Journal of Inequalities in Pure and
Applied Mathematics
http://jipam.vu.edu.au/
Volume 5, Issue 2, Article 47, 2004
ON AN INTEGRAL INEQUALITY
J. PEČARIĆ AND T. PEJKOVIĆ
FACULTY OF T EXTILE T ECHNOLOGY
U NIVERSITY OF Z AGREB
P IEROTTIJEVA 6, 10000 Z AGREB
C ROATIA .
pecaric@mahazu.hazu.hr
pejkovic@student.math.hr
Received 20 October, 2003; accepted 13 March, 2004
Communicated by A. Lupaş
A BSTRACT. In this article we give different sufficient conditions for inequality
Rb
γ
f (x) dx to hold.
a
R
b
a
β
α
f (x) dx ≥
Key words and phrases: Integral inequality, Inequalities between means.
2000 Mathematics Subject Classification. 26D15.
1. I NTRODUCTION
In this paper we wish to investigate some sufficient conditions for the following inequality:
Z b
β Z b
α
(1.1)
f (x) dx ≥
f (x)γ dx.
a
a
This is a generalization of the inequalities that appear in the papers [4, 5, 6, 7, 8, 9].
F. Qi in [7] considered inequality (1.1) for α = n + 2, β = 1/(n + 1), γ = 1, n ∈ N. He
proved that under conditions
f ∈ Cn ([a, b]);
f (i) (a) ≥ 0, 0 ≤ i ≤ n − 1;
f (n) (x) ≥ n!, x ∈ [a, b]
the inequality is valid.
Later, S. Mazouzi and F. Qi gave what appeared to be a simpler proof of the inequality under
the same conditions (Corollary 3.6 in [1]). Unfortunately their proof was incorrect. Namely,
they made a false substitution and arrived at the condition f (x) ≥ (n + 1)(x − a)n which is not
true, e.g. for function f (x) = x − a, whereas this function obviously satisfies the conditions of
the theorem if n = 1.
ISSN (electronic): 1443-5756
c 2004 Victoria University. All rights reserved.
148-03
2
J. P E ČARI Ć AND T. P EJKOVI Ć
K.-W. Yu and F. Qi ([9]) and N. Towghi ([8]) gave other conditions for the inequality (1.1) to
hold under this special choice of constants α, β, γ.
T.K. Pogány in [6], by avoiding the assumption of differentiability used in [7, 8, 9], and
instead using the inequalities due to Hölder, Nehari (Lemma 2.4) and Barnes, Godunova and
Levin (Lemma 2.5) established some inequalities which are a special case of (1.1) when α = 1
or γ = 1.
To obtain some conditions for the inequality (1.1) we will first proceed similarly to T.K.
Pogány ([6]) and in the second part of this article we will be using a method from the paper [4].
2. C ONDITIONS BASED ON I NEQUALITIES B ETWEEN M EANS
We want to transform inequality (1.1) to a form more suitable for us. It can easily be seen
that if f (x) ≥ 0, for all x ∈ [a, b] and γ > 0, inequality (1.1) is equivalent to
 R
! α1  αβ
! γ1
γ
Rb
b
α
γ
β−1
f
(x)
dx
f
(x)
dx
a
 (b − a) γ ≥
 a
(2.1)
.
b−a
b−a
Definition 2.1. Let f be a nonnegative and integrable function on the segment [a, b]. The rmean (or the r-th power mean) of f is defined as
 R
r1
b
r

a f (x) dx

(r 6= 0, +∞, −∞),



b−a
Rb

a ln f (x)dx
(r = 0),
M[r] (f ) := exp
b−a



m
(r = −∞),



M
(r = +∞).
where m = inf f (x) and M = sup f (x) for x ∈ [a, b].
According to the previous definition inequality (2.1) can be written as
αβ
β−1
(2.2)
M[α] (f ) γ (b − a) γ ≥ M[γ] (f )
We will be using the following inequalities:
Lemma 2.1 (power mean inequality, [2]). If f is a nonnegative function on [a, b] and −∞ ≤
r < s ≤ +∞, then
M[r] (f ) ≤ M[s] (f ).
Lemma 2.2 (Berwald inequality, [3, 5]). If f is a nonnegative concave function on [a, b], then
for 0 < r < s we have
(r + 1)1/r [r]
M[s] (f ) ≤
M (f ).
(s + 1)1/s
Lemma 2.3 (Thunsdorff inequality, [3]). If f is a nonnegative convex function on [a, b] with
f (a) = 0, then for 0 < r < s we have
(r + 1)1/r [r]
M (f ) ≥
M (f ).
(s + 1)1/s
[s]
Lemma 2.4 (Nehari inequality, [2]). Let f , g be nonnegative concave functions on [a, b]. Then,
for p, q > 0 such that p−1 + q −1 = 1, we have
1q
Z b
p1 Z b
Z b
q
p
g(x) dx
≤ N(p, q)
f (x)g(x)dx,
f (x) dx
a
J. Inequal. Pure and Appl. Math., 5(2) Art. 47, 2004
a
a
http://jipam.vu.edu.au/
O N AN I NTEGRAL I NEQUALITY
where N(p, q) =
3
6
.
(1+p)1/p (1+q)1/q
Lemma 2.5 (Barnes-Godunova-Levin inequality, [3, 2]). Let f , g be nonnegative concave functions on [a, b]. Then, for p, q > 1 we have
Z
b
p
p1 Z
b
q
f (x) dx
b
Z
≤ B(p, q)
g(x) dx
a
where B(p, q) =
1q
a
f (x)g(x)dx,
a
6(b−a)1/p+1/q−1
.
(1+p)1/p (1+q)1/q
Let us first state our results in a clear table. Each result is an independent set of conditions
that guarantee the inequality (1.1) is valid.
Result
Conditions on
constants α, β, γ, a, b
Conditions on function f
(holds for all x ∈ [a, b])
1.
α ≥ γ > 0, αβ > γ
f (x) ≥ (b − a) αβ−γ
5.
α ≤ γ > 0, αβ < 0
α ≥ γ > 0, αβ ≥ γ,
β−1
(b − a) γ ≥ 1
α ≥ γ > 0, αβ ≤ γ,
β−1
(b − a) γ ≥ 1
0 < α ≤ γ, αβ ≥ γ
β−1
1/α
(b − a) γ ≥ (α+1)
(γ+1)1/γ
0 < α ≤ γ, αβ ≤ γ
β−1
1/α
(b − a) γ ≥ (α+1)
(γ+1)1/γ
0 < α ≤ γ, αβ > γ
6.
0 < γ ≤ α, β < 0
7.
0 < γ ≤ α, αβ > γ
2.
3 (i).
3 (ii).
4 (i).
4 (ii).
−β+1
0 ≤ f (x) ≤ (b − a)
f (x) ≥ 1
Lemma for
the proof
Lemma 2.1
−β+1
αβ−γ
Lemma 2.1
Lemma 2.1
0 ≤ f (x) ≤ 1
Lemma 2.1
f concave
f (x) ≥ 1
f concave
0 ≤ f (x) ≤ 1
f concave, f (x) ≥
γ
αβ−γ
1−β
1/α
(b − a) αβ−γ (α+1)
(γ+1)1/γ
Lemma 2.2
f concave, 0 ≤ f (x) ≤
αβ
αβ−γ
1−β
1/α
(b − a) αβ−γ (α+1)
(γ+1)1/γ
Lemma 2.2
Lemma 2.2
Lemma 2.2
f convex, f (a) = 0, f (x) ≥ Lemma 2.3
α−γ
β
(b−a) α(αβ−γ)
(bα+1 −aα+1 )1/α
·
(α+1) αβ−γ
1
x
(γ+1) αβ−γ
8.
0 < α ≤ γ, β < 0
f convex, f (a) = 0,
0 ≤ f (x) ≤
αβ
1−β
(α+1)1/α αβ−γ
αβ−γ
(b − a)
(γ+1)1/γ
Lemma 2.3
9.
0 < γ < α, β < 0
f concave,
1−β
0 ≤ f (x) ≤ (b − a) αβ−γ
Lemma 2.4
αβ
×
6 γ(γ−αβ)
β(α−γ)
β
γ(γ−αβ) α+γ γ−αβ
( 2α−γ
( γ )
α−γ )
Remark 2.6. Observe that in the results 4(i). and 5. it is enough for the condition on f to hold
in endpoints of segment [a, b] (ie., for f (a) and f (b)).
Remark 2.7. There is only one result in the table obtained with the help of Lemma 2.4 and
none with the Lemma 2.5 because the constants in the conditions are quite complicated.
J. Inequal. Pure and Appl. Math., 5(2) Art. 47, 2004
http://jipam.vu.edu.au/
4
J. P E ČARI Ć AND T. P EJKOVI Ć
Remark 2.8. If we make the substitution γ 7→ 1, β 7→
acquired.
1
β
in Result 1, Theorem 2.1 in [6] is
We will prove only a few results after which the method of proving the others will become
clear.
Proof of Result 1. Lemma 2.1 implies that
M[α] (f ) ≥ M[γ] (f ).
(2.3)
Also
−β+1
M[α] (f ) ≥ M[−∞] (f ) ≥ (b − a) αβ−γ ,
so by raising this inequality to a power, we get
αβ−γ
−β+1
(2.4)
M[α] (f ) γ ≥ (b − a) γ .
Multiplying (2.3) and (2.4) we get (2.2).
Proof of Result 3 (i). M[α] (f ) ≥ 1 because f (x) ≥ 1, so from
αβ
γ
≥ 1 and Lemma 2.1 it follows
αβ
M[α] (f ) γ ≥ M[α] (f ) ≥ M[γ] (f ).
(2.5)
By multiplication of (2.5) and the condition (b − a)
β−1
γ
≥ 1 we get (2.2).
Proof of Result 5. From
M[α] (f ) ≥ M[−∞] (f ) ≥ (b − a)
and
αβ−γ
γ
1−β
αβ−γ
(α + 1)1/α
(γ + 1)1/γ
γ
αβ−γ
> 0 we obtain
αβ−γ
β−1
(α + 1)1/α
M[α] (f ) γ (b − a) γ ≥
.
(γ + 1)1/γ
(2.6)
According to Lemma 2.2:
M[α] (f )
(2.7)
(α + 1)1/α
≥ M[γ] (f ).
(γ + 1)1/γ
From (2.6) and (2.7), by multiplication, we arrive at (2.2).
Proof of Result 7. Since
α−γ
f (x) ≥
(b − a) α(αβ−γ)
β
(α + 1) αβ−γ
x
·
1
(bα+1 − aα+1 )1/α (γ + 1) αβ−γ
by integration it follows that
(2.8)
[α]
M (f ) ≥ (b − a)
1−β
αβ−γ
(α + 1)1/α
(γ + 1)1/γ
γ
αβ−γ
.
However, Lemma 2.3 implies inequality (2.7). Thus, from (2.8) and (2.7) we finally find that
inequality (2.2) is valid.
J. Inequal. Pure and Appl. Math., 5(2) Art. 47, 2004
http://jipam.vu.edu.au/
O N AN I NTEGRAL I NEQUALITY
3. C ONDITIONS A SSOCIATED
WITH THE
5
F UNCTIONS WITH B OUNDED D ERIVATIVE
In this section we will prove inequality (1.1) under different assumptions including the differentiability of f and boundedness of its derivative.
J. Pečarić and W. Janous proved in [4] the following theorem.
Theorem 3.1. Let 1 < p ≤ 2 and r ≥ 3. The differentiable function f : [0, c] → R satisfies
f (0) = 0 and 0 ≤ f 0 (x) ≤ M for all 0 ≤ x ≤ c, c subject to
1
p(p − 1)22−p M p−r r−2p+1
(3.1)
0<c≤
.
r−1
Then
Z
p Z
c
c
f (x)r dx.
≥
f (x)dx
0
0
(If f 0 (x) ≥ M the reverse inequality holds true under the condition that the second inequality
in (3.1) is reversed.)
Remark 3.2. The emphasized words were left out of [4].
The following generalization will be proved:
Theorem 3.3. Let α > 0, 1 < β ≤ 2 and γ ≥ 2α + 1. The differentiable function f : [0, c] → R
satisfies f (0) = 0 and 0 ≤ f 0 (x) ≤ M for all 0 ≤ x ≤ c, c subject to
1
β(β − 1)(α + 1)2−β M αβ−γ γ−αβ−β+1
.
(3.2)
0<c≤
γ−α
Then
Z
Z
β
c
α
f (x) dx
c
f (x)γ dx.
≥
0
0
Remark 3.4. For α = 1, β = p, γ = r, we get Theorem 3.1.
Proof. From f (0) = 0 and 0 ≤ f 0 (x) ≤ M we obtain
Z x
M α xα+1
α
α α
0 ≤ f (x) ≤ M x and 0 ≤
f (t)α dt ≤
α+1
0
Now we define
Z
Z
β
x
f (t)α dt
F (x) :=
0
0
for
0 ≤ x ≤ c.
x
f (t)γ dt.
−
0
α
Then F (0) = 0 and F (x) = f (x) g(x), where
Z x
β−1
α
g(x) := β
f (t) dt
− f (x)γ−α .
0
0
α
Clearly, g(0) = 0 and g (x) = f (x) h(x), where
Z x
β−2
α
h(x) := β(β − 1)
f (t) dt
− (γ − α)f (x)γ−2α−1 f 0 (x).
0
From the conditions of the theorem we have
α α+1 β−2
M x
h(x) ≥ β(β − 1)
− (γ − α)(M x)γ−2α−1 M
α+1
= M γ−2α x(α+1)(β−2) β(β − 1)(α + 1)2−β M αβ−γ − (γ − α)xγ−αβ−β+1
J. Inequal. Pure and Appl. Math., 5(2) Art. 47, 2004
http://jipam.vu.edu.au/
6
J. P E ČARI Ć AND T. P EJKOVI Ć
Thus, since (3.2) is equivalent to
β(β − 1)(α + 1)2−β M αβ−γ ≥ (γ − α)xγ−αβ−β+1 ,
0
x ∈ [0, c],
0
we have h(x) ≥ 0, g (x) ≥ 0, g(x) ≥ 0, F (x) ≥ 0 and finally F (x) ≥ 0. So F (c) ≥ 0.
Substituting c = a − b and translating function f a units to the right (f (x) 7→ f (x − a)) we
obtain the following theorem.
Theorem 3.5. Let α > 0, 1 < β ≤ 2 and γ ≥ 2α + 1. The differentiable function f : [a, b] → R
satisfies f (a) = 0 and 0 ≤ f 0 (x) ≤ M for all a ≤ x ≤ b, where
1
β(β − 1)(α + 1)2−β M αβ−γ γ−αβ−β+1
(3.3)
0<b−a≤
.
γ−α
Then the inequality (1.1) holds.
R EFERENCES
[1] S. MAZOUZI AND F. QI, On an open problem regarding an integral inequality, J. Inequal. Pure
Appl. Math., 4(2)(2003), Art 31. [ONLINE: http://jipam.vu.edu.au/article.php?
sid=269].
[2] D.S. MITRINOVIĆ, Analitičke Nejednakosti, Gradjevinska knjiga, Beograd, 1970. (English edition:
Analyitic Inequalities, Springer-Verlag, Berlin-Heidelberg-New York, 1970.)
[3] J.E. PEČARIĆ, F. PROSCHAN AND Y.L. TONG, Convex Functions, Partial Orderings, and Statistical Applications, Academic Press, Inc., Boston-San Diego-New York, 1992.
[4] J.E. PEČARIĆ AND W. JANOUS, On an integral inequality, El. Math., 46, 1991.
[5] J.E. PEČARIĆ, O jednoj nejednakosti L. Berwalda i nekim primjenama, ANU BIH Radovi, Odj. Pr.
Mat. Nauka, 74(1983), 123–128.
[6] T.K. POGÁNY, On an open problem of F. Qi, J. Inequal. Pure Appl. Math., 3(4) (2002), Art 54.
[ONLINE: http://jipam.vu.edu.au/article.php?sid=206].
[7] F. QI, Several integral inequalities, J. Inequal. Pure Appl. Math., 1(2) (2000), Art 19. [ONLINE:
http://jipam.vu.edu.au/article.php?sid=113].
[8] N. TOWGHI, Notes on integral inequalities, RGMIA Res. Rep. Coll., 4(2) (2001), 277–278.
[9] K.-W. YU
23–25.
AND
F. QI, A short note on an integral inequality, RGMIA Res. Rep. Coll., 4(1) (2001),
J. Inequal. Pure and Appl. Math., 5(2) Art. 47, 2004
http://jipam.vu.edu.au/