91 NEW INEQUALITIES FOR CONVEX FUNCTIONS

91
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NEW INEQUALITIES FOR CONVEX FUNCTIONS
KONVEKS FONKSİYONLAR İÇİN YENİ EŞİTSİZLİKLER
Mevlüt TUNÇ1* ve S. Uğur KIRMACI2
1 Kilis
2
7 Aralık Üniversitesi, Fen Edebiyat Fakültesi Matematik Bölümü, Kilis
Atatürk Üniversitesi, KKEF, Matematik Eğitimi A.B.D., 25240, Erzurum
Geliş Tarihi: 28 Nisan 2010
Kabul Tarihi: 25 Mayıs 2010
ABSTRACT
In the present paper we establish some new integral inequalities
analogous to the well known Hadamard's inequality by using a fairly
elementary analysis.
ÖZET
Bu makalede biz temel analiz işlemlerini kullanarak literatürde iyi
bilinen Hadamard eşitsizliğine benzer yeni integral eşitsizlikleri kurduk.
Key Words: Hadamard's inequality, convex function, concave function,
special means.
1.INTRODUCTION
The following inequality [see Dragomir, (1992)]
b
f ( a ) + f (b )
1
⎛a+b⎞
≤
f⎜
f ( x ) dx ≤
⎟
∫
2
⎝ 2 ⎠ b−a a
(1.1)
Which holds for all convex functions f : [ a , b ] → is known
in the literature as Hadamard's inequality. Since its discovery in 1893,
Hadamard's inequality [see Hadamard, (1893)] has been proven to be
one of the most useful inequalities in mathematical analysis. A
number of papers have been written on this inequality providing new
proofs, noteworthy extensions, generalizations and numerous
*
Sorumlu Yazar: [email protected]
92
Tunç & Kırmacı
applications, see the references cited therein. The main purpose of
this paper is to establish some new integral inequalities analogous to
that of Hadamard's inequality given in (1.1) involving two convex
functions. The analysis used in the proof is elementary.
2. MAIN RESULT
We need the following Lemma proved in [see, Pecaric and
Dragomir, 104pp, (1991) ] which deals with the simple
characterization of convex functions.
Lemma A: The following statements are equivalent to a mapping:
f : [a, b ] → ;
i) f is convex on [ a, b ] ,
[ a, b ]
ii) for all x, y in
the mapping g : [ 0,1] → , defined by
g ( t ) = f ( tx + (1 − t ) y ) is convex on [ 0,1] .
For the proof of this Lemma, see [Pecaric and Dragomir,
104pp, (1991)].
Our main result is given in the following theorem.
f : [a, b ] → Theorem 2.1: Let
be a nonnegative and convex
function. Then one has the inequality:
2 f (a)
(b − a )
b
2
∫ ( b − x ) f ( x ) dx +
a
b
1
≤
f 2 ( x)
b − a ∫a
2 f (b )
b
( x − a ) f ( x ) dx
2
( b − a ) ∫a
M ( a, b )
dx +
where M ( a, b ) = f
(2.1)
3
( a ) + f ( a ) f (b) + f 2 (b) .
Proof: Since f is a convex function on [ a, b ] , then we have that
f ( ta + (1 − t ) b ) ≤ tf ( a ) + (1 − t ) f ( b )
2
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New Inequalıtıes For Convex Functıons
for
t ∈ [ 0,1] .
all
Using
the
elementary
inequality
G ( a, b ) ≤ A ( a, b ) , 0 ≤ a, b , we can conclude that
{
}
2 tf ( a ) f ( ta + (1 − t ) b ) + (1 − t ) f ( b ) f ( ta + (1 − t ) b )
≤ f 2 ( ta + (1 − t ) b ) + t 2 f 2 ( a )
+ 2t (1 − t ) f ( a ) f ( b ) + (1 − t ) f 2 ( b )
2
(
)
By the Lemma A f ta + (1 − t ) b is convex on [ 0,1] , it is integrable
on [ 0,1] . Integrating both sides of the above inequality over t on
[0,1] we get
1
1
2 f ( a ) ∫ tf ( ta + (1 − t ) b ) dt + 2 f ( b ) ∫ (1 − t ) f ( ta + (1 − t ) b ) dt
0
0
1
1
≤ ∫ f 2 ( ta + (1 − t ) b ) dt + f 2 ( a ) ∫ t 2 dt
0
0
1
+ 2 f ( a ) f ( b ) ∫ t (1 − t ) dt + f
0
1
2
( b ) ∫ (1 − t )
2
dt
0
By substituting x = ta + (1 − t ) b , it is easy to observe that
1
∫ tf ( ta + (1 − t ) b ) dt
0
1
x −b
1
=
f ( x ) dx =
2
∫
b−a a a −b
(b − a )
b
b
∫ ( b − x ) f ( x ) dx
a
and
1
∫ (1 − t ) f ( ta + (1 − t ) b ) dt
0
1
a−x
1
=
f ( x ) dx =
2
∫
b−a a a−b
(b − a )
b
b
∫ ( x − a ) f ( x ) dx
a
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It can be easily checked that
1
1
1
2
2
∫0 t dt = ∫0 (1 − t ) dt = 3 ,
1
∫
f 2 ( ta + (1 − t ) b ) dt =
0
1
1
∫ t (1 − t ) dt = 6
0
b
1
f 2 ( x ) dx
∫
b−a a
When above equalities are taken into account, the proof is complete.
Teorem 2.2: Let f : [ a , b ] → be a nonnegative and convex function.
Then one has the inequality:
1
1 ⎛a+b⎞
f ( x ) dx ≤ f ⎜
⎟
∫
b−a a
2 ⎝ 2 ⎠
b
b
N ( a, b )
1
f 2 ( x ) dx +
+
∫
⎛a+b⎞
⎛a+b⎞
a
4f ⎜
24 f ⎜
⎟ (b − a )
⎟
⎝ 2 ⎠
⎝ 2 ⎠
where N ( a, b ) = f 2 ( a ) + 4 f ( a ) f ( b ) + f 2 ( b ) .
(2.2)
Proof: Since f is a convex function on [ a, b ] , then we have that
⎛ ta + (1 − t ) b (1 − t ) a + tb ⎞
⎛ a+b⎞
f⎜
+
⎟
⎟= f ⎜
2
2
⎝ 2 ⎠
⎝
⎠
f ( ta + (1 − t ) b ) + f ( (1 − t ) a + tb )
≤
2
for all t ∈ [ 0,1] . Similarly as explained in the proof of inequality (2.1)
given above, we obtain
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New Inequalıtıes For Convex Functıons
⎛ a+b⎞
⎛ a+b⎞
f⎜
⎟ f ( ta + (1 − t ) b ) + f ⎜
⎟ f ( (1 − t ) a + tb )
⎝ 2 ⎠
⎝ 2 ⎠
f 2 ( ta + (1 − t ) b )
2⎛ a+b⎞
≤ f ⎜
⎟+
4
⎝ 2 ⎠
f ( ta + (1 − t ) b ) f ( (1 − t ) a + tb ) f 2 ( (1 − t ) a + tb )
+
+
2
4
2
2
⎛ a + b ⎞ f ( ta + (1 − t ) b ) + f ( (1 − t ) a + tb )
≤ f2⎜
⎟+
4
⎝ 2 ⎠
⎡tf ( a ) + (1 − t ) f ( b ) ⎤⎦ ⎡⎣(1 − t ) f ( a ) + tf ( b ) ⎤⎦
+⎣
2
2
2
⎛ a + b ⎞ f ( ta + (1 − t ) b ) + f ( (1 − t ) a + tb )
= f2⎜
+
⎟
4
⎝ 2 ⎠
2
t (1 − t ) ⎡⎣ f 2 ( a ) + f 2 ( b ) ⎤⎦ + ⎡t 2 + (1 − t ) ⎤ f ( a ) f ( b )
⎣
⎦
+
2
Likewise as explained in the proof of inequality (2.1) given above, we
integrate both sides of the above inequality over t on [ 0,1] , we get
⎛ a+b⎞
⎛ a+b⎞
f⎜
⎟ ∫ f ( ta + (1 − t ) b ) dt + f ⎜
⎟∫ f
⎝ 2 ⎠0
⎝ 2 ⎠0
1
1
( (1 − t ) a + tb ) dt
1
⎛ a+b⎞
2
2
≤ f2⎜
⎟ ∫ dt + ∫ f ( ta + (1 − t ) b ) + f ( (1 − t ) a + tb ) dt
2
4
⎝
⎠0
0
1
+
1
{
}
f 2 ( a ) + f 2 (b ) 1
f ( a ) f (b ) 1 2
2
1
t
t
dt
t + (1 − t ) dt
−
+
(
)
∫
∫
2
2
0
0
(
)
By substituting ta + (1 − t ) b = x and (1 − t ) a + tb = y it can be easily
obtained,
1
∫
0
1
f ( ta + (1 − t ) b ) dt = ∫ f
0
b
( (1 − t ) a + tb ) dt = b −1 a ∫ f ( x ) dx
a
1
∫ t (1 − t ) dt = 6 , ∫ ( t + (1 − t )
1
1
2
0
0
2
) dt = 32
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From the above equalities it is easily obtained that
⎛ a+b⎞
⎛ a+b⎞
⎛ a+b⎞
f⎜
f⎜
f⎜
⎟1
⎟1
⎟1
⎝ 2 ⎠ f x dx + ⎝ 2 ⎠ f x dx = 2 ⎝ 2 ⎠ f x dx
( )
( )
( )
b − a ∫0
b − a ∫0
b − a ∫0
1
f 2 ( a ) + f 2 (b ) f ( a ) f (b )
1
⎛ a+b⎞
2
≤ f2⎜
+
+
+
f
x
dx
(
)
⎟
∫
12
3
⎝ 2 ⎠ 2 (b − a ) 0
⎛ a+b⎞
Now dividing both sides of the above inequality by 2 f ⎜
⎟ we
⎝ 2 ⎠
get the desired inequality in (2.2). The proof is complete.
Remark: If we choose a = 0 and b = 1 and the convex functions
f ( x ) = x , then it is easy to observe that the inequalities obtained (2.1)
and (2.2) are certain in the sense that we hold equalities in (2.1) and
(2.2).
Theorem 2.3: Let
f : [a, b ] → be a nonnegative and convex
function. Then
b b 1
f ( tx + (1 − t ) y ) ( tf ( x ) + (1 − t ) f ( y ) ) dtdydx
2
( b − a ) ∫a ∫a ∫0
2
≤
b b 1
f ( tx + (1 − t ) y ) dtdydx
(b − a ) ∫ ∫ ∫
1
2
2
(2.3)
a a 0
b
ψ ( a, b )
2
2
+
+
f
x
dx
(
)
3 ( b − a ) ∫a
12
where ψ ( a, b ) = f 2 ( a ) + 2 f ( a ) f ( b ) + f 2 ( b ) .
Proof: Since f is a convex function on [ a, b ] , then we have that
f ( tx + (1 − t ) y ) ≤ tf ( x ) + (1 − t ) f ( y )
for all x, y ∈ [ a, b ] and t ∈ [ 0,1] . Using the elementary inequality
G ( a, b ) ≤ A ( a, b ) , 0 ≤ a, b , we get,
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New Inequalıtıes For Convex Functıons
2 f ( tx + (1 − t ) y ) ( tf ( x ) + (1 − t ) f ( y ) )
≤ f 2 ( tx + (1 − t ) y ) + t 2 f 2 ( x )
+ 2t (1 − t ) f ( x ) f ( y ) + (1 − t ) f 2 ( y )
2
As explained in the proof of inequality (2.1) given above, we integrate
both sides of that above inequality over t on [ 0,1] , and we obtain
1
2 ∫ f ( tx + (1 − t ) y ) ( tf ( x ) + (1 − t ) f ( y ) ) dt
0
1
≤∫ f
1
2
0
( tx + (1 − t ) y ) dt + f ( x ) ∫ t dt
2
2
(2.3.1)
0
1
1
+ 2 f ( x ) f ( y ) ∫ t (1 − t ) dt + f 2 ( y ) ∫ (1 − t ) dt
2
0
0
1
= ∫ f 2 ( tx + (1 − t ) y ) dt +
0
1 2
1
1
f ( x) + f ( x) f ( y) + f 2 ( y)
3
3
3
Integrating both sides of that above inequality (2.3.1) over x and y
on [ a, b ] we obtain
2
b b 1
2 ∫ ∫ ∫ f ( tx + (1 − t ) y ) ( tf ( x ) + (1 − t ) f ( y ) ) dtdydx
a a 0
b b 1
≤ ∫ ∫ ∫ f 2 ( tx + (1 − t ) y ) dtdydx
a a 0
b b
b b
b b
1
1
1
+ ∫ ∫ f 2 ( x ) dydx + ∫ ∫ f ( x ) f ( y ) dydx + ∫ ∫ f 2 ( y ) dydx
3aa
3aa
3aa
(2.3.2)
b b 1
≤ ∫ ∫ ∫ f 2 ( tx + (1 − t ) y ) dtdydx
a a 0
+
b
b
2 (b − a ) b 2
1
f
x
dx
f
x
dx
+
(
)
(
)
∫a
∫a f ( y ) dy
3
3 ∫a
By using the right half of the Hadamard's inequality given in (1.1) on
the right side of (2.3.2) we write
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b b 1
2 ∫ ∫ ∫ f ( tx + (1 − t ) y ) ( tf ( x ) + (1 − t ) f ( y ) ) dtdydx
a a 0
b b 1
≤ ∫∫∫
a a 0
2 (b − a ) b 2
f ( tx + (1 − t ) y ) dtdydx +
∫a f ( x ) dx
3
2
1 f ( a ) + f (b) f ( a ) + f (b)
2
(b − a )
3
2
2
b b 1
2 (b − a ) b 2
= ∫ ∫ ∫ f 2 ( tx + (1 − t ) y ) dtdydx +
f ( x ) dx
∫
3
a a 0
a
+
+
1
2
f 2 ( a ) + 2 f ( a ) f (b ) + f 2 (b )) (b − a )
(
12
Now multiplying both sides of the above inequality by 1 ( b − a ) we
2
get desired inequality in (2.3). The proof is complete.
Theorem 2.4: Let
f : [a, b ] → be a nonnegative and convex
function. Then
b 1
2
b − a ∫a ∫0
a + b ⎞⎛
⎛
f ⎜ tx + (1 − t )
⎟ tf ( x ) + (1 − t )
2 ⎠ ⎜⎝
⎝
⎛ a + b ⎞⎞
f⎜
⎟ ⎟ dtdx
⎝ 2 ⎠⎠
ψ ( a, b )
1
a+b⎞
⎛
f 2 ⎜ tx + (1 − t )
≤
(b − a + 2)
⎟ dtdx +
∫
∫
2 ⎠
12
b−a a 0 ⎝
b 1
where ψ ( a, b ) = f 2 ( a ) + 2 f ( a ) f ( b ) + f 2 ( b ) .
Proof: Since f is a convex function on [ a, b ] , then we have that
a+b⎞
⎛
⎛ a+b⎞
f ⎜ tx + (1 − t )
⎟ ≤ tf ( x ) + (1 − t ) f ⎜
⎟
2 ⎠
⎝
⎝ 2 ⎠
for all x, y ∈ [ a, b ] and t ∈ [ 0,1] , we get
a + b ⎞⎛
⎛
2 f ⎜ tx + (1 − t )
⎟ tf ( x ) + (1 − t )
2 ⎠ ⎜⎝
⎝
⎛ a + b ⎞⎞
f⎜
⎟⎟
⎝ 2 ⎠⎠
a+b⎞ 2 2
⎛
≤ f 2 ⎜ tx + (1 − t )
⎟ + t f ( x)
2 ⎠
⎝
2
⎛ a+b⎞
2⎛ a+b⎞
+ 2t (1 − t ) f ( x ) f ⎜
⎟ + (1 − t ) f ⎜
⎟
⎝ 2 ⎠
⎝ 2 ⎠
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(2.4)
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New Inequalıtıes For Convex Functıons
As explained in the proof of inequality (2.1) given above, we integrate
both sides of that above inequality over t on [ 0,1] , and we obtain
a + b ⎞⎛
⎛
2∫ f ⎜ tx + (1 − t )
⎟ tf ( x ) + (1 − t )
2 ⎠ ⎜⎝
⎝
0
1
⎛ a + b ⎞⎞
f⎜
⎟ ⎟ dt
⎝ 2 ⎠⎠
a+b⎞
⎛
2
2
≤ ∫ f 2 ⎜ tx + (1 − t )
⎟ dt + f ( x ) ∫ t dt
2 ⎠
⎝
0
0
1
1
(2.4.1)
2
⎛ a+b⎞
2⎛ a+b⎞
+2f ⎜
⎟ f ( x ) ∫ t (1 − t ) dt + f ⎜
⎟ ∫ (1 − t ) dt
⎝ 2 ⎠
⎝ 2 ⎠0
0
1
1
1 2
1 ⎛ a+b⎞
1 2⎛ a+b⎞
a+b⎞
⎛
= ∫ f 2 ⎜ tx + (1 − t )
⎟ dt + f ( x ) + f ⎜
⎟ f ( x) + f ⎜
⎟
2
3
3
2
3 ⎝ 2 ⎠
⎝
⎠
⎝
⎠
0
As explained in the proof of inequality (2.3) given above, integrating
1
both sides of that above inequality (2.4.1) over x on [ a, b ] , and using
the right half of the Hadamard's inequality given in (1.1) and
convexity of f we obtain
a + b ⎞⎛
⎛
2 ∫ ∫ f ⎜ tx + (1 − t )
⎟ tf ( x ) + (1 − t )
2 ⎠ ⎜⎝
⎝
a 0
b 1
⎛ a + b ⎞⎞
f⎜
⎟ ⎟ dtdx
⎝ 2 ⎠⎠
a+b⎞
⎛
≤ ∫ ∫ f 2 ⎜ tx + (1 − t )
⎟ dtdx
2 ⎠
⎝
a 0
b 1
(b − a ) f 2 ⎛ a + b ⎞
1
1 ⎛a+b⎞
+ ∫ f 2 ( x ) dx + f ⎜
⎟ ∫ f ( x ) dx +
⎜
⎟
3a
3 ⎝ 2 ⎠a
3
⎝ 2 ⎠
b
b
a+b⎞
⎛
≤ ∫ ∫ f 2 ⎜ tx + (1 − t )
⎟ dtdx
2 ⎠
⎝
a 0
b 1
1
2 ⎛ f ( a ) + f (b ) ⎞
+ (b − a ) ⎜
⎟
3
2
⎝
⎠
2
f ( a ) + f (b ) (b − a ) ⎛ f ( a ) + f (b ) ⎞
1 f ( a ) + f (b )
+
+
(b − a )
⎜
⎟
3
2
2
3 ⎝
2
⎠
2
a+b⎞
⎛
= ∫ ∫ f 2 ⎜ tx + (1 − t )
⎟ dtdx
2 ⎠
⎝
a 0
b 1
+
ψ ( a, b )
12
((b − a )
2
+ (b − a ) + (b − a )
)
(2.4.2)
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Now dividing both sides of (2.4.2) by ( b − a ) and rewriting (2.4.2) we
get the required inequality in (2.4). The proof is complete.
3. APPLICATIONS FOR SPECIAL MEANS
As in [Kırmacı, 2004], we shall consider the means as arbitrary
real numbers a, b, a ≠ b . In the resources there includes
A ( a, b ) =
a+b
,
2
a, b ∈ +
(arithmetic mean)
a 2 + b2
, a, b ∈ + (quadratic mean)
2
b−a
L ( a, b ) =
a ≠ b , ab ≠ 0, a, b ∈ +
,
ln b − ln a
K ( a, b ) =
(logarithmic
mean)
G ( a, b ) = ab ,
a, b ∈ +
(geometric mean).
Now, using the results of Section 2, we illustrate some
applications of special means of real numbers.
Proposition 3.1: Let 0 < a < b . Then one has the inequality,
4 A ( a , b ) 2 K 2 ( a , b ) 10
≤
+
L ( a, b ) 3 G 2 ( a, b ) 3
Proof: The proof is immediate from Theorem 2.1 as applied for
1
and the details are omitted.
x
Proposition 3.2: Let 0 < a < b . Then one has the inequality,
A ( a, b ) ( 2 K 2 ( a, b ) + 4G 2 ( a, b ) )
A ( a, b )
1
1
≤
+
+
L ( a, b ) 2 A ( a, b ) 4G 2 ( a, b )
24G 2 ( a, b )
f ( x) =
Proof: The assertion follows from Theorem 2.2 as applied to
f ( x) =
1
and the details are omitted.
x
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New Inequalıtıes For Convex Functıons
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