, the idea of this proof can be used to prove many Hardy inequalities on the
real line!
It is proved that, see for example [3], for all functions f ∈ L1 (T) we have
∞
X
|fˆ(n)|2
n=1
n
≤
Ckf k21
+C
∞
X
|fˆ(−n)|2
n=1
n
.
Using an argument similar to that of Theorem 3.1 we can prove:
Real Hardy Inequality
Mohammad Sababheh
Theorem 3.2. There exists a constant C > 0 such that for all functions f ∈ L1 (R)
we have
Z ∞ ˆ 2
Z ∞ ˆ
|f (ξ)|
|f (−ξ)|2
2
dξ ≤ Ckf k1 + C
dξ.
ξ
ξ
0
0
vol. 10, iss. 4, art. 104, 2009
Title Page
Contents
This modifies the result proved in [8].
JJ
II
J
I
Page 9 of 18
Go Back
Full Screen
Close
4.
Another Hardy Inequality
In this section we prove (1.1) on the real line under a certain condition on the signs
of the Fourier coefficients.
We should remark here that the method used to prove this inequality is standard
and all recent articles use this idea; we need a bounded function whose Fourier
coefficients obey some desired decay conditions.
One last remark before proceeding, although the given proof is for a real Hardy
inequality, we can imitate the given steps to prove a similar inequality on the circle.
For j ≥ 1 put
Z j
1 4 ixξ
fj (x) = j
e dξ, x ∈ R.
4 4j−1
Then we have our first lemma:
Lemma 4.1. Let fj be as above, then
( 2π
, 4j−1 < ξ < 4j ,
4j
ˆ
fj (ξ) =
0, otherwise.
Real Hardy Inequality
Mohammad Sababheh
vol. 10, iss. 4, art. 104, 2009
Title Page
Contents
JJ
II
J
I
Page 10 of 18
Proof. Let
Go Back
(
gj (ξ) =
2
2π
,
4j
4
0,
otherwise,
j−1
j
<ξ<4 ,
Full Screen
Close
2
then gj ∈ L . Therefore, gj is the Fourier transform of some function hj ∈ L and
Z
1
g(ξ)eixξ dξ
hj (x) =
2π R
= f (x).
However, ĥj = gj , which implies fˆj = gj as claimed.
Observe that the above proof implies that fj ∈ L2 and
Lemma 4.2. For fj as above, we have
√
kfj k2 =
6π
4j/2
.
Real Hardy Inequality
2
Proof. Since fj ∈ L we have
Mohammad Sababheh
vol. 10, iss. 4, art. 104, 2009
1
kfj k2 = √ kfˆj k2
2π
1 2π
=√
2π 4j
√
6π
= j/2 .
4
Z
! 12
4j
Title Page
dξ
4j−1
Contents
JJ
II
J
I
Page 11 of 18
Lemma 4.3. For fj as above and for M ∈ N, let
M X
FM (x) =
fj (x) − fj (x) ,
j=1
then kFM k∞ ≤ C where C is some absolute constant (independent of M ).
Go Back
Full Screen
Close
Proof. Observe first that
Z j
4
2
sin(xξ)dξ fj (x) − fj (x) = j 4 4j−1
2 cos(4j x) − cos(4j−1 x) = j
4
x
2 2 sin2 (4j−1 x/2) [−1 + 16 cos2 (4j−1 x) cos2 (4j−1 x/2)] = j
4
x
60 sin2 (4j−1 x/2)
≤ j
.
4
|x|
Real Hardy Inequality
Mohammad Sababheh
vol. 10, iss. 4, art. 104, 2009
Title Page
Note that FM is an even function, so it suffices to consider only the case x > 0.
Now, fix x > 0 and observe that
∞
X
1 sin2 (4j−1 x/2)
FM (x) ≤ 60
4j
x
j=1


2
2
j−1
j−1
X 1 sin (4 x/2)
X 1 sin (4 x/2)
,
= 60 
+
j
j
4
x
4
x
−j
−j
4
≤x
4
Contents
JJ
II
J
I
Page 12 of 18
Go Back
>x
Full Screen
where, by convention, the second sum is zero if 4−j ≤ x for all j ≥ 1.
Now
∞
X 1 sin2 (4j−1 x/2)
1X 1
≤
,
4j
x
x j=k 4j
−j
4
≤x
x
−j
where kx is the smallest positive integer such that 4
≤ x for all j ≥ kx .
Close
Consequently
X 1 sin2 (4j−1 x/2)
14 1
≤
j
4
x
x 3 4kx
−j
4
≤x
14
4
x= .
x3
3
≤
On the other hand, if
P
4−j >x
1 sin2 (4j−1 x/2)
4j
x
Real Hardy Inequality
is not zero, we get
Mohammad Sababheh
4 (1/x)
X 1 sin2 (4j−1 x/2) logX
1 sin2 (4j−1 x/2)
=
4j
x
4j
x
−j
j=1
4
>x
log4 (1/x)
≤
X
j=1
1
4j x
1 j−1
4 x
2
log4 (1/x)
=
X
1
x
4j
64
j=1
log4 (1/x)
1 4
x
64
1
≤
.
192
≤
−4
FM (x) ≤ 60
4
1
+
3 192
Title Page
2
Contents
JJ
II
J
I
Page 13 of 18
Go Back
3
Full Screen
Close
Thus, we have
vol. 10, iss. 4, art. 104, 2009
:= C.
Now let X be the set of all f ∈ L1 such that the sign of fˆ(ξ) is constant in the
block (4j−1 , 4j ). Then we have our main result:
Theorem 4.4. There is an absolute constant K > 0 such that
Z ∞ ˆ
Z ∞ ˆ
|f (ξ)|
|f (−ξ)|
(4.1)
dξ ≤ Kkf k1 + K
dξ
ξ
ξ
1
1
Real Hardy Inequality
for all f ∈ X.
Mohammad Sababheh
Proof. Let f ∈ X be such that fˆ is compactly supported, let fj be as above and let
M ∈ N be such that the support of fˆ is contained in [−M, M ]. Denote the sign of fˆ
in the block (4j−1 , 4j ) by σj and put
F (x) =
M
X
σj fj (x) − fj (x) .
j=1
Then kF k∞ ≤ C, where C is the constant of Lemma 4.3. Moreover, F̂ (ξ) = 2π
σ,
4j j
where j is the unique index such that ξ ∈ [4j−1 , 4j ]∪[−4j , −4j−1 ] if −4M ≤ ξ ≤ 4M
and F̂ (ξ) = 0 otherwise.
Moreover, since fˆ is of compact support, we have f ∈ L2 . Now we apply a
standard duality argument:
Z
Ckf k1 ≥ f (x)F (x)dx
R
Z
1 ˆ
=
f (ξ)F̂ (ξ)dξ (see Lemma 2.1)
2π R
vol. 10, iss. 4, art. 104, 2009
Title Page
Contents
JJ
II
J
I
Page 14 of 18
Go Back
Full Screen
Close
1
≥
2π
Z 1
∞ Z 4j
X
1
fˆ(ξ)F̂ (ξ)dξ −
fˆ(ξ)F̂ (ξ)dξ 2π −1
j−1
j=1 4
∞ Z j
1 X 4 ˆ
f (−ξ)F̂ (−ξ)dξ .
−
2π 4j−1
j=1
That is,
1
(4.2)
2π
Real Hardy Inequality
∞ Z j
Z 1
X 4
1
ˆ(ξ)F̂ (ξ)dξ f
fˆ(ξ)F̂ (ξ)dξ ≤ Ckf k1 +
2π −1
j−1
j=1 4
∞ Z j
1 X 4 ˆ
+
f (−ξ)F̂ (−ξ)dξ .
2π j−1
4
Mohammad Sababheh
vol. 10, iss. 4, art. 104, 2009
Title Page
Contents
j=1
σ , where σj is
However, when ξ ∈ (4j−1 , 4j ) ∪ (−4j , 4j−1 ), we have F̂ (ξ) = 2π
4j j
the sign of fˆ in (4j , 4j−1 ). Thus, when ξ ∈ (4j−1 , 4j ) we have fˆ(ξ)F̂ (ξ) = |fˆ(ξ)|.
Moreover
Z 1
Z 1
ˆ(ξ)F̂ (ξ)dξ ≤ kf k1
f
F̂
(ξ)
dξ
−1
−1
Z
1
12 Z
≤ kf k1
dξ
−1
√
≤ 2kf k1 kF̂ k2
√
√
= 2kf k1 2πkF k2
1
−1
|F̂ (ξ)|2 dξ
12
JJ
II
J
I
Page 15 of 18
Go Back
Full Screen
Close
∞
X
√
≤ 2 πkf k1 × 2
kfj k2
√
= 4 6πkf k1 ,
j=1
where we have used the facts
F (x) =
M
X
σj fj (x) − fj (x)
and kfj k2 =
√
6π4−j/2 .
j=1
Real Hardy Inequality
Mohammad Sababheh
vol. 10, iss. 4, art. 104, 2009
Consequently, (4.2) becomes
∞ Z 4j
∞ Z 4j
X
X
√
|fˆ(−ξ)|
|fˆ(ξ)|
6kf
k
+
dξ.
(4.3)
dξ
≤
Ckf
k
+
2
1
1
4j
4j
j−1
j−1
j=1 4
j=1 4
However, when ξ ∈ [4j−1 , 4j ] we have
1
1
≤
4j
ξ
and
1
1
≥
.
4j
4ξ
Therefore, (4.3) boils down to
Z ∞ ˆ
Z ∞ ˆ
|f (ξ)|
|f (−ξ)|
≤ Kkf k1 + K
dξ,
ξ
ξ
1
1
√
where K = 4(C + 2 6) and where C is the constant of Lemma 4.3.
This completes the proof for f ∈ X with the property that fˆ is compactly supported. Now for general f ∈ X, consider the convolution f ∗ Kλ where Kλ is the
Fejer Kernel of order λ. A standard limiting process yields the result.
Title Page
Contents
JJ
II
J
I
Page 16 of 18
Go Back
Full Screen
Close
Remark:
1. We note that the above proof can be adopted to prove that: There is an absolute constant C 0 > 0 such that for any function f ∈ L1 (T) whose Fourier
coefficients have the same sign on the block [4j−1 , 4j ) we have
∞
X
|fˆ(n)|
n=1
n
0
≤ C kf k1 + C
0
∞
X
|fˆ(−n)|
n=1
n
.
2. Observe that the condition that fˆ has the same sign on the block (4j−1 , 4j ) is
flexible. This is because functions in Lp are in fact equivalent classes. Therefore, even if fˆ obeys our condition but for a set of measure zero, then we may
modify our choice by changing the values of fˆ so that the new fˆ satisfies our
condition.
We should remark that this process does not effect the proof above.
Real Hardy Inequality
Mohammad Sababheh
vol. 10, iss. 4, art. 104, 2009
Title Page
Contents
JJ
II
J
I
Page 17 of 18
Go Back
Full Screen
Close
References
[1] Y. KATZNELSON, An Introduction To Harmonic Analysis, John Wiley and
Sons, Inc., 1968.
[2] I. KLEMES, A note on Hardy’s inequality, Canad. Math. Bull., 36(4) (1993),
442–448.
[3] P. KOOSIS, Conference On Harmonic Analysis in Honour of Antoni Zygmund,
Volume 2, Wadsworth International Group, Belmont, California, A Division of
Wadsworth, Inc., (1981), 740–748.
[4] O.C. McGEHEE, L. PIGNO AND B. SMITH, Hardy’s inequality and the L1
norm of exponential sums, Annals of Math., 113 (1981), 613–618.
[5] W. RUDIN, Real and Complex Analysis, 3rd edition, McGraw-Hill Book Co.,
New York, 1987.
[6] M. SABABHEH, Two-sided probabilistic versions of Hardy’s inequality, J.
Fourier Anal. and Applics., (2007), 577–587.
[7] M. SABABHEH, On an argument of Körner and Hardy’s inequality, Analysis
Mathematica, 34 (2008), 51–57.
[8] M. SABABHEH, Hardy-type inequalities on the real line, J. Inequal. in Pure
and Appl. Math., 9(3) (2008), Art. 72. [ONLINE: http://jipam.vu.edu.
au/article.php?sid=1009].
Real Hardy Inequality
Mohammad Sababheh
vol. 10, iss. 4, art. 104, 2009
Title Page
Contents
JJ
II
J
I
Page 18 of 18
Go Back
Full Screen
Close