Journal of Inequalities in Pure and
Applied Mathematics
http://jipam.vu.edu.au/
Volume 1, Issue 1, Article 2, 2000
ON HADAMARD’S INEQUALITY ON A DISK
S.S. DRAGOMIR
S CHOOL OF C OMMUNICATIONS AND I NFORMATICS , V ICTORIA U NIVERSITY OF T ECHNOLOGY,
PO B OX 14428, M ELBOURNE C ITY MC 8001, AUSTRALIA
Sever.Dragomir@vu.edu.au
URL: http://rgmia.vu.edu.au/SSDragomirWeb.html
Received 17 September, 1999; accepted 16 December, 1999
Communicated by T. Mills
A BSTRACT. 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.
Key words and phrases: Convex functions of double variable, Jensen’s inequality, Hadamard’s inequality.
2000 Mathematics Subject Classification. 26D07, 26D15.
1. I NTRODUCTION
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
The following properties are also proved:
(i) H is convex and monotonic nondecreasing.
(ii) One has the bounds
Z b
1
sup H (t) = H (1) =
f (x) dx
b−a a
t∈[0,1]
ISSN (electronic): 1443-5756
c 2000 Victoria University. All rights reserved.
003-99
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D RAGOMIR
and
a+b
inf H (t) = H (0) = f
.
t∈[0,1]
2
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
convex
on
[0,
1]
and
monotonic
nonincreasing
on
0, 2 and nondecreasing on
1 ,1 .
2
(ii) F is symmetric about 12 . That is,
F (t) = F (1 − t) ,
for all t ∈ [0, 1] .
(iii) One has the bounds
1
sup F (t) = F (0) = F (1) =
b−a
t∈[0,1]
Z
b
f (x) dx
a
and
Z bZ b 1
1
x+y
a+b
inf F (t) = F
=
dxdy ≥ f
.
f
t∈[0,1]
2
2
2
(b − a)2 a a
(iv) The following inequality holds
F (t) ≥ max {H (t) , H (1 − t)} ,
for all t ∈ [0, 1] .
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].
2. H ADAMARD ’ S I NEQUALITY ON THE D ISK
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)
where S (C, R) is the circle centered at the point C with radius R. The above inequalities are
sharp.
Proof. Consider the transformation of the plane R2 in itself given by
h : R2 → R 2 ,
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)
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H ADAMARD ’ S I NEQUALITY ON
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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)
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) .
D(C,R)
In addition, as
ZZ
ZZ
f (−x + 2a, −y + 2b) dxdy,
f (x, y) dxdy =
D(C,R)
D(C,R)
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
g:
g1 (r, θ) = r cos θ + a,
g2 (r, θ) = r sin θ + b,
r ∈ [0, R] , θ ∈ [0, 2π] .
Then we have
∂ (g1 , g2 ) cos θ
sin θ =
= r.
−r sin θ r cos θ ∂ (r, θ)
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
f (r cos θ + a, r sin θ + b) = f
(R cos θ + a, R sin θ + b) + 1 −
(a, b)
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
for all (r, θ) ∈ [0, R] × [0, 2π].
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D RAGOMIR
Integrating on [0, R] × [0, 2π] we get
ZZ
Z R 2 Z 2π
r
f (x, y) dxdy ≤
dr
f (R cos θ + a, R sin θ + b) dθ
D(C,R)
0 R
0
Z 2π Z R r
(2.3)
+ f (a, b)
dθ
r 1−
dr
R
0
0
Z
πR2
R2 2π
=
f (R cos θ + a, R sin θ + b) dθ +
f (a, b) .
3 0
3
Now, consider the curve γ : [0, 2π] → R2 given by
x (θ) := R cos θ + a,
γ:
y (θ) := R sin θ + b,
θ ∈ [0, 2π] .
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
2π
f (R cos θ + a, R sin θ + b) dθ.
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
2
1
1
1
(2.4)
f
(x,
y)
dxdy
≤
·
f
(γ)
dl
(γ)
+
f (a, b) .
πR2
3 2πR S(C,R)
3
D(C,R)
As we proved that
ZZ
1
f (C) ≤
f (x, y) dxdy,
πR2
D(C,R)
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
1
2
1
1
f (x, y) dxdy ≤
·
f (γ) dl (γ) + f (C)
2
πR
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 (C) = 1,
f0 (x, y) dxdy = 1 and
f0 (γ) dl (γ) = 1,
πR2
2πR S(C,R)
D(C,R)
which shows us the inequalities (2.1) are sharp.
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H ADAMARD ’ S I NEQUALITY ON
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3. S OME M APPINGS C ONNECTED TO H ADAMARD ’ S I NEQUALITY ON THE D ISK
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
H (t) :=
f (t (x, y) + (1 − t) C) dxdy,
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:
(i) The mapping H is convex on [0, 1].
(ii) One has the bounds
inf H (t) = H (0) = f (C)
(3.1)
t∈[0,1]
and
(3.2)
1
sup H (t) = H (1) =
πR2
t∈[0,1]
ZZ
f (x, y) dxdy.
D(C,R)
(iii) The mapping H is monotonic nondecreasing on [0, 1].
Proof.
(i) Let t1 , t2 ∈ [0, 1] and α, β ≥ 0 with α + β = 1. Then we have
ZZ
1
H (αt1 + βt2 ) =
f (α (t1 (x, y) + (1 − t1 ) C)
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 ) ,
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
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).
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D RAGOMIR
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)
since ∂(ψ,η)
= t2 , which gives us the equality (3.3).
∂(x,y) 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).
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
t
1−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
ZZ
1
f (x, y) dxdy,
H (1) =
πR2
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
H (t2 ) − H (t1 )
H (t1 ) − H (0)
≥
≥0
t2 − t1
t1
as H (t1 ) ≥ H (0) for all t1 ∈ [0, 1]. This proves the monotonicity of the mapping H in
the interval [0, 1].
Further on, we shall introduce another mapping connected to Hadamard’s inequality
Z

1

f (γ) dl (γ (t)) , t ∈ (0, 1] ,
h : [0, 1] → R, 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.
Theorem 3.2. With the above assumptions one has:
(i) The mapping h : [0, 1] → R is convex on [0, 1].
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H ADAMARD ’ S I NEQUALITY ON
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(ii) One has the bounds
inf h (t) = h (0) = f (C)
(3.4)
t∈[0,1]
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].
(iv) We have the inequality
H (t) ≤ h (t)
for all t ∈ [0, 1] .
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
Z
1
f (γ) dl (γ)
2πtR 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 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
≤ α·
f (t1 (R cos θ, R sin θ) + (a, b)) dθ
2π 0
Z 2π
1
+β ·
f (t2 (R cos θ, R sin θ) + (a, b)) dθ
2π 0
= αh (t1 ) + βh (t2 )
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)
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D RAGOMIR
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
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.
(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
Z 2π
1
1
≤ t·
f (R cos θ + a, R sin θ + b) dθ + (1 − t) f (a, b)
dθ
2π 0
2π 0
Z 2π
1
≤ t·
f (R cos θ + a, R sin θ + b) dθ
2π 0
Z 2π
1
+ (1 − t) ·
f (R cos θ + a, R sin θ + b) dθ
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.
For a convex mapping f defined on the disk D (C, R) we can also consider the mapping
ZZ
1
g (t, (x, y)) :=
f (t (x, y) + (1 − t) (z, u)) dzdu
πR2
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].
(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
f (α [t1 (x, y) + (1 − t1 ) (z, u)]
g (αt1 + βt2 , (x, y)) =
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)
= αg (t1 , (x, y)) + βg (t2 , (x, y)) ,
for all (x, y) ∈ D (C, R), and the statement is proved.
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H ADAMARD ’ S I NEQUALITY ON
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(ii) Let (x1 , y1 ) , (x2 , y2 ) ∈ D (C, R) and α, β ≥ 0 with α + β = 1. Then
ZZ
1
g (t, α (x1 , y1 ) + β (x2 , y2 )) =
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)
= αg (t, (x1 , y1 )) + βg (t, (x2 , y2 )) ,
for all t ∈ [0, 1], and the statement is proved.
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, 12
1
1
G s+
=G
−s ,
2
2
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 ≥ f (C)
2
2
(πR2 )2
D(C,R)×D(C,R)
and
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)} ,
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.
(iii) By (i) and (ii) we have
G (t) + G (1 − t)
1
G (t) =
≥G
, for all t ∈ [0, 1]
2
2
which proves the first bound in (iii).
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D RAGOMIR
Note that the inequality
1
G
≥ f (C)
2
and taking into account that H
1
follows by (3.6) for t = 12
≥ f (C).
2
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)
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) =
g (t, (x, y)) dxdy
πR2
D(C,R)
ZZ
1
≥ g (t, (a, b)) =
f (t (x, y) + (1 − t) (a, b)) dxdy = H (t)
πR2
D(C,R)
for all t ∈ [0, 1].
As G (t) = G (1 − t) ≥ H (1 − t), we obtain the desired inequality (3.6).
The theorem is thus proved.
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