Journal of Inequalities in Pure and
Applied Mathematics
ON HADAMARD’S INEQUALITY ON A DISK
S.S. DRAGOMIR
School of Communications and Informatics (F019)
Victoria University of Technology
PO BOX 14428
Melbourne City MC 8001
AUSTRALIA
EMail: sever.dragomir@vu.edu.au
URL: http://rgmia.vu.edu.au/SSDragomirWeb.html
volume 1, issue 1, article 2,
2000.
Received 17 September 1999;
accepted 16 December 1999.
Communicated by: T. Mills
Abstract
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2000
Victoria University
ISSN (electronic): 1443-5756
003-99
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Abstract
In this paper an inequality of Hadamard type for convex functions defined on
a disk in the plane is proved. Some mappings naturally connected with this
inequality and related results are also obtained.
2000 Mathematics Subject Classification: 26D07, 26D15
Key words: Convex functions of double variable, Jensen’s inequality, Hadamard’s
inequality.
Contents
1
2
3
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
Hadamard’s Inequality on the Disk . . . . . . . . . . . . . . . . . . . . . . 6
Some Mappings Connected to Hadamard’s Inequality on
the Disk . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
References
On Hadamard’s Inequality on a
Disk
S.S. Dragomir
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1.
Introduction
Let f : I ⊆ R → R be a convex mapping defined on the interval I of real
numbers and a, b ∈ I with a < b. 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 for convex mappings. Note
that some of the classical inequalities for means can be derived from (1.1) for
appropriate particular selections of the mapping f.
In the paper [4] (see also [6] and [7]) the following mapping naturally connected with Hadamard’s result is considered
Z b 1
a+b
H : [0, 1] → R, H (t) :=
f tx + (1 − t)
dx.
b−a a
2
On Hadamard’s Inequality on a
Disk
S.S. Dragomir
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The following properties are also proved:
(i) H is convex and monotonic nondecreasing.
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(ii) One has the bounds
1
sup H (t) = H (1) =
b−a
t∈[0,1]
Z
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b
f (x) dx
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a
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and
inf H (t) = H (0) = f
t∈[0,1]
a+b
2
.
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Another mapping also closely connected with Hadamard’s inequality is the
following one [6] (see also [7])
Z bZ b
1
F : [0, 1] → R, F (t) :=
f (tx + (1 − t) y) dxdy.
(b − a)2 a a
The properties of this mapping are itemized below:
1
(i) F is convexon [0,
1]
and
monotonic
nonincreasing
on
0, 2 and nonde
creasing on 21 , 1 .
(ii) F is symmetric about 12 . That is,
F (t) = F (1 − t) ,
On Hadamard’s Inequality on a
Disk
S.S. Dragomir
for all t ∈ [0, 1] .
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(iii) One has the bounds
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1
sup F (t) = F (0) = F (1) =
b−a
t∈[0,1]
Z
b
f (x) dx
a
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and
Z bZ b 1
x+y
1
a+b
inf F (t) = F
=
f
dxdy ≥ f
.
t∈[0,1]
2
2
2
(b − a)2 a a
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(iv) The following inequality holds
F (t) ≥ max {H (t) , H (1 − t)} ,
for all t ∈ [0, 1] .
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In this paper we will point out a similar inequality to Hadamard’s that applies to convex mappings defined on a disk embedded in the plane R2 . We will
also consider some mappings similar in a sense to the mappings H and F and
establish their main properties.
For recent refinements, counterparts, generalizations and new Hadamard’s
type inequalities, see the papers [1]-[11] and [14]-[15] and the book [13].
On Hadamard’s Inequality on a
Disk
S.S. Dragomir
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2.
Hadamard’s Inequality on the Disk
Let us consider a point C = (a, b) ∈ R2 and the disk D (C, R) centered at the
point C and having the radius R > 0. The following inequality of Hadamard
type holds.
Theorem 2.1. If the mapping f : D (C, R) → R is convex on D (C, R), then
one has the inequality
ZZ
Z
1
1
(2.1)
f (C) ≤
f (x, y) dxdy ≤
f (γ) dl (γ)
πR2
2πR S(C,R)
D(C,R)
On Hadamard’s Inequality on a
Disk
where S (C, R) is the circle centered at the point C with radius R. The above
inequalities are sharp.
S.S. Dragomir
Proof. Consider the transformation of the plane R2 in itself given by
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h : R2 → R2 ,
h = (h1 , h2 ) and h1 (x, y) = −x + 2a, h2 (x, y) = −y + 2b.
Then h (D (C, R)) = D (C, R) and since
∂ (h1 , h2 ) −1 0
=
0 −1
∂ (x, y)
= 1,
we have the change of variable
ZZ
ZZ
∂ (h1 , h2 ) dxdy
f (x, y) dxdy =
f (h1 (x, y) , h2 (x, y)) ∂
(x,
y)
D(C,R)
D(C,R)
ZZ
=
f (−x + 2a, −y + 2b) dxdy.
D(C,R)
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Now, by the convexity of f on D (C, R) we also have
1
[f (x, y) + f (−x + 2a, −y + 2b)] ≥ f (a, b)
2
which gives, by integration on the disk D (C, R), that
Z Z
ZZ
1
(2.2)
f (x, y) dxdy +
f (−x + 2a, −y + 2b) dxdy
2
D(C,R)
D(C,R)
ZZ
≥ f (a, b)
dxdy = πR2 f (a, b) .
On Hadamard’s Inequality on a
Disk
D(C,R)
S.S. Dragomir
In addition, as
ZZ
ZZ
f (−x + 2a, −y + 2b) dxdy,
f (x, y) dxdy =
D(C,R)
D(C,R)
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then by the inequality (2.2) we obtain the first part of (2.1).
Now, consider the transformation
g = (g1 , g2 ) : [0, R] × [0, 2π] → D (C, R)
given by
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g:
Then we have
g1 (r, θ) = r cos θ + a,
g2 (r, θ) = r sin θ + b,
r ∈ [0, R] , θ ∈ [0, 2π] .
∂ (g1 , g2 ) cos θ
sin θ
=
−r
sin
θ
r
cos θ
∂ (r, θ)
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= r.
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Thus, we have the change of variable
ZZ
Z R Z 2π
∂ (g1 , g2 ) drdθ
f (x, y) dxdy =
f (g1 (r, θ) , g2 (r, θ)) ∂ (r, θ) D(C,R)
0
0
Z R Z 2π
=
f (r cos θ + a, r sin θ + b) rdrdθ.
0
0
Note that, by the convexity of f on D (C, R), we have
r
r
(R cos θ + a, R sin θ + b) + 1 −
(a, b)
f (r cos θ + a, r sin θ + b) = f
R
R
r
r
≤ f (R cos θ + a, R sin θ + b) + 1 −
f (a, b) ,
R
R
which yields that
r2
r
f (r cos θ + a, r sin θ + b) r ≤ f (R cos θ + a, R sin θ + b)+r 1 −
f (a, b)
R
R
S.S. Dragomir
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for all (r, θ) ∈ [0, R] × [0, 2π].
Integrating on [0, R] × [0, 2π] we get
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(2.3)
ZZ
Z
f (x, y) dxdy ≤
D(C,R)
On Hadamard’s Inequality on a
Disk
0
R
r2
dr
R
Z
2π
f (R cos θ + a, R sin θ + b) dθ
0
Z
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2π
Z
R
r
r 1−
dr
R
+ f (a, b)
dθ
0
0
2 Z 2π
R
πR2
=
f (R cos θ + a, R sin θ + b) dθ +
f (a, b) .
3 0
3
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Now, consider the curve γ : [0, 2π] → R2 given by
x (θ) := R cos θ + a,
γ:
θ ∈ [0, 2π] .
y (θ) := R sin θ + b,
Then γ ([0, 2π]) = S (C, R) and we write (integrating with respect to arc
length)
Z
Z 2π
1
f (γ) dl (γ) =
f (x (θ) , y (θ)) [ẋ (θ)]2 + [ẏ (θ)]2 2 dθ
0
S(C,R)
Z
= R
On Hadamard’s Inequality on a
Disk
2π
f (R cos θ + a, R sin θ + b) dθ.
S.S. Dragomir
0
By the inequality (2.3) we obtain
ZZ
Z
R
πR2
f (x, y) dxdy ≤
f (γ) dl (γ) +
f (a, b)
3 S(C,R)
3
D(C,R)
which gives the following inequality which is interesting in itself
ZZ
Z
1
2 1
1
(2.4)
f (x, y) dxdy ≤ ·
f (γ) dl (γ) + f (a, b) .
2
πR
3 2πR S(C,R)
3
D(C,R)
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As we proved that
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f (C) ≤
1
πR2
ZZ
f (x, y) dxdy,
D(C,R)
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then by the inequality (2.4) we deduce the inequality
Z
1
(2.5)
f (C) ≤
f (γ) dl (γ) .
2πR S(C,R)
Finally, by (2.5) and (2.4) we have
ZZ
Z
2
1
1
1
f
(x,
y)
dxdy
≤
·
f
(γ)
dl
(γ)
+
f (C)
πR2
3 2πR S(C,R)
3
D(C,R)
Z
1
f (γ) dl (γ)
≤
2πR S(C,R)
and the second part of (2.1) is proved.
Now, consider the map f0 : D (C, R) → R, f0 (x, y) = 1. Thus
1 = f0 (λ (x, y) + (1 − λ) (u, z))
= λf0 (x, y) + (1 − λ) f0 (u, z) = 1.
Therefore f0 is convex on D (C, R) → R. We also have
ZZ
Z
1
1
f0 (x, y) dxdy = 1 and
f0 (γ) dl (γ) = 1,
f0 (C) = 1,
πR2
2πR S(C,R)
D(C,R)
On Hadamard’s Inequality on a
Disk
S.S. Dragomir
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which shows us the inequalities (2.1) are sharp.
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3.
Some Mappings Connected to Hadamard’s Inequality on the Disk
As above, assume that the mapping f : D (C, R) → R is a convex mapping on
the disk centered at the point C = (a, b) ∈ R2 and having the radius R > 0.
Consider the mapping H : [0, 1] → R associated with the function f and given
by
ZZ
1
f (t (x, y) + (1 − t) C) dxdy,
H (t) :=
πR2
D(C,R)
which is well-defined for all t ∈ [0, 1].
The following theorem contains the main properties of this mapping.
Theorem 3.1. With the above assumption, we have:
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(ii) One has the bounds
inf H (t) = H (0) = f (C)
t∈[0,1]
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and
(3.2)
S.S. Dragomir
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(i) The mapping H is convex on [0, 1].
(3.1)
On Hadamard’s Inequality on a
Disk
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sup H (t) = H (1) =
t∈[0,1]
1
πR2
ZZ
f (x, y) dxdy.
D(C,R)
(iii) The mapping H is monotonic nondecreasing on [0, 1].
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Proof. (i) Let t1 , t2 ∈ [0, 1] and α, β ≥ 0 with α + β = 1. Then we have
ZZ
1
f (α (t1 (x, y) + (1 − t1 ) C)
H (αt1 + βt2 ) =
πR2
D(C,R)
+ β (t2 (x, y) + (1 − t2 ) C)) dxdy
ZZ
1
≤ α·
f (t1 (x, y) + (1 − t1 ) C) dxdy
πR2
D(C,R)
ZZ
1
+β ·
f (t2 (x, y) + (1 − t2 ) C) dxdy
πR2
D(C,R)
= αH (t1 ) + βH (t2 ) ,
On Hadamard’s Inequality on a
Disk
S.S. Dragomir
which proves the convexity of H on [0, 1].
(ii) We will prove the following identity
ZZ
1
(3.3)
H (t) = 2 2
f (x, y) dxdy
πt R
D(C,tR)
for all t ∈ (0, 1].
Fix t in (0, 1] and consider the transformation g = (ψ, η) : R2 → R2 given
by
ψ (x, y) := tx + (1 − t) a,
g:
(x, y) ∈ R2 ;
η (x, y) := ty + (1 − t) b,
then g (D (C, R)) = D (C, tR).
Indeed, for all (x, y) ∈ D (C, R) we have
(ψ − a)2 + (η − b)2 = t2 (x − a)2 + (y − b)2 ≤ (tR)2
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which shows that (ψ, η) ∈ D (C, tR), and conversely, for all (ψ, η) ∈
D (C, tR) , it is easy to see that there exists (x, y) ∈ D (C, R) so that
g (x, y) = (ψ, η).
We have the change of variable
ZZ
ZZ
∂ (ψ, η) dxdy
f (ψ, η) dψdη =
f (ψ (x, y) , η (x, y)) ∂
(x,
y)
D(C,tR)
D(C,R)
ZZ
=
f (t (x, y) + (1 − t) (a, b)) t2 dxdy
D(C,R)
2 2
= πR t H(t)
∂(ψ,η) since ∂(x,y) = t2 , which gives us the equality (3.3).
Now, by the inequality (2.1), we have
ZZ
1
f (x, y) dxdy ≥ f (C)
πt2 R2
D(C,tR)
which gives us H (t) ≥ f (C) for all t ∈ [0, 1] and since H (0) = f (C),
we obtain the bound (3.1).
On Hadamard’s Inequality on a
Disk
S.S. Dragomir
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By the convexity of f on the disk D (C, R) we have
ZZ
1
[tf (x, y) + (1 − t) f (C)] dxdy
H (t) ≤
πR2
D(C,R)
ZZ
t
=
f (x, y) dxdy + (1 − t) f (C)
πR2
D(C,R)
ZZ
ZZ
1−t
t
≤
f (x, y) dxdy +
f (x, y) dxdy
πR2
πR2
D(C,R)
D(C,R)
ZZ
1
f (x, y) dxdy.
=
πR2
D(C,R)
As we have
1
H (1) =
πR2
On Hadamard’s Inequality on a
Disk
S.S. Dragomir
ZZ
f (x, y) dxdy,
D(C,R)
then the bound (3.2) holds.
(iii) Let 0 ≤ t1 < t2 ≤ 1. Then, by the convexity of the mapping H we have
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H (t2 ) − H (t1 )
H (t1 ) − H (0)
≥
≥0
t2 − t1
t1
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as H (t1 ) ≥ H (0) for all t1 ∈ [0, 1]. This proves the monotonicity of the
mapping H in the interval [0, 1].
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Further on, we shall introduce another mapping connected to Hadamard’s
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inequality
h : [0, 1] → R,
Z

 1
f (γ) dl (γ (t)) , t ∈ (0, 1] ,
h (t) := 2πtR S(C,tR)

f (C) ,
t = 0,
where f : D (C, R) → R is a convex mapping on the disk D (C, R) centered at
the point C = (a, b) ∈ R2 and having the same radius R.
The main properties of this mapping are embodied in the following theorem.
On Hadamard’s Inequality on a
Disk
Theorem 3.2. With the above assumptions one has:
S.S. Dragomir
(i) The mapping h : [0, 1] → R is convex on [0, 1].
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(ii) One has the bounds
inf h (t) = h (0) = f (C)
(3.4)
t∈[0,1]
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and
(3.5)
1
sup h (t) = h (1) =
2πR
t∈[0,1]
Z
f (γ) dl (γ) .
S(C,R)
(iii) The mapping h is monotonic nondecreasing on [0, 1].
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(iv) We have the inequality
H (t) ≤ h (t)
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for all t ∈ [0, 1] .
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Proof. For a fixed t in [0, 1] consider the curve
x (θ) = tR cos θ + a,
γ:
y (θ) = tR sin θ + b,
θ ∈ [0, 2π] .
Then γ ([0, 2π]) = S (C, tR) and
1
2πtR
Z
f (γ) dl (γ)
S(C,tR)
Z 2π
q
1
=
f (tR cos θ + a, tR sin θ + b) (ẋ (θ))2 + (ẏ (θ))2 dθ
2πtR 0
Z 2π
1
f (tR cos θ + a, tR sin θ + b) dθ.
=
2π 0
We note that, then
Z 2π
1
h (t) =
f (tR cos θ + a, tR sin θ + b) dθ
2π 0
Z 2π
1
=
f (t (R cos θ, R sin θ) + (a, b)) dθ
2π 0
for all t ∈ [0, 1].
(i) Let t1 , t2 ∈ [0, 1] and α, β ≥ 0 with α + β = 1. Then, by the convexity of
On Hadamard’s Inequality on a
Disk
S.S. Dragomir
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f we have that
Z 2π
1
h (αt1 + βt2 ) =
f (α [t1 (R cos θ, R sin θ) + (a, b)]
2π 0
+ β [t2 (R cos θ, R sin θ) + (a, b)]) dθ
Z 2π
1
f (t1 (R cos θ, R sin θ) + (a, b)) dθ
≤ α·
2π 0
Z 2π
1
f (t2 (R cos θ, R sin θ) + (a, b)) dθ
+β ·
2π 0
= αh (t1 ) + βh (t2 )
On Hadamard’s Inequality on a
Disk
S.S. Dragomir
which proves the convexity of h on [0, 1].
(iv) In the above theorem we showed that
ZZ
1
H (t) = 2 2
f (x, y) dxdy for all t ∈ (0, 1] .
πt R
D(C,tR)
By Hadamard’s inequality (2.1) we can state that
ZZ
Z
1
1
f (x, y) dxdy ≤
f (γ) dl (γ)
πt2 R2
2πtR S(C,tR)
D(C,tR)
which gives us that
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H (t) ≤ h (t)
for all t ∈ (0, 1] .
As it is easy to see that H (0) = h (0) = f (C), then the inequality embodied in (iv) is proved.
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(ii) The bound (3.4) follows by the above considerations and we shall omit the
details.
By the convexity of f on the disk D (C, R) we have
Z 2π
1
h (t) =
f (t [(R cos θ, R sin θ) + (a, b)] + (1 − t) (a, b)) dθ
2π 0
Z 2π
1
≤ t·
f (R cos θ + a, R sin θ + b) dθ
2π 0
Z 2π
1
dθ
+ (1 − t) f (a, b)
2π 0
Z 2π
1
f (R cos θ + a, R sin θ + b) dθ
≤ t·
2π 0
Z 2π
1
f (R cos θ + a, R sin θ + b) dθ
+ (1 − t) ·
2π 0
Z 2π
1
=
f (R cos θ + a, R sin θ + b) dθ = h (1) ,
2π 0
for all t ∈ [0, 1], which proves the bound (3.5).
(iii) Follows by the above considerations as in the Theorem 3.1. We shall omit
the details.
On Hadamard’s Inequality on a
Disk
S.S. Dragomir
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For a convex mapping f defined on the disk D (C, R) we can also consider
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the mapping
1
g (t, (x, y)) :=
πR2
ZZ
f (t (x, y) + (1 − t) (z, u)) dzdu
D(C,R)
which is well-defined for all t ∈ [0, 1] and (x, y) ∈ D (C, R).
The main properties of the mapping g are enclosed in the following proposition.
Proposition 3.3. With the above assumptions on the mapping f one has:
(i) For all (x, y) ∈ D (C, R), the map g (·, (x, y)) is convex on [0, 1].
On Hadamard’s Inequality on a
Disk
S.S. Dragomir
(ii) For all t ∈ [0, 1], the map g (t, ·) is convex on D (C, R).
Proof. (i) Let t1 , t2 ∈ [0, 1] and α, β ≥ 0 with α + β = 1. By the convexity
of f we have
ZZ
1
g (αt1 + βt2 , (x, y)) =
f (α [t1 (x, y) + (1 − t1 ) (z, u)]
πR2
D(C,R)
+ β [t2 (x, y) + (1 − t2 ) (z, u)]) dzdu
ZZ
1
≤α·
f (t1 (x, y) + (1 − t1 ) (z, u)) dzdu
πR2
D(C,R)
ZZ
1
+β·
f (t2 (x, y) + (1 − t2 ) (z, u)) dzdu
πR2
D(C,R)
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= αg (t1 , (x, y)) + βg (t2 , (x, y)) ,
J. Ineq. Pure and Appl. Math. 1(1) Art. 2, 2000
for all (x, y) ∈ D (C, R), and the statement is proved.
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(ii) Let (x1 , y1 ) , (x2 , y2 ) ∈ D (C, R) and α, β ≥ 0 with α + β = 1. Then
g (t, α (x1 , y1 ) +β (x2 , y2 ))
ZZ
1
=
f [α (t (x1 , y1 ) + (1 − t) (z, u))
πR2
D(C,R)
+β (t (x2 , y2 ) + (1 − t) (z, u))] dzdu
ZZ
1
≤α 2
f (t (x1 , y1 ) + (1 − t) (z, u)) dzdu
πR
D(C,R)
ZZ
1
+β
f (t (x2 , y2 ) + (1 − t) (z, u)) dzdu
πR2
D(C,R)
On Hadamard’s Inequality on a
Disk
S.S. Dragomir
= αg (t, (x1 , y1 )) + βg (t, (x2 , y2 )) ,
for all t ∈ [0, 1], and the statement is proved.
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By the use of this mapping we can introduce the following application as
well
ZZ
1
G : [0, 1] → R, G (t) :=
g (t, (x, y)) dxdy
πR2
D(C,R)
where g is as above.
The main properties of this mapping are embodied in the following theorem.
Theorem 3.4. With the above assumptions we have:
(i) For all s ∈ 0, 21
1
1
=G
−s ,
G s+
2
2
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and for all t ∈ [0, 1] one has
G (1 − t) = G (t) .
(ii) The mapping G is convex on the interval [0, 1].
(iii) One has the bounds
1
inf G (t) = G
t∈[0,1]
2
ZZZZ
1
x+z y+u
=
f
,
dxdydzdu
2
2
(πR2 )2
D(C,R)×D(C,R)
≥ f (C)
On Hadamard’s Inequality on a
Disk
S.S. Dragomir
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and
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1
sup G (t) = G (0) = G (1) =
πR2
t∈[0,1]
ZZ
f (x, y) dxdy.
D(C,R)
(iv) The mapping
G is monotonic nonincreasing on 0, 12 and nondecreasing
on 12 , 1 .
(v) We have the inequality
(3.6)
G (t) ≥ max {H (t) , H (1 − t)} ,
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for all t ∈ [0, 1] .
Proof. The statements (i) and (ii) are obvious by the properties of the mapping
g defined above and we shall omit the details.
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(iii) By (i) and (ii) we have
G (t) + G (1 − t)
G (t) =
≥G
2
1
,
2
for all t ∈ [0, 1]
which proves the first bound in (iii).
Note that the inequality
1
G
≥ f (C)
2
follows by (3.6) for t =
1
2
On Hadamard’s Inequality on a
Disk
and taking into account that H
1
2
≥ f (C).
S.S. Dragomir
We also have
Z Z
ZZ
1
G (t) =
f (t (x, y) + (1 − t) (z, u)) dzdu dxdy
(πR2 )2
D(C,R)
D(C,R)
1
≤
(πR2 )2
ZZ
ZZ
2
×
tf (x, y) πR + (1 − t)
f (z, u) dzdu dxdy
D(C,R)
D(C,R)
1
(πR2 )2
ZZ
ZZ
2
2
× tπR
f (x, y) dxdy + (1 − t) πR
f (x, y) dxdy
D(C,R)
D(C,R)
ZZ
1
=
f (x, y) dxdy
πR2
D(C,R)
=
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for all t ∈ [0, 1], and the second bound in (iii) is also proved.
(iv) The argument is similar to the proof of Theorem 3.1 (iii) (see also [6]) and
we shall omit the details.
(v) By Theorem 2.1 we have that
ZZ
1
g (t, (x, y)) dxdy
G (t) =
πR2
D(C,R)
ZZ
1
≥ g (t, (a, b)) =
f (t (x, y) + (1 − t) (a, b)) dxdy
πR2
D(C,R)
= H (t)
On Hadamard’s Inequality on a
Disk
S.S. Dragomir
for all t ∈ [0, 1].
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As G (t) = G (1 − t) ≥ H (1 − t), we obtain the desired inequality (3.6).
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The theorem is thus proved.
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References
[1] S.S. DRAGOMIR, Two refinements of Hadamard’s inequalities, Coll. of
Sci. Pap. of the Fac. of Sci. Kragujevać (Yugoslavia), 11 (1990), 23-26.
ZBL No. 729:26017.
[2] S.S. DRAGOMIR, Some refinements of Hadamard’s inequalities, Gaz.
Mat. Metod. (Romania), 11 (1990), 189-191.
[3] S.S. DRAGOMIR, J. E. PEČARIĆ AND J. SÁNDOR, A note on the
Jensen-Hadamard inequality, L’ Anal. Num. Theor. L’Approx. (Romania),
19 (1990), 21-28. MR 93C:26016, ZBL No. 743:26016.
[4] S.S. DRAGOMIR, A mapping in connection to Hadamard’s inequality, An
Öster. Akad. Wiss. (Wien), 128 (1991), 17-20. MR 93h:26032, ZBL No.
747:26015.
[5] S.S. DRAGOMIR, Some integral inequalities for differentiable convex
functions, Contributions, Macedonian Acad. of Sci. and Arts (Scopie), 16
(1992), 77-80.
[6] S.S. DRAGOMIR, Two mappings in connection to Hadamard’s inequality, J. Math. Anal, Appl., 167 (1992), 49-56. MR 93m:26038, ZBL No.
758:26014.
[7] S.S. DRAGOMIR, On Hadamard’s inequality for convex functions, Mat.
Balkanica, 6 (1992), 215-222. MR 934:26033.
On Hadamard’s Inequality on a
Disk
S.S. Dragomir
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J. Ineq. Pure and Appl. Math. 1(1) Art. 2, 2000
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[8] S.S. DRAGOMIR AND N.M. IONESCU, Some integral inequalities for
differentiable convex functions, Coll. Pap. of the Fac. of Sci. Kragujevać
(Yugoslavia), 13 (1992), 11-16. ZBL No. 770:26009.
[9] S.S. DRAGOMIR AND N.M. IONESCU, Some remarks in convex functions, L’Anal. Num. Theór. L’Approx. (Romania), 21 (1992), 31-36.
MR:93m:26023, ZBL No. 770:26008.
[10] S.S. DRAGOMIR, A refinement of Hadamard’s inequality for isotonic linear functionals, Tamkang J. of Math. (Taiwan), 24 (1993), 101-106.
[11] S.S. DRAGOMIR, D. BARBU AND C. BUSE, A Probabilistic argument
for the convergence of some sequences associated to Hadamard’s inequality, Studia Univ. Babes-Bolyai, Mathematica, 38 (1993), (1), 29-34.
[12] S.S. DRAGOMIR, D.M. MILOŚEVIĆ AND J. SÁNDOR, On some refinements of Hadamard’s inequalities and applications, Univ. Beograd, Publ.
Elektrotelm. Fak, Ser. Mat., 4 (1993), 3-10.
[13] D.S. MITRINOVIĆ, J.E. PEČARIĆ AND A. M. FINK, Classical and New
Inequalities in Analysis, Kluwer Academic Publishers, 1993.
On Hadamard’s Inequality on a
Disk
S.S. Dragomir
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[14] J.E. PEČARIĆ AND S.S. DRAGOMIR, On some integral inequalities for
convex functions, Bull. Mat. Inst. Pol. Iasi, 36 (1-4) (1990), 19-23. MR
93i:26019. ZBL No. 765:26008.
[15] J.E. PEČARIĆ AND S.S. DRAGOMIR, A generalisation of Hadamard’s
inequality for isotonic linear functionals, Rodovi Mat. (Sarajevo), 7 (1991),
103-107. MR 93k:26026. ZBL No. 738:26006.
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