Journal of Inequalities in Pure and
Applied Mathematics
http://jipam.vu.edu.au/
Volume 5, Issue 3, Article 74, 2004
NOTE ON SOME HADAMARD-TYPE INEQUALITIES
M. KLARIČIĆ BAKULA AND J. PEČARIĆ
D EPARTMENT OF M ATHEMATICS
FACULTY OF NATURAL S CIENCES , M ATHEMATICS AND E DUCATION
U NIVERSITY OF S PLIT , T ESLINA 12
21 000 S PLIT, C ROATIA .
milica@pmfst.hr
FACULTY OF T EXTILE T ECHNOLOGY
U NIVERSITY OF Z AGREB
P IEROTTIJEVA 6, 10000 Z AGREB
C ROATIA .
pecaric@hazu.hr
Received 20 January, 2004; accepted 05 April, 2004
Communicated by C. Giordano
A BSTRACT. Some Hadamard-type inequalities involving the product of two convex functions
are obtained. Our results generalize the corresponding results of B.G.Pachpatte.
Key words and phrases: Integral inequalities, Hadamard’s inequality.
2000 Mathematics Subject Classification. 26D15, 26D20.
1. I NTRODUCTION
Let f be a convex function on [a, b] ⊂ R. The following double inequality:
Z b
a+b
1
f (a) + f (b)
(1.1)
f
≤
f (x)dx ≤
2
b−a a
2
is known in the literature as Hadamard’s inequality [1, p. 137], [2, p. 10] for convex functions.
Recently B.G.Pachpatte [3] considered some new integral inequalities, analogous to that of
Hadamard, involving the product of two convex functions. In [3] the following theorem has
been proved:
Theorem 1.1. Let f and g be nonnegative, convex functions on [a, b] ⊂ R. Then
(i)
Z b
1
1
1
(1.2)
f (x) g (x) dx ≤ M (a, b) + N (a, b) ,
b−a a
3
6
ISSN (electronic): 1443-5756
c 2004 Victoria University. All rights reserved.
018-04
2
M. K LARI ČI Ć BAKULA AND J. P E ČARI Ć
(ii)
(1.3)
2f
a+b
2
Z b
a+b
1
1
1
g
≤
f (x) g (x) dx + M (a, b) + N (a, b) ,
2
b−a a
6
3
where M (a, b) = f (a) g (a) + f (b) g (b) and N (a, b) = f (a) g (b) + f (b) g (a). Inequalities (1.2) and (1.3) are sharp in the sense that equalities hold for some f (x) and
g (x) on [a, b].
In the following Theorem 1.2 we give a variant of the corresponding Theorem 2 in [3].
Theorem 1.2. Let f and g be nonnegative, convex functions on [a, b] ⊂ R. Then
(i)
3
(1.4)
2 (b − a)2
Z bZ bZ
1
f (tx + (1 − t) y) g (tx + (1 − t) y) dtdxdy
a
a
0
1
≤
b−a
Z
b
f (x) g (x) dx +
a
1
[M (a, b) + N (a, b)] ;
8
(ii)
3
(1.5)
b−a
Z bZ
1
f
a
0
a+b
g tx + (1 − t)
dtdx
tx + (1 − t)
2
Z b
1
1
f (x) g (x) dx + [M (a, b) + N (a, b)] ,
≤
b−a a
2
a+b
2
where M (a, b) and N (a, b) are as in Theorem 1.1.
It should be noted that in [3, Theorem 2] inequalities (3) and (4) are established. Inequality
(3) from [3, Theorem 2] is a variant of our inequality (1.4) in which
1 M (a, b) + N (a, b)
8
(b − a)2
stands in place of the term 81 [M (a, b) + N (a, b)]. Analogously, inequality (4) from [3, Theorem 2] is a variant of our inequality (1.5) in which
1 1+b−a
[M (a, b) + N (a, b)]
4
b−a
stands in place of the term 12 [M (a, b) + N (a, b)] .
However, one can compare inequalities (3) and (4) with (1.4) and (1.5), respectively, to find
out that estimates given by (1.4) and (1.5) are better (worse) than those given by (3) and (4) in
[3, Theorem 2] in case of b − a < 1 (b − a > 1).
But on careful inspection of the proof in [3, Theorem 2], the reader can notice some errors in
Pachpatte’s calculation, so inequalities (3) and (4) in [3, Theorem 2] are in fact incorrect.
The aim of this paper is to prove some simple generalizations of Theorem 1.1 and Theorem
1.2, which additionally involve weight functions and also nonlinear transformations of the base
interval [a, b]. Those generalizations are established in Theorem 2.1 and Theorem 2.3. The
above cited Theorem 1.1 is a special case of Theorem 2.1, while the above Theorem 1.2 is a
special case of our Theorem 2.3 (see Remark 2.4).
J. Inequal. Pure and Appl. Math., 5(3) Art. 74, 2004
http://jipam.vu.edu.au/
H ADAMARD - TYPE INEQUALITIES
3
2. R ESULTS
Throughout the rest of the paper we shall use the following notation
[h; x, y] =
h (y) − h (x)
,
y−x
x 6= y
e
h (t) = th (α + β − t) ,
b
h (t) = th (t)
where h : [α, β] → R is a function, [α, β] ⊂ R, x, y, t ∈ [α, β]. Note that from the above
equalities we get
h
i βh (α) − αh (β)
e
h; α, β =
,
β−α
h
i βh (β) − αh (α)
b
h; α, β =
,
β−α
and, by simple calculation,
h
i h
i
b
e
(2.1)
h; α, β − h; α, β = (α + β) [h; α, β] .
The following results are valid:
Theorem 2.1. Let f be a nonnegative convex function on [m1 , M1 ], g a nonnegative convex
function on [m2 , M2 ], u : [a, b] → [m1 , M1 ] and v : [a, b] → [m2 , M2 ] continuous functions,
and p : [a, b] → R a positive integrable function. Then
(i)
Z
1 b
p (x) f (u (x)) g (v (x)) dx
(2.2)
P a
Z
1 b
≤ [f ; m1 , M1 ] [g; m2 , M2 ]
p (x) u (x) v (x) dx
P a
Z
1 b
+ [f ; m1 , M1 ] [e
g ; m2 , M2 ]
p (x) u (x) dx
P a
Z b
h
i
1
+ fe; m1 , M1 [g; m2 , M2 ]
p (x) v (x) dx
P a
h
i
e
+ f ; m1 , M1 [e
g ; m2 , M2 ] .
(ii)
(2.3) f
Z b
m 2 + M2
1
g
≤
p (x) f (u (x)) g (v (x)) dx
2
4P a
Z b
+
p (x) f (M1 + m1 − u (x)) g (M2 + m2 − v (x)) dx
a
Z b
1
+
−2 [f ; m1 , M1 ] [g; m2 , M2 ]
p (x) u (x) v (x) dx
4P
a
Z b
+ ([b
g ; m2 , M2 ] − [e
g ; m2 , M2 ]) [f ; m1 , M1 ]
p (x) u (x) dx
m 1 + M1
2
a
J. Inequal. Pure and Appl. Math., 5(3) Art. 74, 2004
http://jipam.vu.edu.au/
4
M. K LARI ČI Ć BAKULA AND J. P E ČARI Ć
+
h
i
h
fb; m1 , M1 − fe; m1 , M1
i
Z
[g; m2 , M2 ]
b
p (x) v (x) dx
a
i
h
i
1 h e
f ; m1 , M1 [b
g ; m2 , M2 ] + fb; m1 , M1 [e
g ; m2 , M2 ] ,
4
+
where P =
Rb
a
p (x) dx.
Proof. For any x ∈ [a, b] we can write
(2.4)
u (x) =
u (x) − m1
M1 − u (x)
m1 +
M1
M1 − m 1
M1 − m 1
v (x) =
v (x) − m2
M2 − v (x)
m2 +
M2 .
M2 − m 2
M2 − m 2
and
(2.5)
Since f and g are convex functions we have
u (x) − m1
M1 − u (x)
f (m1 ) +
f (M1 )
M1 − m 1
M1 − m 1
u (x)
M1 f (m1 ) − m1 f (M1 )
=
(f (M1 ) − f (m1 )) +
M1 − m 1
M1 − m 1
h
i
= [f ; m1 , M1 ] u (x) + fe; m1 , M1
f (u (x)) ≤
(2.6)
and
M2 − v (x)
v (x) − m2
g (m2 ) +
g (M2 )
M2 − m 2
M2 − m 2
v (x)
M2 g (m2 ) − m2 g (M2 )
=
(g (M2 ) − g (m2 )) +
M2 − m 2
M2 − m 2
= [g; m2 , M2 ] v (x) + [e
g ; m2 , M2 ] .
g (v (x)) ≤
(2.7)
Functions f and g are nonnegative by assumption, so after multiplying (2.6) and (2.7) we obtain
(2.8) f (u (x)) g (v (x))
≤ [f ; m1 , M1 ] [g; m2 , M2 ] u (x) v (x) + [f ; m1 , M1 ] [e
g ; m2 , M2 ] u (x)
h
i
h
i
e
e
+ [g; m2 , M2 ] f ; m1 , M1 v (x) + f ; m1 , M1 [e
g ; m2 , M2 ] .
Now, multiplying (2.8) by weight p (x), integrating over [a, b] and dividing by P > 0 we get
(i).
To obtain (ii) we can write
m 1 + M1
1 M1 − u (x)
u (x) − m1
=
m1 +
M1
2
2 M1 − m 1
M1 − m 1
u (x) − m1
M1 − u (x)
+
m1 +
M1 ,
M1 − m 1
M1 − m 1
m 2 + M2
1 M2 − v (x)
v (x) − m2
m2 +
M2
=
2
2 M2 − m 2
M2 − m 2
v (x) − m2
M2 − v (x)
+
m2 +
M2 .
M2 − m 2
M2 − m 2
J. Inequal. Pure and Appl. Math., 5(3) Art. 74, 2004
http://jipam.vu.edu.au/
H ADAMARD - TYPE INEQUALITIES
5
Using the Hadamard inequality (1.1) and the convexity of functions f and g, we get
m 2 + M2
m 1 + M1
f
g
2
2
M1 − u (x)
u (x) − m1
u (x) − m1
M1 − u (x)
1
≤
f
m1 +
M1 + f
m1 +
M1
4
M1 − m 1
M1 − m 1
M1 − m 1
M1 − m 1
v (x) − m2
v (x) − m2
M2 − v (x)
M2 − v (x)
× g
m2 +
M2 + g
m2 +
M2 .
M2 − m 2
M2 − m 2
M2 − m 2
M2 − m 2
According to (2.4) and (2.5) , after some simple calculus we obtain
m 1 + M1
m 2 + M2
(2.9) f
g
2
2
1
≤ [f (u (x)) g (v (x)) + f (M1 + m1 − u (x)) g (M2 + m2 − v (x))]
4
1
M1 − u (x)
u (x) − m1
v (x) − m2
M2 − v (x)
+
f
m1 +
M1 × g
m2 +
M2
4
M1 − m 1
M1 − m 1
M2 − m 2
M2 − m 2
u (x) − m1
M1 − u (x)
M2 − v (x)
v (x) − m2
+f
m1 +
M1 ×g
m2 +
M2 .
M1 − m 1
M1 − m 1
M2 − m 2
M2 − m 2
Using the convexity of functions f and g, from inequality (2.9) we get
m 1 + M1
m 2 + M2
(2.10) f
g
2
2
1
≤ [f (u (x)) g (v (x)) + f (M1 + m1 − u (x)) g (M2 + m2 − v (x))]
4
1
M1 − u (x)
u (x) − m1
+
f (m1 ) +
f (M1 )
4
M1 − m 1
M1 − m 1
v (x) − m2
M2 − v (x)
×
g (m2 ) +
g (M2 )
M2 − m 2
M2 − m 2
u (x) − m1
M1 − u (x)
+
f (m1 ) +
f (M1 )
M1 − m 1
M1 − m 1
M2 − v (x)
v (x) − m2
g (m2 ) +
g (M2 ) .
×
M2 − m 2
M2 − m 2
With respect to the notation introduced at the beginning of this section, inequality (2.10) becomes
m 1 + M1
m 2 + M2
f
g
2
2
o
1n
≤
f (u (x)) g (v (x)) + f (M1 + m1 − u (x)) g (M2 + m2 − v (x))
4 (
h
i 1 +
[f ; m1 , M1 ] u (x) + fe; m1 , M1
[b
g ; m2 , M2 ] − [g; m2 , M2 ] v (x)
4
)
h
i
+ fb; m1 , M1 − [f ; m1 , M1 ] u (x)
[g; m2 , M2 ] v (x) + [e
g ; m2 , M2 ]
J. Inequal. Pure and Appl. Math., 5(3) Art. 74, 2004
http://jipam.vu.edu.au/
6
M. K LARI ČI Ć BAKULA AND J. P E ČARI Ć
o
1n
=
f (u (x)) g (v (x)) + f (M1 + m1 − u (x)) g (M2 + m2 − v (x))
4 (
1
+
− 2 [f ; m1 , M1 ] [g; m2 , M2 ] u (x) v (x)
4
+ [b
g ; m2 , M2 ] − [e
g ; m2 , M2 ] [f ; m1 , M1 ] u (x)
h
i h
i
b
e
+ f ; m1 , M1 − f ; m1 , M1 [g; m2 , M2 ] v (x)
)
h
i
h
i
+ fe; m1 , M1 [b
g ; m2 , M2 ] + fb; m1 , M1 [e
g ; m2 , M2 ]
(2.11)
Now we multiply both sides of (2.11) by p (x), integrate over [a, b] and divide by P . We thus
obtain (ii) and the proof is completed.
Remark 2.2. Pachpatte’s results (1.2) and (1.3) can be obtained from (2.2) and (2.3) respectively if we put p (x) = 1, u (x) = v (x) = x for all x ∈ [a, b] (then we have m1 = m2 = a and
M1 = M2 = b). In the case of g (x) ≡ 1 inequality (i) becomes the right side of Hadamard’s
inequality (1.1) .
Theorem 2.3. Let f be a nonnegative convex function on [m1 , M1 ], g a nonnegative convex
function on [m2 , M2 ], u : [a, b] → [m1 , M1 ] and v : [a, b] → [m2 , M2 ] continuous functions,
and p, q : [a, b] → R positive integrable functions. Then
(i)
Z bZ bZ
1
PQ
a
a
≤
0
1
3P Q
1
p (x) q (y) f (tu (x) + (1 − t) u (y)) × g (tv (x) + (1 − t) v (y)) dtdxdy
Z b
Z b
Q
f (u (x)) g (v (x)) p (x) dx + P
f (u (y)) g (v (y)) q (y) dy
a
a
1
+
3P Q
Z
b
Z
p (x) f (u (x)) dx
a
b
q (y) g (v (y)) dy;
a
(ii)
1
P
Z bZ
a
1
p (x) f (tu (x) + (1 − t) u) g (tv (x) + (1 − t) v) dtdx
Z b
1
1
≤
p (x) f (u (x)) g (v (x)) dx + f (u) g (v)
3P a
3
Z b
Z b
1
+
g (v)
p (x) f (u (x)) dx + f (u)
p (x) g (v (x)) dx ,
6P
a
a
Rb
Rb
where u = P1 a p (x) u (x) dx, v = Q1 a q (x) v (x) dx.
0
Proof. Since f and g are convex functions, for t ∈ [0, 1] we have
(2.12)
f (tu (x) + (1 − t) u (y)) ≤ tf (u (x)) + (1 − t) f (u (y))
(2.13)
g (tv (x) + (1 − t) v (y)) ≤ tg (v (x)) + (1 − t) g (v (y)) .
J. Inequal. Pure and Appl. Math., 5(3) Art. 74, 2004
http://jipam.vu.edu.au/
H ADAMARD - TYPE INEQUALITIES
7
Functions f and g are nonnegative, so multiplying (2.12) and (2.13) we get
(2.14) f (tu (x) + (1 − t) u (y)) g (tv (x) + (1 − t) v (y))
≤ t2 f (u (x)) g (v (x)) + (1 − t)2 f (u (y)) g (v (y))
+ t (1 − t) [f (u (x)) g (v (y)) + f (u (y)) g (v (x))] .
Integrating (2.14) over [0, 1] we obtain
Z 1
(2.15)
f (tu (x) + (1 − t) u (y)) g (tv (x) + (1 − t) v (y)) dt
0
1
[f (u (x)) g (v (x)) + f (u (y)) g (v (y))]
3
1
+ [f (u (x)) g (v (y)) + f (u (y)) g (v (x))] .
6
Now we multiply (2.15) by p (x) q (y), integrate over [a, b] × [a, b] and divide by P Q, where
Z b
Z b
P =
p (x) dx, Q =
q (x) dx,
≤
a
a
so we get
1
PQ
Z bZ bZ
1
p (x) q (y) f (tu (x) + (1 − t) u (y))
a
a
0
× g (tv (x) + (1 − t) v (y)) dtdxdy
Z b
Z b
1
≤
p (x) f (u (x)) g (v (x)) dx
q (y) dy
3P Q a
a
Z b
Z b
+
q (y) f (u (y)) g (v (y)) dy
p (x) dx
a
a
Z b
Z b
1
+
p (x) f (u (x)) dx
q (y) g (v (y)) dy
6P Q a
a
Z b
Z b
+
p (y) f (u (y)) dy
q (x) g (v (x)) dx
a
a
Z b
Z b
1
=
p (x) f (u (x)) g (v (x)) dx
q (y) dy
3P Q a
a
Z b
Z b
+
q (y) f (u (y)) g (v (y)) dy
p (x) dx
a
a
Z
b
Z
b
1
p (x) f (u (x)) dx
q (y) g (v (y)) dy.
3P Q a
a
This is the desired inequality (i).
To prove inequality (ii), in (2.12) and (2.13) we substitute u (y) and v (y) with u and v
respectively.
Then we obtain
(2.16)
+
(2.17) f (tu (x) + (1 − t) u) g (tv (x) + (1 − t) v)
≤ t2 f (u (x)) g (v (x)) + (1 − t)2 f (u) g (v)
+ t (1 − t) [f (u (x)) g (v) + f (u) g (v (x))] .
J. Inequal. Pure and Appl. Math., 5(3) Art. 74, 2004
http://jipam.vu.edu.au/
8
M. K LARI ČI Ć BAKULA AND J. P E ČARI Ć
Integrating (2.17) in respect to t over [0, 1] we obtain
1
Z
f (tu (x) + (1 − t) u) g (tv (x) + (1 − t) v) dt
(2.18)
0
1
1
[f (u (x)) g (v (x)) + f (u) g (v)] + [f (u (x)) g (v) + f (u) g (v (x))] .
6
3
≤
Similarly as before, from (2.18) we get
1
P
Z bZ
a
1
0
p (x) f (tu (x) + (1 − t) u) g (tv (x) + (1 − t) v) dtdx
Z b
1
1
≤
p (x) f (u (x)) g (v (x)) dx + f (u) g (v)
3P a
3
Z b
Z b
1
+
g (v)
p (x) f (u (x)) dx + f (u)
p (x) g (v (x)) dx .
6P
a
a
This completes the proof.
Remark 2.4. If in (i) we put u (x) = v (x) = x for all x ∈ [a, b], it becomes
1
(2.19)
PQ
Z bZ bZ
a
a
≤
1
0
1
3P Q
p (x) q (y) f (tx + (1 − t) y) g (tx + (1 − t) y) dtdxdy
Z b
Z b
Q
p (x) f (x) g (x) dx + P
q (y) f (y) g (y) dy
a
a
1
+
3P Q
Z
b
Z
p (x) f (x) dx
a
b
q (y) g (y) dy,
a
so by using a generalization of Hadamard’s inequality [1, p.138]
Z
a+b
1 b
f (a) + f (b)
(2.20)
f
≤
p (x) f (x) dx ≤
2
P a
2
which holds for p (a + t) = p (b − t) , 0 ≤ t ≤ 12 (a + b) , we obtain from (2.19) the following
inequality
Z bZ bZ 1
1
p (x) q (y) f (tx + (1 − t) y) g (tx + (1 − t) y) dtdxdy
PQ a a 0
Z b
Z b
1
≤
Q
p (x) f (x) g (x) dx + P
q (y) f (y) g (y) dy
3P Q
a
a
1 f (a) + f (b) g (a) + g (b)
+
3
2
2
Z b
Z b
1
=
Q
p (x) f (x) g (x) dx + P
q (y) f (y) g (y) dy
3P Q
a
a
1
(2.21)
[M (a, b) + N (a, b)] .
+
12
Now it is easy to observe that if p (x) = q (x) = 1 for all x ∈ [a, b] inequality (2.21) becomes
the corrected Pachpatte’s result (1.4).
J. Inequal. Pure and Appl. Math., 5(3) Art. 74, 2004
http://jipam.vu.edu.au/
H ADAMARD - TYPE INEQUALITIES
9
If we do the same in (ii) we get
Z Z
a+b
1 b 1
a+b
p (x) f tx + (1 − t)
g tx + (1 − t)
dtdx
P a 0
2
2
Z b
a+b
a+b
1
1
≤
p (x) f (x) g (x) dx + f
g
3P a
3
2
2
Z b
Z b
1
a+b
a+b
+
g
p (x) f (x) dx + f
p (x) g (x) dx .
6P
2
2
a
a
Using again (2.20) we obtain
Z Z
a+b
1 b 1
a+b
p (x) f tx + (1 − t)
g tx + (1 − t)
dtdx
P a 0
2
2
Z b
1
1 f (a) + f (b) g (a) + g (b)
≤
p (x) f (x) g (x) dx +
3P a
3
2
2
1 g (a) + g (b) f (a) + f (b) f (a) + f (b) g (a) + g (b)
+
+
6
2
2
2
2
Z b
1
1
=
p (x) f (x) g (x) dx + (M (a, b) + N (a, b))
3P a
6
Furthermore, in the case p (x) = 1 for all x ∈ [a, b] we get
Z bZ 1 a+b
a+b
1
f tx + (1 − t)
g tx + (1 − t)
dtdx
b−a a 0
2
2
Z b
1
1
≤
f (x) g (x) dx + (M (a, b) + N (a, b)) ,
3 (b − a) a
6
which is the corrected Pachpatte’s result (1.5).
R EFERENCES
[1] J.E. PEČARIĆ, F. PROSCHAN AND Y.L. TONG, Convex Functions, Partial Orderings, and Statistical Applications, Academic Press, Inc. (1992).
[2] D.S. MITRINOVIĆ, J.E. PEČARIĆ
Kluwer Academic Publishers, 1993.
AND
A.M. FINK, Classical and New Inequalities in Analysis,
[3] B.G. PACHPATTE, On some inequalities for convex functions, RGMIA Res. Rep. Coll., 6(E) (2003).
[ONLINE http://rgmia.vu.edu.au/v6(E).html].
J. Inequal. Pure and Appl. Math., 5(3) Art. 74, 2004
http://jipam.vu.edu.au/