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volume 1, issue 1, article 2, 2000.

Received 17 September 1999;

accepted 16 December 1999.

Communicated by:T. Mills

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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:[email protected]

URL:http://rgmia.vu.edu.au/SSDragomirWeb.html

c

2000Victoria University ISSN (electronic): 1443-5756 003-99

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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

http://jipam.vu.edu.au

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.

1. Introduction

Let f : I ⊆ R → R be a convex mapping defined on the interval I of real numbers anda, b∈I witha < b. The following double inequality

(1.1) f

a+b 2

≤ 1 b−a

Z b a

f(x)dx≤ f(a) +f(b) 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 mappingf.

In the paper [4] (see also [6] and [7]) the following mapping naturally connected with Hadamard’s result is considered

H: [0,1]→R, H(t) := 1 b−a

Z b a

f

tx+ (1−t)a+b 2

dx.

The following properties are also proved:

(i) His convex and monotonic nondecreasing.

(ii) One has the bounds sup

t∈[0,1]

H(t) =H(1) = 1 b−a

Z b a

f(x)dx

and

t∈[0,1]inf H(t) =H(0) =f

a+b 2

.

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Another mapping also closely connected with Hadamard’s inequality is the following one [6] (see also [7])

F : [0,1]→R, F (t) := 1 (b−a)²

Z b a

f(tx+ (1−t)y)dxdy.

The properties of this mapping are itemized below:

(i) F is convex on[0,1]and monotonic nonincreasing on 0,¹₂

and nondecreasing on₁

2,1 .

(ii) F is symmetric about ¹₂.That is,

F(t) =F (1−t), for allt∈[0,1]. (iii) One has the bounds

sup

t∈[0,1]

F (t) = F (0) =F(1) = 1 b−a

Z b a

f(x)dx

and

t∈[0,1]inf F (t) = F 1

2

= 1

(b−a)² Z b

a

Z b a

f

x+y 2

dxdy ≥f

a+b 2

.

(iv) The following inequality holds

F (t)≥max{H(t), H(1−t)}, for allt∈[0,1].

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In this paper we will point out a similar inequality to Hadamard’s that ap- plies to convex mappings defined on a disk embedded in the planeR². We will also consider some mappings similar in a sense to the mappingsH andF 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].

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2. Hadamard’s Inequality on the Disk

Let us consider a pointC = (a, b) ∈ R² and the diskD(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) → Ris convex onD(C, R), then one has the inequality

(2.1) f(C)≤ 1 πR²

Z Z

D(C,R)

f(x, y)dxdy≤ 1 2πR

Z

S(C,R)

f(γ)dl(γ) where S(C, R)is the circle centered at the pointC with radius R. The above inequalities are sharp.

Proof. Consider the transformation of the planeR² in itself given by

h:R² →R², h= (h₁, h₂) and h₁(x, y) =−x+ 2a, h₂(x, y) = −y+ 2b.

Thenh(D(C, R)) =D(C, R)and since

∂(h₁, h₂)

∂(x, y) =

−1 0 0 −1

= 1, we have the change of variable

Z Z

D(C,R)

f(x, y)dxdy= Z Z

D(C,R)

f(h₁(x, y), h₂(x, y))

∂(h₁, h₂)

∂(x, y)

dxdy

= Z Z

D(C,R)

f(−x+ 2a,−y+ 2b)dxdy.

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Now, by the convexity off onD(C, R)we also have 1

2[f(x, y) +f(−x+ 2a,−y+ 2b)]≥f(a, b) which gives, by integration on the diskD(C, R), that

(2.2) 1 2

Z Z

D(C,R)

f(x, y)dxdy+ Z Z

D(C,R)

f(−x+ 2a,−y+ 2b)dxdy

≥f(a, b) Z Z

D(C,R)

dxdy=πR²f(a, b). In addition, as

Z Z

D(C,R)

f(x, y)dxdy= Z Z

D(C,R)

f(−x+ 2a,−y+ 2b)dxdy, 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) =rcosθ+a,

g₂(r, θ) = rsinθ+b, r∈[0, R], θ∈[0,2π]. Then we have

∂(g₁, g₂)

∂(r, θ) =

cosθ sinθ

−rsinθ rcosθ

=r.

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Thus, we have the change of variable Z Z

D(C,R)

f(x, y)dxdy = Z R

0

Z 2π 0

f(g₁(r, θ), g₂(r, θ))

∂(g1, g2)

∂(r, θ)

drdθ

= Z R

0

Z 2π 0

f(rcosθ+a, rsinθ+b)rdrdθ.

Note that, by the convexity off onD(C, R), we have f(rcosθ+a, rsinθ+b) =f

r

R (Rcosθ+a, Rsinθ+b) +

1− r R

(a, b)

≤ r

Rf(Rcosθ+a, Rsinθ+b) +

1− r R

f(a, b), which yields that

f(rcosθ+a, rsinθ+b)r≤ r²

Rf(Rcosθ+a, Rsinθ+b)+r 1− r

R

f(a, b) for all(r, θ)∈[0, R]×[0,2π].

Integrating on[0, R]×[0,2π]we get

Z Z

D(C,R)

f(x, y)dxdy ≤ Z R

0

r² Rdr

Z 2π 0

f(Rcosθ+a, Rsinθ+b)dθ

+f(a, b) Z 2π

0

dθ Z R

0

r 1− r

R

dr

= R² 3

Z 2π 0

f(Rcosθ+a, Rsinθ+b)dθ+ πR²

3 f(a, b). (2.3)

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Now, consider the curveγ : [0,2π]→R²given by γ :

x(θ) := Rcosθ+a,

y(θ) :=Rsinθ+b, θ ∈[0,2π].

Then γ([0,2π]) = S(C, R) and we write (integrating with respect to arc length)

Z

S(C,R)

f(γ)dl(γ) = Z 2π

0

f(x(θ), y(θ)) [ ˙x(θ)]²+ [ ˙y(θ)]²¹₂ dθ

= R

Z 2π 0

f(Rcosθ+a, Rsinθ+b)dθ.

By the inequality (2.3) we obtain Z Z

D(C,R)

f(x, y)dxdy≤ R 3

Z

S(C,R)

f(γ)dl(γ) + πR²

3 f(a, b) which gives the following inequality which is interesting in itself (2.4) 1

πR² Z Z

D(C,R)

f(x, y)dxdy ≤ 2 3· 1

2πR Z

S(C,R)

f(γ)dl(γ) +1

3f(a, b). As we proved that

f(C)≤ 1 πR²

Z Z

D(C,R)

f(x, y)dxdy,

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then by the inequality (2.4) we deduce the inequality

(2.5) f(C)≤ 1

2πR Z

S(C,R)

f(γ)dl(γ).

Finally, by (2.5) and (2.4) we have 1

πR² Z Z

D(C,R)

f(x, y)dxdy ≤ 2 3· 1

2πR Z

S(C,R)

f(γ)dl(γ) + 1 3f(C)

≤ 1 2πR

Z

S(C,R)

f(γ)dl(γ)

and the second part of (2.1) is proved.

Now, consider the mapf0 :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.

Thereforef₀ is convex onD(C, R)→R. We also have f₀(C) = 1, 1

πR² Z Z

D(C,R)

f₀(x, y)dxdy= 1 and 1 2πR

Z

S(C,R)

f₀(γ)dl(γ) = 1,

which shows us the inequalities (2.1) are sharp.

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3. Some Mappings Connected to Hadamard’s In- equality on the Disk

As above, assume that the mappingf : D(C, R)→Ris a convex mapping on the disk centered at the point C = (a, b) ∈ R² and having the radius R > 0.

Consider the mappingH : [0,1]→Rassociated with the functionf and given by

H(t) := 1 πR²

Z Z

D(C,R)

f(t(x, y) + (1−t)C)dxdy, which is well-defined for allt∈[0,1].

The following theorem contains the main properties of this mapping.

Theorem 3.1. With the above assumption, we have:

(i) The mappingH is convex on[0,1].

(ii) One has the bounds

(3.1) inf

t∈[0,1]H(t) =H(0) =f(C) and

(3.2) sup

t∈[0,1]

H(t) =H(1) = 1 πR²

Z Z

D(C,R)

f(x, y)dxdy.

(iii) The mappingH is monotonic nondecreasing on[0,1].

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Proof. (i) Lett₁, t₂ ∈[0,1]andα, β ≥0withα+β = 1.Then we have H(αt₁+βt₂) = 1

πR² Z Z

D(C,R)

f(α(t₁(x, y) + (1−t₁)C) +β(t₂(x, y) + (1−t₂)C))dxdy

≤ α· 1 πR²

Z Z

D(C,R)

f(t₁(x, y) + (1−t₁)C)dxdy

+β· 1 πR²

Z Z

D(C,R)

f(t₂(x, y) + (1−t₂)C)dxdy

= αH(t₁) +βH(t₂), which proves the convexity ofHon[0,1].

(ii) We will prove the following identity

(3.3) H(t) = 1

πt²R² Z Z

D(C,tR)

f(x, y)dxdy for allt∈(0,1].

Fixtin(0,1]and consider the transformationg = (ψ, η) :R² →R²given by

g :

ψ(x, y) :=tx+ (1−t)a,

η(x, y) :=ty+ (1−t)b, (x, y)∈R²; theng(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 Z Z

D(C,tR)

f(ψ, η)dψdη = Z Z

D(C,R)

f(ψ(x, y), η(x, y))

∂(ψ, η)

∂(x, y)

dxdy

= Z Z

D(C,R)

f(t(x, y) + (1−t) (a, b))t²dxdy

=πR²t²H(t) since

∂(ψ,η)

∂(x,y)

=t², which gives us the equality (3.3).

Now, by the inequality (2.1), we have 1

πt²R² Z Z

D(C,tR)

f(x, y)dxdy ≥f(C)

which gives usH(t) ≥ f(C)for all t ∈ [0,1]and sinceH(0) = f(C), we obtain the bound (3.1).

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By the convexity off on the diskD(C, R)we have H(t) ≤ 1

πR² Z Z

D(C,R)

[tf(x, y) + (1−t)f(C)]dxdy

= t

πR² Z Z

D(C,R)

f(x, y)dxdy+ (1−t)f(C)

≤ t πR²

Z Z

D(C,R)

f(x, y)dxdy+ 1−t πR²

Z Z

D(C,R)

f(x, y)dxdy

= 1

πR² Z Z

D(C,R)

f(x, y)dxdy.

As we have

H(1) = 1 πR²

Z Z

D(C,R)

f(x, y)dxdy, then the bound (3.2) holds.

(iii) Let0≤t₁ < t₂ ≤1. Then, by the convexity of the mappingHwe have H(t₂)−H(t₁)

t₂−t₁ ≥ H(t₁)−H(0)

t₁ ≥0

asH(t₁) ≥ H(0) for allt₁ ∈ [0,1]. This proves the monotonicity of the mappingHin the interval[0,1].

Further on, we shall introduce another mapping connected to Hadamard’s

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inequality

h: [0,1]→R, h(t) :=





 1 2πtR

Z

S(C,tR)

f(γ)dl(γ(t)), t∈(0,1],

f(C), t= 0,

wheref :D(C, R)→Ris a convex mapping on the diskD(C, R)centered at the pointC = (a, b)∈R²and having the same radiusR.

The main properties of this mapping are embodied in the following theorem.

Theorem 3.2. With the above assumptions one has:

(i) The mappingh: [0,1]→Ris convex on[0,1].

(ii) One has the bounds

(3.4) inf

t∈[0,1]h(t) =h(0) =f(C) and

(3.5) sup

t∈[0,1]

h(t) =h(1) = 1 2πR

Z

S(C,R)

f(γ)dl(γ).

(iii) The mappinghis monotonic nondecreasing on[0,1].

(iv) We have the inequality

H(t)≤h(t) for allt∈[0,1].

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Proof. For a fixedtin[0,1]consider the curve γ :

x(θ) =tRcosθ+a,

y(θ) = tRsinθ+b, θ ∈[0,2π]. Thenγ([0,2π]) =S(C, tR)and

1 2πtR

Z

S(C,tR)

f(γ)dl(γ)

= 1

2πtR Z 2π

0

f(tRcosθ+a, tRsinθ+b) q

( ˙x(θ))²+ ( ˙y(θ))²dθ

= 1 2π

Z 2π 0

f(tRcosθ+a, tRsinθ+b)dθ.

We note that, then

h(t) = 1 2π

Z 2π 0

f(tRcosθ+a, tRsinθ+b)dθ

= 1

2π Z 2π

0

f(t(Rcosθ, Rsinθ) + (a, b))dθ for allt ∈[0,1].

(i) Lett₁, t₂ ∈[0,1]andα, β ≥0withα+β = 1.Then, by the convexity of

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f we have that

h(αt₁+βt₂) = 1 2π

Z 2π 0

f(α[t₁(Rcosθ, Rsinθ) + (a, b)]

+β[t₂(Rcosθ, Rsinθ) + (a, b)])dθ

≤ α· 1 2π

Z 2π 0

f(t1(Rcosθ, Rsinθ) + (a, b))dθ +β· 1

2π Z 2π

0

f(t₂(Rcosθ, Rsinθ) + (a, b))dθ

= αh(t₁) +βh(t₂) which proves the convexity ofhon[0,1].

(iv) In the above theorem we showed that H(t) = 1

πt²R² Z Z

D(C,tR)

f(x, y)dxdyfor allt∈(0,1]. By Hadamard’s inequality (2.1) we can state that

1 πt²R²

Z Z

D(C,tR)

f(x, y)dxdy≤ 1 2πtR

Z

S(C,tR)

f(γ)dl(γ) which gives us that

H(t)≤h(t) for allt ∈(0,1].

As it is easy to see thatH(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 off on the diskD(C, R)we have h(t) = 1

2π Z 2π

0

f(t[(Rcosθ, Rsinθ) + (a, b)] + (1−t) (a, b))dθ

≤ t· 1 2π

Z 2π 0

+ (1−t)f(a, b) 1 2π

Z 2π 0

dθ

≤ t· 1 2π

Z 2π 0

+ (1−t)· 1 2π

Z 2π 0

= 1

2π Z 2π

0

f(Rcosθ+a, Rsinθ+b)dθ=h(1), for allt∈[0,1], which proves the bound (3.5).

(iii) Follows by the above considerations as in the Theorem3.1. We shall omit the details.

For a convex mappingf defined on the diskD(C, R)we can also consider

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the mapping

g(t,(x, y)) := 1 πR²

Z Z

D(C,R)

f(t(x, y) + (1−t) (z, u))dzdu

which is well-defined for allt∈[0,1]and(x, y)∈D(C, R).

The main properties of the mappingg are enclosed in the following proposition.

Proposition 3.3. With the above assumptions on the mappingf one has:

(i) For all(x, y)∈D(C, R), the mapg(·,(x, y))is convex on[0,1].

(ii) For allt∈[0,1], the mapg(t,·)is convex onD(C, R).

Proof. (i) Lett₁, t₂ ∈ [0,1]andα, β ≥ 0withα+β = 1. By the convexity off we have

g(αt₁+βt₂,(x, y)) = 1 πR²

Z Z

D(C,R)

f(α[t₁(x, y) + (1−t₁) (z, u)]

+β[t₂(x, y) + (1−t₂) (z, u)])dzdu

≤α· 1 πR²

Z Z

D(C,R)

f(t1(x, y) + (1−t1) (z, u))dzdu +β· 1

πR² Z Z

D(C,R)

f(t₂(x, y) + (1−t₂) (z, u))dzdu

=αg(t1,(x, y)) +βg(t2,(x, y)), 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)andα, β ≥0withα+β = 1. Then g(t, α(x₁, y₁) +β(x₂, y₂))

= 1

πR² Z Z

D(C,R)

f[α(t(x₁, y₁) + (1−t) (z, u)) +β(t(x₂, y₂) + (1−t) (z, u))]dzdu

≤α 1 πR²

Z Z

D(C,R)

f(t(x₁, y₁) + (1−t) (z, u))dzdu

+β 1 πR²

Z Z

D(C,R)

f(t(x2, y2) + (1−t) (z, u))dzdu

=αg(t,(x₁, y₁)) +βg(t,(x₂, y₂)), for allt∈[0,1], and the statement is proved.

By the use of this mapping we can introduce the following application as well

G: [0,1]→R, G(t) := 1 πR²

Z Z

D(C,R)

g(t,(x, y))dxdy wheregis 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 alls∈ 0,¹₂

G

s+ 1 2

=G 1

2 −s

,

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and for allt∈[0,1]one has

G(1−t) = G(t). (ii) The mappingGis convex on the interval[0,1].

(iii) One has the bounds inf

t∈[0,1]G(t) =G 1

2

= 1

(πR²)²

Z Z Z Z

D(C,R)×D(C,R)

f

x+z

2 ,y+u 2

dxdydzdu

≥f(C) and

sup

t∈[0,1]

G(t) = G(0) =G(1) = 1 πR²

Z Z

D(C,R)

f(x, y)dxdy.

(iv) The mappingGis monotonic nonincreasing on 0,¹₂

and nondecreasing on₁

2,1 .

(v) We have the inequality

(3.6) G(t)≥max{H(t), H(1−t)}, for allt∈[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(t) +G(1−t)

2 ≥G

1 2

, for allt∈[0,1]

which proves the first bound in(iii).

Note that the inequality G

1 2

≥f(C)

follows by (3.6) fort = ¹₂ and taking into account thatH ¹₂

≥f(C).

We also have G(t) = 1

(πR²)² Z Z

D(C,R)

Z Z

D(C,R)

f(t(x, y) + (1−t) (z, u))dzdu

dxdy

≤ 1 (πR²)²

× Z Z

D(C,R)

tf(x, y)πR²+ (1−t) Z Z

D(C,R)

f(z, u)dzdu

dxdy

= 1

(πR²)²

×

tπR² Z Z

D(C,R)

f(x, y)dxdy+ (1−t)πR² Z Z

D(C,R)

f(x, y)dxdy

= 1

πR² Z Z

D(C,R)

f(x, y)dxdy

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for allt∈[0,1], and the second bound in(iii)is also proved.

(iv) The argument is similar to the proof of Theorem3.1(iii)(see also [6]) and we shall omit the details.

(v) By Theorem2.1we have that G(t) = 1

πR² Z Z

D(C,R)

g(t,(x, y))dxdy

≥g(t,(a, b)) = 1 πR²

Z Z

D(C,R)

f(t(x, y) + (1−t) (a, b))dxdy

=H(t) for allt∈[0,1].

AsG(t) = G(1−t)≥H(1−t), we obtain the desired inequality (3.6).

The theorem is thus proved.

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References

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[5] S.S. DRAGOMIR, Some integral inequalities for differentiable convex functions, Contributions, Macedonian Acad. of Sci. and Arts (Scopie), 16 (1992), 77-80.

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Balkanica, 6 (1992), 215-222. MR 934:26033.

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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. Kragujeva´c (Yugoslavia), 13 (1992), 11-16. ZBL No. 770:26009.

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