Journal of Inequalities in Pure and
Applied Mathematics
http://jipam.vu.edu.au/
Volume 4, Issue 1, Article 19, 2003
SOME INTEGRAL INEQUALITIES RELATED TO HILBERT’S
T.C. PEACHEY
S CHOOL OF C OMPUTER S CIENCE AND S OFTWARE E NGINEERING
M ONASH U NIVERSITY
C LAYTON , VIC 3168, AUSTRALIA .
tcp@csse.monash.edu.au
Received 9 October, 2002; accepted 29 January, 2003
Communicated by T.M. Mills
A BSTRACT. We prove some integral inequalities involving the Laplace transform. These are
sharper than some known generalizations of the Hilbert integral inequality including a recent
result of Yang.
Key words and phrases: Hilbert inequality, Laplace transform.
2000 Mathematics Subject Classification. 26D15.
1. I NTRODUCTION
By the Hilbert integral inequality we mean the following well-known result.
Theorem 1.1. Suppose p > 1, q = p/(p − 1), functions f and g to be non-negative and
Lebesgue measurable on (0, ∞) then
Z ∞
p1 Z ∞
1q
Z ∞Z ∞
f (u)g(v)
1 1
p
q
(1.1)
dudv ≤ B
,
f (u) du
g(v) dv ,
u+v
p q
0
0
0 0
where B is the beta function. The inequality is strict unless f or g are null.
The constant B p1 , 1q = π cosec πp is known to be the best possible. See [3, Chapter 9],
for the history of this result.
Many authors on integral inequalities follow the practice of Hardy, Littlewood and Polya [3]
of implicitly assuming all functions mentioned are measurable and non-negative. We intend to
make such conditions explicit. Further, if one of the integrals on the right of (1.1) is infinite
then the strict inequality gives the impression that the
R ∞left side is finite. Recent
R ∞authors such as
Yang [7] avoid this by adding the conditions 0 < 0 f p (u)du < ∞, 0 < 0 g q (v)dv < ∞
to give the strict inequality. Hence it becomes convenient to introduce a class of functions that
satisfy all the required conditions. We define P (E) to be the class of functions f : E → R such
that on E:
ISSN (electronic): 1443-5756
c 2003 Victoria University. All rights reserved.
105-02
2
T.C. P EACHEY
(1)
(2)
(3)
(4)
f
f
f
f
is measurable,
is non-negative,
R
is not null (henceRpositive on a set of positive measure so E f > 0) and
is integrable (so E f < ∞).
Yang [7], [8] has recently found various inequalities related to that above. One of these (in
our notation) is
Theorem 1.2. If p > 1, q = p/(p − 1), λ > 2 − min(p, q) and (u − a)1−λ f p (u) and (v −
a)1−λ g q (v) are in P (a, ∞) then
Z ∞Z ∞
f (u)g(v)
dudv
(u + v − 2a)λ
a a
Z ∞
p1 Z ∞
1q
p+λ−2 q+λ−2
1−λ
p
1−λ
q
<B
,
(u − a) f (u) du
(v − a) g(v) dv .
p
q
a
a
We will specifically write out the case a = 0 since this case is equivalent to the complete
theorem.
Theorem 1.3. If p > 1, q = p/(p − 1), λ > 2 − min(p, q), u1−λ f p (u) and v 1−λ g q (v) are in
P (0, ∞) then
Z ∞Z ∞
f (u)g(v)
dudv
(u + v)λ
0 0
1q
Z ∞
p1 Z ∞
p+λ−2 q+λ−2
<B
,
u1−λ f (u)p du
v 1−λ g(v)q dv .
p
q
0
0
To show that Theorem 1.3 implies Theorem 1.2, change the variables u and v in Theorem 1.3
to s = u + a and t = v + a and then replace f (s − a), g(t − a) by φ(s) and ψ(t) respectively.
Thus Theorem 1.3 is equivalent to Theorem 1.2. Note also that for the case λ = 1, Theorem 1.3
reduces to Theorem 1.1.
This paper will consider a further generalization of this inequality.
Theorem 1.4. If p > 1, q = p/(p − 1), b > − p1 , c > − 1q and up−pb−2 f p (u) and v q−qc−2 g q (v)
are in P (0, ∞) then
Z ∞Z ∞
f (u)g(v)
1
1 u1−b−2/p f (u) v 1−c−2/q g(v) .
(1.2)
<
B
b
+
,
c
+
p
q
(u + v)b+c+1 dudv
p
q
0 0
This reduces to Theorem 1.3 for the case b = (λ + p − 3)/p, c = (λ + q − 3)/q.
Theorem 1.4 is itself a special case of the Hardy-Littlewood-Polya inequality, Proposition 319
of [3]. Recall that a function K(u, v) is homogeneous of degree −1 if K(λu, λv) = λ−1 K(u, v)
for allowed u, v, λ. The inequality is
Theorem 1.5. If p > 1, q = p/(p − 1), φp and ψ q are in P (0, ∞) and K(u, v) is positive on
(0, ∞) × (0, ∞), homogeneous of degree −1 with
Z ∞
1
K(u, 1)u− p du = k
0
then
Z ∞Z
∞
K(u, v)φ(u)ψ(v) dudv < k kφkp kψkq
(1.3)
0
0
J. Inequal. Pure and Appl. Math., 4(1) Art. 19, 2003
http://jipam.vu.edu.au/
H ILBERT ’ S I NTEGRAL I NEQUALITY
3
and
Z
(1.4)
∞
0
K(u, v)φ(u) du
< k kφkp .
p
The constant k is the best possible.
To obtain Theorem 1.4 substitute
2
2
ub+ p −1 v c+ q −1
K(u, v) =
(u + v)b+c+1
2
2
and φ(u) = u1−b− p f (u), ψ(v) = v 1−c− q g(v).
In summary, we have a chain of generalizations:
Th.1.1 (Hilbert) ⇐ Th.1.2 (Yang) ⇔ Th.1.3 ⇐ Th.1.4 ⇐ Th.1.5 (HLP).
This paper presents a new inequality Theorem 2.1, sharper than Theorem 1.4, involving the
Laplace transform. Logically the implications are
Th.2.2
⇒ Th.2.1 ⇒ Th.1.4.
Th.1.5 (HLP) ⇒ Th.2.3
In Section 2 we state and prove this result. Section 3 considers the extension of our theory to
the case of non-conjugate p and q. In that section, Theorem 1.1 generalizes to Theorem 3.1,
Theorem 1.5 generalizes to Theorem 3.2 and Theorem 2.1 to Theorem 3.4. The structure of the
section is
Th.2.2
⇒ Th.3.4.
Th.3.2 (Bonsall) ⇒ Th.3.3
2. A S HARPER I NEQUALITY
Theorem 2.1. Suppose p > 1, q = p/(p − 1), b > − p1 , c > − 1q , suppose up−pb−2 f p (u) and
v q−qc−2 g q (v) are P (0, ∞) and F , G are the respective Laplace transforms of f and g, then
Z ∞Z ∞
b
f (u)g(v)
1
s F (s) ksc G(s)k
(2.1)
dudv ≤
q
p
b+c+1
(u + v)
Γ(b + c + 1)
0 0
1
1 u1−b−2/p f (u) v 1−c−2/q g(v) .
(2.2)
< B b + ,c +
p
q
p
q
To show that (2.1) gives a considerably sharper bound than (1.2) consider the case p = q = 2,
b = c = 0 and f (x) = g (x) = e−x . Then F (s) = G (s) = 1/(s + 1) and
b
s F (s) = ksc G (s)k = 1
p
q
while
1
kf kp = kgkq = √ .
2
The right side of (2.1) is thus equal to 1 while that of (1.2) is π2 .
The case b = 1 − p2 , c = 1 − 2q for Theorem 2.1 has been considered by the author in [6,
Theorem 15].
The proof uses the following identity
Theorem 2.2. If a > −1, f and g are non-negative and measurable on (0, ∞) with respective
Laplace transforms F and G then
Z ∞Z ∞
Z ∞
f (u)g(v)
1
dudv =
sa F (s) G (s) ds.
(2.3)
a+1
(u
+
v)
Γ(a
+
1)
0
0 0
J. Inequal. Pure and Appl. Math., 4(1) Art. 19, 2003
http://jipam.vu.edu.au/
4
T.C. P EACHEY
The proof is a simple application of Fubini’s theorem,
Z ∞
Z ∞
Z ∞
Z ∞
a
a
−su
s F (s)G(s)ds =
s ds
e f (u) du
e−sv g(v) dv
0
0
Z0 ∞
Z ∞
Z ∞0
=
f (u) du
g(v) dv
sa e−s(u+v) ds
0
0
Z0 ∞Z ∞
Γ(a + 1)f (u)g(v)
=
dudv
(u + v)a+1
0 0
which is equation (2.3).
Despite the simplicity of the proof of (2.3) and the obvious relation with Hilbert’s inequality,
we can find no explicit reference to this identity in the literature before the case a = 0 in [6].
That case appears implicitly in Hardy’s proof of Widder’s inequality, [2]. Mulholland [5] used
the discrete analogue, namely
∞ X
∞
X
m=0 n=0
am b n
=
m+n+1
Z
∞
1X
0 m=0
am x
m
∞
X
an xn dx,
n=0
to prove the series case of Hilbert’s inequality.
Theorem 2.1 also depends on the following bound on Laplace transforms.
Theorem 2.3. Suppose p > 1, α +
transform of f , then
1
p
> 0, xp−pα−2 f p (x) is P (0, ∞) and F is the Laplace
1
ks F (s)kp < Γ α +
p
α
1−α− p2
f
(x)
x
.
p
u
2
2
This is also a corollary of Theorem 1.5. For if we make K(u, v) = e− v uα+ p −1 v −α− p and
2
φ(u) = u−α− p +1 f (u) then we obtain from (1.4)
−α− 2
1
1
1−α− p2
v
<Γ α+
pF
x
f
(x)
v p
p
p
and a substitution s = 1/v in the left side gives the theorem. We can find no direct mention of
this result although the last two formulae in Propositions 350 of [3] are special cases.
We may now prove Theorem 2.1. Since b + c > −1, Theorem 2.2 gives
Z ∞Z ∞
Z ∞
f (u)g(v)
1
dudv =
sb+c F (s) G (s) ds.
b+c+1
(u
+
v)
Γ(b
+
c
+
1)
0 0
0
Then by Hölder’s inequality
Z ∞Z ∞
b
f (u)g(v)
1
s F (s) ksc G(s)k
dudv
≤
q
p
(u + v)b+c+1
Γ(b + c + 1)
0 0
which is equation (2.1). Applying Theorem 2.3 to each norm, the right side of this is less than
Γ(b + p1 )Γ(c + 1q ) 1−c− 2q
1−b− p2
f (u) v
g(v)
u
Γ(b + c + 1)
p
q
and this is (2.2).
J. Inequal. Pure and Appl. Math., 4(1) Art. 19, 2003
http://jipam.vu.edu.au/
H ILBERT ’ S I NTEGRAL I NEQUALITY
5
3. N ON -C ONJUGATE PARAMETERS
Here we consider inequalities of the type considered in Section 2 but where q 6= p/(p −
1). (Henceforth we will use p0 and q 0 for the respective conjugates of p and q.) The earliest
investigation of this type seems to be Theorem 340 of Hardy, Littlewood and Polya, [3]. In our
notation this is
Theorem 3.1. Suppose p > 1, q > 1, p1 + 1q ≥ 1 and λ = 2 − p1 − 1q (so 0 < λ ≤ 1); suppose
f p and g q are in P (0, ∞) then
Z ∞Z ∞
f (u)g(v)
dudv < C kf kp kgkq ,
(u + v)λ
0 0
where C depends on p and q only.
Hardy, Littlewood and Polya did not give a specific
value
for the constant C. An alternative
1
1
λ
proof by Levin, [4] established that C = B λp0 , λq0 suffices but the paper did not decide
whether this was the best possible constant. This question remains open.
Just as Theorem 1.5 generalized Theorem 1.1 to a general kernel, Bonsall [1] has generalized
the above result.
Theorem 3.2. Suppose p > 1, q > 1, p1 + 1q ≥ 1, λ = 2 − p1 − 1q ; φp and ψ q are in P (0, ∞) and
K(u, v) is positive on (0, ∞) × (0, ∞), homogeneous of degree −1 with
Z ∞
0
K(x, 1)x−1/(λq ) dx = k
0
then
Z ∞Z
∞
K λ (u, v)φ(u)ψ(v) dudv < k λ kφkp kψkq
(3.1)
0
0
and
Z
(3.2)
0
∞
λ
K (u, v)φ(u) du
0 < k kφkp .
λ
q
Since this theorem is more than that claimed by Bonsall we will repeat and extend his proof,
using the standard methods of [3] Section 9.3.
Since p0 λ and q 0 λ are conjugate,
K λ (u, v)φ(u)ψ(v)
u 0−10 v 0−10 q
p
1
1
p q λ
p q λ
0
0
0
0
K(u, v) q φ q (u)
= K(u, v) p ψ p (v)
v
u
× φp(1−λ) (u)ψ q(1−λ) (v)
= F1 F2 F3 say.
1
p0
1
q0
+ (1 − λ) = 1, Holder’s inequality gives
Z ∞Z ∞
10 Z ∞Z
Z ∞Z ∞
p
p0
λ
K(u, v) φ(u)ψ(v) dudv ≤
F1 dudv
Then since
0
+
0
0
0
0
Z ∞Z
×
0
1
0
∞
1/(1−λ)
F3
∞
0
F2q
10
q
dudv
0
(1−λ)
dudv
0
1
0
= I1p I2q I31−λ .
J. Inequal. Pure and Appl. Math., 4(1) Art. 19, 2003
http://jipam.vu.edu.au/
6
T.C. P EACHEY
Here
Z ∞Z
∞
q
I1 =
K(u, v)ψ (v)
0
u −
∞
K(x, 1)ψ q (v)x
=
0
dudv
v
0
Z ∞Z
1
q0 λ
− q10 λ
dxdv = k kψkqq .
0
Similarly
Z ∞Z
∞
K(1, x)φp (v)x
I2 =
0
and
∞
φ(u)p ψ(v)q dudv = kφkpp kψkqq .
I3 =
0
0
dxdu = k kφkpp
0
Z ∞Z
Substituting these Ii , gives
Z ∞Z
(3.3)
− p10 λ
0
∞
K λ (u, v)φ(u)ψ(v)dudv ≤ k λ kφkp kψkq .
0
Note that equality in (3.3) can only occur if one of the Fi is null or all are effectively proportional, see [3, Proposition 188]. The first possibility would contradict one of the hypotheses;
the alternative implies that for almost all v,
v − 10
u − 10
p λ
q λ
K(u, v)φp (u)
= K(u, v)ψ q (v)
u
v
for almost all u. For such a v, φp (u) = Au−1 for some positive constant A, contradicting
kφkp < ∞. Thus the inequality in (3.3) is strict giving (3.1).
Finally we note that
Z ∞
Z ∞
ψ(v) dv
K λ (u, v)φ(u) du < k λ kφkp kψkq
0
0
for all ψ ∈ Lq . Equation (3.2) follows by the converse of Hölder’s inequality, [3, Proposition
191].
We next extend Theorem 2.3 to non-conjugate p and q.
Theorem 3.3. Suppose p > 1, q > 1, p1 + 1q ≥ 1, λ = 2 − p1 − 1q and α + q10 > 0; suppose
0
up(λ−α−2/q ) f p (u) is in P (0, ∞) and F is the Laplace transform of f , then
α
1 α
λ−α− q20
α+ q10 λ
Γ
+ 0 u
f (u) .
s F (s) 0 < λ
λ λq
p
q
This follows from Theorem 3.2 by substituting
0
0
K(u, v) = e−u/v u(α/λ)+(2/λq )−1 v −(α/λ)−(2/λq )
and
0
φ(u) = u−α−(2/q )+λ f (u)
giving k = Γ
α
λ
+
1
λq 0
and
Z
∞
λ
K (u, v)φ(u) du = v
0
−α− q20
λ
F
v
and then equation (3.2) gives the required inequality.
J. Inequal. Pure and Appl. Math., 4(1) Art. 19, 2003
http://jipam.vu.edu.au/
H ILBERT ’ S I NTEGRAL I NEQUALITY
7
The case q = p0 gives λ = 1 and reduces to Theorem 2.3. Another known case occurs if
q = p (which requires 1 < p ≤ 2) and α = 0 giving λ = 2/p0 and
10
2π p
kF kp0 <
kf kp ,
p0
which is Proposition 352 of [3].
A third case of interest can be obtained by substituting α = p10 −
with r ≥ p. Then Theorem 3.3 gives
1
1
1− p1 − r1
λ
0
kf kp
F (x) < λ p Γ
x
p0 λ
r
1
q0
and introducing r = q 0
1
where λ = p10 + 1r . This is given in [3] as Proposition 360 except that the constant λ p0 Γλ p10 λ
is not specified there.
We can now extend Theorem 2.1 to the case of non-conjugate parameters.
Theorem 3.4. Suppose p > 1, q > 1, p1 + 1q ≥ 1, q ≤ r ≤ p0 , b + r10 > 0 and c + 1r > 0;
0
0
0
up(1/p −1/r −b) f p (u) and v q(1/q −1/r−c) g q (v) are in P (0, ∞) and suppose F , G are the Laplace
transforms of f and g respectively, then
Z ∞Z ∞
b
1
f (u)g(v)
s F (s) 0 ksc G(s)k
(3.4)
dudv
≤
r
r
(u + v)b+c+1
Γ(b + c + 1)
0 0
1 1
1 1
− −b
− −c
< C u p0 r0 f (u) v q0 r g(v) ,
(3.5)
p
where β =
1
p0
+ r10 , γ =
1
q0
+
q
1
r
and
1
c
1
b
b+ r10 c+ r1 β
γ
C=β
γ Γ
+ 0
Γ
+
Γ−1 (b + c + 1).
β rβ
γ rγ
The proof of (3.4) proceeds as that of (2.1) except that p and q are replaced by r0 and r
respectively. We then apply Theorem 3.3 to ksb F (s)kr0 . In that theorem replace α by b, q by r
and λ = p10 + q10 by β = p10 + r10 , giving
b
1 p10 − r10 −b
b
b+ r10 β
ks F (s)kr0 < β
Γ
+ 0 u
f (u) .
β βr
p
The condition p1 + 1q ≥ 1 is satisfied because r ≤ p0 . Alternatively if in Theorem 3.3 we replace
α by c, p by q and q 0 by r with r ≥ q we obtain
c
1 q10 − r1 −c
c
c+ r1 γ
g(u)
ks G(s)kr < γ
Γ
+
u
.
γ γr
q
Applying these to (3.4) gives (3.5) and completes the proof.
The question arises as to whether the constant C in (3.5) is better than Levin’s B λ λp1 0 , λq1 0
− r10 , c =
1
q0
+ r10 and γ =
1
q0
in the case b =
where β =
better one.
1
p0
1
p0
− 1r . For that case
1
1
1
1
β
γ
0
0
C = βp γq Γ
Γ
Γ−1 (λ)
0
0
βp
γq
+ 1r . Experiments with Maple suggest that Levin’s constant is the
J. Inequal. Pure and Appl. Math., 4(1) Art. 19, 2003
http://jipam.vu.edu.au/
8
T.C. P EACHEY
R EFERENCES
[1] F.F. BONSALL, Inequalities with non-conjugate parameters, Quart. J. Math., 2 (1951), 135–150.
[2] G.H. HARDY, Remarks in addition to Dr Widder’s note on inequalities, J. London Math. Soc., 4
(1929), 199–202.
[3] G.H. HARDY, J.E. LITTLEWOOD AND G. POLYA, Inequalities, (Second Edition) CUP, 1952.
[4] V. LEVIN, On the two parameter extension and analogue of Hilbert’s inequality, J. London Math.
Soc., 11 (1936), 119–124.
[5] H.P. MULHOLLAND, Note on Hilbert’s double series theorem, J. London Math. Soc., 3 (1928),
197–199.
[6] T.C. PEACHEY, A framework for proving Hilbert’s inequality and related results, RGMIA Research
Report Collection, Victoria University of Technology, 2(3) (1999), 265–274.
[7] BICHENG YANG, On new generalizations of Hilbert’s inequality, J .Math. Anal. Appl., 248 (2000),
29–40.
[8] BICHENG YANG, On Hardy-Hilbert’s integral inequality, J .Math. Anal. Appl., 261 (2001), 295–
306.
J. Inequal. Pure and Appl. Math., 4(1) Art. 19, 2003
http://jipam.vu.edu.au/