Volume 10 (2009), Issue 4, Article 104, 8 pp.
A STUDY OF THE REAL HARDY INEQUALITY
MOHAMMAD SABABHEH
D EPARTMENT OF S CIENCE AND A RTS
P RINCESS S UMAYA U NIVERSITY FOR T ECHNOLOGY
A MMAN 11941 J ORDAN
sababheh@psut.edu.jo
URL: http://www.psut.edu.jo/sites/sababheh
Received 11 November, 2008; accepted 20 October, 2009
Communicated by L. Pick
A BSTRACT. 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.
Key words and phrases: Hardy’s inequality, inequalities involving Fourier transforms.
2000 Mathematics Subject Classification. 42A16, 42A05, 42B30.
1. I NTRODUCTION
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
X
|fˆ(n)|
|fˆ(−n)|
(1.1)
≤ ckf k1 + c
n
n
n=1
n=1
for all f ∈ L1 (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
0
307-08
∞
|fˆ(ξ)|2
dξ ≤ ckf k21 + c
ξ
Z
0
∞
|fˆ(−ξ)|2
dξ.
ξ
2
M OHAMMAD S ABABHEH
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).
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.
2. S ETUP
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
fˆ(ξ) =
f (x)e−ixξ dξ.
R
If fˆ ∈ L then the inversion formula for f holds:
Z
1
f (x) =
fˆ(ξ)eiξx dξ.
2π R
1
If f ∈ L2 then the Fourier transform of f is defined to be the L2 − limit:
Z n
fˆ(ξ) = lim
f (x)e−iξx dx.
n→∞
−n
In this case fˆ ∈ L2 and
1
f (x) = lim
n→∞ 2π
The Plancheral theorem then says:
Z
n
fˆ(ξ)eixξ dξ,
in the L2 sense.
−n
1
kf k2 = √ kfˆk2 .
2π
Lemma 2.1. For f, g ∈ L2 we have
Z
Z
1
ˆ
fˆ(ξ)g(ξ)dξ.
f (x)g(x)dx =
2π R
R
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
∞
X
(2.1)
ϕN (t) = 2π
fN (t + 2πj),
where fN (x) = N f (N x).
j=−∞
Then we have
n
ϕN ∈ L1 (T), ϕ̂N (n) = fˆ
and
lim kϕN kL1 (T) = kf kL1 (R) .
N →∞
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
(2.2)
J. Inequal. Pure and Appl. Math., 10(4) (2009), Art. 104, 8 pp.
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A S TUDY O F T HE R EAL H ARDY I NEQUALITY
3
to prove some Hardy-type inequalities on the real line using only a corresponding inequality on
the circle.
3. F ROM T HE C IRCLE T O T HE L INE
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)|
≤ Ckf k1 .
n
n=1
As a matter of notation, let H 1 (R) be defined by
n
o
1
1
ˆ
H (R) = f ∈ L (R) : f (ξ) = 0, ∀ξ < 0 .
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
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),
lim kϕN kL1 (T) = kf kL1 (R) and ϕ̂N (n) = fˆ(n/N ).
N →∞
Clearly ϕ̂N (n) = 0, ∀n < 0, hence Hardy’s inequality applies and we have
∞
X
|ϕ̂N (n)|
≤ CkϕN kL1 (T) .
n
n=1
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
(3.1)
0
1
|fˆ(ξ)|
≤ Ckf k1 .
ξ
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
Z 1 ˆ
|ĥ(ξ)|
|f (M ξ)|
≤ Ckhk1 ⇒
dξ ≤ Ckf k1 .
ξ
ξ
0
0
J. Inequal. Pure and Appl. Math., 10(4) (2009), Art. 104, 8 pp.
http://jipam.vu.edu.au/
4
M OHAMMAD S ABABHEH
Put M ξ = t to get
Z
M
(3.2)
0
|fˆ(ξ)|
dξ ≤ Ckf k1 .
ξ
Now, letting M tend to ∞, we obtain the result.
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:
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
This modifies the result proved in [8].
4. A NOTHER H ARDY I NEQUALITY
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
(
fˆj (ξ) =
2π
,
4j
4j−1 < ξ < 4j ,
0,
otherwise.
2π
,
4j
4j−1 < ξ < 4j ,
0,
otherwise,
Proof. Let
(
gj (ξ) =
2
then gj ∈ L . Therefore, gj is the Fourier transform of some function hj ∈ L2 and
Z
1
hj (x) =
g(ξ)eixξ dξ
2π R
= f (x).
However, ĥj = gj , which implies fˆj = gj as claimed.
Observe that the above proof implies that fj ∈ L2 and
J. Inequal. Pure and Appl. Math., 10(4) (2009), Art. 104, 8 pp.
http://jipam.vu.edu.au/
A S TUDY O F T HE R EAL H ARDY I NEQUALITY
5
Lemma 4.2. For fj as above, we have
√
kfj k2 =
6π
4j/2
.
Proof. Since fj ∈ L2 we have
1
kfj k2 = √ kfˆj k2
2π
1 2π
=√
2π 4j
√
6π
= j/2 .
4
Z
! 12
4j
dξ
4j−1
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 ).
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|
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
>x
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
where kx is the smallest positive integer such that 4−j ≤ x for all j ≥ kx .
J. Inequal. Pure and Appl. Math., 10(4) (2009), Art. 104, 8 pp.
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6
M OHAMMAD S ABABHEH
Consequently
X 1 sin2 (4j−1 x/2)
14 1
≤
4j
x
x 3 4kx
−j
4
≤x
14
4
x= .
x3
3
≤
On the other hand, if
1 sin2 (4j−1 x/2)
4−j >x 4j
x
P
is not zero, we get
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
2
log4 (1/x)
X
1
= x
4j
64
j=1
1 4log4 (1/x) − 4
x
64
3
1
≤
.
192
≤
Thus, we have
FM (x) ≤ 60
4
1
+
3 192
:= 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
for all f ∈ X.
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 , where j
4j j
j−1 j
j
j−1
M
M
is the unique index such that ξ ∈ [4 , 4 ] ∪ [−4 , −4 ] if −4 ≤ ξ ≤ 4 and F̂ (ξ) = 0
otherwise.
J. Inequal. Pure and Appl. Math., 10(4) (2009), Art. 104, 8 pp.
http://jipam.vu.edu.au/
A S TUDY O F T HE R EAL H ARDY I NEQUALITY
7
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
∞ Z j
Z
1 1 ˆ
1 X 4 ˆ
f (ξ)F̂ (ξ)dξ f (ξ)F̂ (ξ)dξ −
≥
2π −1
2π j=1 4j−1
∞ Z j
1 X 4 ˆ
−
f (−ξ)F̂ (−ξ)dξ .
2π 4j−1
j=1
That is,
1
(4.2)
2π
∞ Z j
Z
X 4
1 1 ˆ
ˆ
f (ξ)F̂ (ξ)dξ f (ξ)F̂ (ξ)dξ ≤ Ckf k1 +
2π −1
j−1
j=1 4
∞ Z j
4
X
1 ˆ
+
f (−ξ)F̂ (−ξ)dξ .
2π 4j−1
j=1
However, when ξ ∈ (4j−1 , 4j ) ∪ (−4j , 4j−1 ), we have F̂ (ξ) = 2π
σ , where σj is the sign of fˆ
4j j
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
12 Z
1
1
|F̂ (ξ)|2 dξ
≤ kf k1
dξ
−1
−1
√
≤ 2kf k1 kF̂ k2
√
√
= 2kf k1 2πkF k2
∞
X
√
≤ 2 πkf k1 × 2
kfj k2
√
= 4 6πkf k1 ,
where we have used the facts
M
X
F (x) =
σj fj (x) − fj (x)
12
j=1
and kfj k2 =
√
6π4−j/2 .
j=1
Consequently, (4.2) becomes
∞ Z 4j
∞ Z 4j
X
X
√
|fˆ(ξ)|
|fˆ(−ξ)|
(4.3)
dξ
≤
Ckf
k
+
2
6kf
k
+
dξ.
1
1
j
j
4
4
j−1
j−1
4
4
j=1
j=1
However, when ξ ∈ [4j−1 , 4j ] we have
1
1
≤
j
4
ξ
J. Inequal. Pure and Appl. Math., 10(4) (2009), Art. 104, 8 pp.
and
1
1
≥ .
j
4
4ξ
http://jipam.vu.edu.au/
8
M OHAMMAD S ABABHEH
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.
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
X
|fˆ(−n)|
|fˆ(n)|
0
0
≤ C kf k1 + C
.
n
n
n=1
n=1
(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.
R EFERENCES
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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
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[5] W. RUDIN, Real and Complex Analysis, 3rd edition, McGraw-Hill Book Co., New York, 1987.
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J. Inequal. Pure and Appl. Math., 10(4) (2009), Art. 104, 8 pp.
http://jipam.vu.edu.au/