Journal of Inequalities in Pure and
Applied Mathematics
A REFINEMENT OF JENSEN’S INEQUALITY
J. ROOIN
Department of Mathematics
Institute for Advanced Studies in Basic Sciences
Zanjan, Iran
volume 6, issue 2, article 38,
2005.
Received 27 August, 2004;
accepted 16 March, 2005.
Communicated by: C.E.M. Pearce
EMail: rooin@iasbs.ac.ir
Abstract
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2000
Victoria University
ISSN (electronic): 1443-5756
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Abstract
We refine Jensen’s inequality as
Z Z
Z
Z
f(x)ω(x, y)dµ(x) dλ(y) ≤ (ϕ ◦ f)dµ,
ϕ
fdµ ≤ ϕ
X
Y
X
X
where (X, A, µ) and (Y, B, λ) are two probability measure spaces, ω : X ×
Y → [0, ∞) is a weight function on X × Y , I is an interval of the real line,
f ∈ L1 (µ), f(x) ∈ I for all x ∈ X and ϕ is a real-valued convex function on I.
A Refinement of Jensen’s
Inequality
J. Rooin
2000 Mathematics Subject Classification: Primary: 26D15, 28A35.
Key words: Product measure, Fubini’s Theorem, Jensen’s inequality.
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Contents
1
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2
Refinement . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
References
Contents
3
4
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1.
Introduction
The classical integral form of Jensen’s inequality states that
Z
Z
(1.1)
ϕ
f dµ ≤
(ϕ ◦ f )dµ,
X
X
where (X, A, µ) is a probability measure space, I is an interval of the real line,
f ∈ L1 (µ), f (x) ∈ I for all x ∈ X and ϕ is a real-valued convex function on I;
see e.g. [2, p. 202] or [4, p. 62]. Now suppose that (X, A, µ) and (Y, B, λ) are
two probability measure spaces. By a (separately) weight function on X × Y
we mean a product- measurable mapping ω : X × Y → [0, ∞), see e.g. [4, p.
160], such that
Z
(1.2)
ω(x, y)dµ(x) = 1
(for each y in Y ),
X
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and
Z
ω(x, y)dλ(y) = 1
(1.3)
A Refinement of Jensen’s
Inequality
(for each x in X).
Y
For example, if we take X and Y as the unit interval [0, 1] with Lebesgue measure, then ω(x, y) = 1+(sin 2πx)(sin 2πy) is a weight function on [0, 1]×[0, 1].
In this paper, using a weight function ω, we refine Jensen’s inequality (1.1)
as in the following section. For some applications in the discrete case, see e.g.
[3].
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2.
Refinement
In this section, using the terminologies of the introduction, we refine the integral
form of Jensen’s inequality (1.1) via a weight function ω.
Theorem 2.1. Let (X, A, µ) and (Y, B, λ) be two probability measure spaces
and ω : X × Y → [0, ∞) be a weight function on X × Y . If I is an interval
of the real line, f ∈ L1 (µ), f (x) ∈ I for all x ∈ X, and ϕ is a real convex
function on I, then
Z
Z
ϕ
f (x)ω(x, y)dµ(x) dλ(y)
Y
X
J. Rooin
has meaning and we have
Z
Z
Z
Z
(2.1) ϕ
f dµ ≤
ϕ
f (x)ω(x, y)dµ(x) dλ(y) ≤
(ϕ ◦ f )dµ.
X
A Refinement of Jensen’s
Inequality
Y
X
X
Proof. The functions ω and (x, y) → f (x), and so
(x, y) → f (x)ω(x, y)
is product-measurable on X × Y . Now since
Z Z
(2.2)
|f (x)|ω(x, y)dλ(y)dµ(x)
X Y
Z
Z
=
|f (x)|
ω(x, y)dλ(y) dµ(x)
Y
ZX
=
|f (x)|dµ(x) = kf kL1 (µ) < ∞,
X
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by Fubini’s theorem, the real-valued function (x, y) → f (x)ω(x, y) on X × Y
belongs to L1 (µ × λ). Therefore for λ-almost all y ∈ Y , the function x →
f (x)ω(x, y) belongs to L1 (µ). Fix an arbitrary α ∈ I. Define F : Y → R, by
Z
f (x)ω(x, y)dµ(x)
F (y) =
X
if the integral exists, and F (y) = α otherwise. By Fubini’s theorem, we have
F ∈ L1 (λ). It is easy to show that F (y) ∈ I (y ∈ Y ). So,
Z
Z
Z
f (x)ω(x, y)dµ(x) dλ(y) := (ϕ ◦ F )(y)dλ(y)
ϕ
Y
X
A Refinement of Jensen’s
Inequality
J. Rooin
Y
has meaning and is an extended real number belonging to (−∞, +∞]; see e.g.
[4, p. 62]. Now, since (x, y) → f (x)ω(x, y) belongs to L1 (µ × λ), by (1.1) and
Fubini’s theorem, we have
Z
Z
Z
ϕ
f (x)ω(x, y)dµ(x) dλ(y) = (ϕ ◦ F )(y)dλ(y)
Y
X
Y
Z
≥ϕ
F (y)dλ(y)
Y
Z Z
=ϕ
f (x)ω(x, y)dµ(x)dλ(y)
Y X
Z
Z
=ϕ
f (x)
ω(x, y)dλ(y) dµ(x)
X
Y
Z
=ϕ
f dµ ,
X
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and the left-hand side inequality (2.1) is obtained.
R
For the right-hand side inequality in (2.1), we consider two cases: If X (ϕ ◦
f )dµ = +∞, the assertion is trivial. Suppose then, ϕ ◦ f ∈ L1 (µ). Take an
arbitrary y ∈ Y such that x → f (x)ω(x, y) belongs to L1 (µ), and put
dν y = ω y dµ,
where
ω y (x) = ω(x, y)
(x ∈ X).
y
1
y
Trivially, (X, A, ν ) is a probability measure space, f ∈ L (ν ) and
Z
Z
F (y) =
f (x)ω(x, y)dµ(x) =
f (x)dν y (x).
X
X
X
Since ϕ ◦ f ∈ L1 (µ),
Z Z
|(ϕ ◦ f )(x)|ω(x, y)dλ(y)dµ(x)
X Y
Z
Z
(2.4)
=
|(ϕ ◦ f )(x)|dµ(x) ω(x, y)dλ(y)
Y
ZX
=
|(ϕ ◦ f )(x)|dµ(x) < ∞,
X
J. Rooin
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Thus, by Jensen’s inequality (1.1), we have
Z
Z
y
(2.3)
(ϕ ◦ F )(y) = ϕ
f (x)dν (x) ≤
(ϕ ◦ f )dν y .
X
A Refinement of Jensen’s
Inequality
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and so for λ-almost all y ∈ Y , the function x → (ϕ ◦ f )(x)ω(x, y) belongs to
L1 (µ) and for these y’s, we have
Z
Z
(2.5)
(ϕ ◦ f )(x)ω(x, y)dµ(x) =
(ϕ ◦ f )(x)dν y (x).
X
X
Thus, by (2.3) and (2.5), for λ-almost all y ∈ Y
Z
(2.6)
(ϕ ◦ F )(y) ≤
(ϕ ◦ f )(x)ω(x, y)dµ(x).
X
Denote temporarily the right-hand side of (2.6) by ψ(y) (put ψ(y) = 0, if the
integral does not exist).R Since by (2.4), ψR ∈ L1 (λ), from (ϕ ◦ F )+ ≤ ψ + (λ+
a.e.), we conclude that Y (ϕ ◦ F )+ dλ
R ≤ Y ψ −dλ < ∞.
On the other hand, we know that Y (ϕ ◦ F ) dλ < ∞. Thus ϕ ◦ F ∈ L1 (λ),
and so by (2.6), (2.4) and Fubini’s theorem,
Z
Z
Z
ϕ
f (x)ω(x, y)dµ(x) dλ(y) = (ϕ ◦ F )(y)dλ(y)
Y
X
ZY
≤
ψ(y)dλ(y)
Y
Z Z
=
(ϕ ◦ f )(x)ω(x, y)dµ(x)dλ(y)
ZY X
Z
=
(ϕ ◦ f )(x)dµ(x) ω(x, y)dλ(y)
Y
ZX
=
(ϕ ◦ f )dµ.
X
This completes the proof.
A Refinement of Jensen’s
Inequality
J. Rooin
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Corollary 2.2. If ϕ is a real convex function on a closed interval [a, b], then we
have Hermite-Hadamard inequalities [1]:
Z b
a+b
1
ϕ(a) + ϕ(b)
(2.7)
ϕ
≤
ϕ(t)dt ≤
.
2
b−a a
2
Proof. Put X = {0, 1} with A = 2X and µ{0} = µ{1} = 12 , and Y = [0, 1]
with Lebesgue measure λ. Now, (2.7) follows from (2.1) by taking ω(0, y) =
2(1 − y), ω(1, y) = 2y (0 ≤ y ≤ 1), I = [a, b], f (0) = a, f (1) = b, and
considering the change of variables t = (1 − y)a + yb.
We conclude this paper by the following open problem:
Open problem. Characterize all weight functions. Actually, if ω(x, y) is a
weight function, then θ(x, y) = ω(x, y) − 1 satisfies the following relations:
A Refinement of Jensen’s
Inequality
J. Rooin
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Z
θ(x, y)dµ(x) = 0
(2.8)
(for each y in Y ),
X
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Z
θ(x, y)dλ(y) = 0
(2.9)
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(for each x in X).
Y
So precisely, the weight functions are of the form 1 + θ(x, y) with nonnegative values such that θ(x, y) is product-measurable and satisfies (2.8) and (2.9).
Therefore, it is sufficient only to characterize these θ’s.
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References
[1] S.S. DRAGOMIR AND C.E.M. PEARCE, Selected Topics on HermiteHadamard Inequalities and Applications, RGMIA Monographs, Victoria University, 2000. [ONLINE: http://rgmia.vu.edu.au/
monographs/]
[2] E. HEWITT AND K. STROMBERG, Real and Abstract Analysis, SpringerVerlag, New York, 1965.
[3] J. ROOIN, Some aspects of convex functions and their applications, J. Ineq.
Pure and Appl. Math., 2(1) (2001), Art. 4. [ONLINE http://jipam.
vu.edu.au/article.php?sid=120]
[4] W. RUDIN, Real and Complex Analysis, 3rd ed., McGraw-Hill, New York,
1974.
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Inequality
J. Rooin
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