Journal of Inequalities in Pure and
Applied Mathematics
A VARIANT OF JENSEN’S INEQUALITY
A.McD. MERCER
Department of Mathematics and Statistics,
University of Guelph,
Guelph, Ontario N1G 2W1,
Canada.
EMail: amercer@reach.net
volume 4, issue 4, article 73,
2003.
Received 14 August, 2003;
accepted 27 August, 2003.
Communicated by: P.S. Bullen
Abstract
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Abstract
If f is a convex function the following variant of the classical Jensen’s Inequality
is proved
X
X
f x1 + xn − wk kk ≤ f(x1 ) + f(xn ) − wk f(xk ).
A Variant of Jensen’s Inequality
2000 Mathematics Subject Classification: 26D15.
Key words: Jensen’s inequality, Convex functions.
A.McD. Mercer
Contents
Title Page
1
Main Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2
The Proofs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3
Two Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
References
3
4
5
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1.
Main Theorem
Let 0 < x1 ≤ x2 ≤ · · · ≤ xn and let wk (1 ≤ k ≤ n) be positive weights
associated with these xk and whose sum is unity. Then Jensen’s inequality [2]
reads :
Theorem 1.1. If f is a convex function on an interval containing the xk then
X
X
(1.1)
f
wk xk ≤
wk f (xk ).
A Variant of Jensen’s Inequality
P
Pn
Note: Here and, in all that follows,
means 1 .
Our purpose in this note is to prove the following variant of (1.1).
Theorem 1.2. If f is a convex function on an interval containing the xk then
X
X
f x1 + xn −
wk xk ≤ f (x1 ) + f (xn ) −
wk f (xk ).
Towards proving this theorem we shall need the following lemma:
Lemma 1.3. For f convex we have:
(1.2)
f (x1 + xn − xk ) ≤ f (x1 ) + f (xn ) − f (xk ),
A.McD. Mercer
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(1 ≤ k ≤ n).
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2.
The Proofs
Proof of Lemma 1.3. Write yk = x1 + xn − xk . Then x1 + xn = xk + yk so that
the pairs x1 , xn and xk , yk possess the same mid-point. Since that is the case
there exists λ such that
xk = λx1 + (1 − λ)xn ,
yk = (1 − λ)x1 + λxn ,
where 0 ≤ λ ≤ 1 and 1 ≤ k ≤ n.
Hence, applying (1.1) twice we get
f (yk ) ≤ (1 − λ)f (x1 ) + λf (xn )
= f (x1 ) + f (xn ) − [λf (x1 ) + (1 − λ)f (xn )]
≤ f (x1 ) + f (xn ) − f (λx1 + (1 − λ)xn )
= f (x1 ) + f (xn ) − f (xk )
and since yk = x1 + xn − xk this concludes the proof of the lemma.
Proof of Theorem 1.2. We have
X
X
f (x1 + xn −
wk xk ) = f
wk (x1 + xn − xk )
X
≤
wk f (x1 + xn − xk )
by (1.1)
X
≤
wk [f (x1 ) + f (xn ) − f (xk )] by (1.2)
X
= f (x1 ) + f (xn ) −
wf (xk )
A Variant of Jensen’s Inequality
A.McD. Mercer
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J. Ineq. Pure and Appl. Math. 4(4) Art. 73, 2003
and this concludes the proof.
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3.
Two Examples
e = x1 + xn − A and G
e = x1 xn , where A and G denote the usual
Let us write A
G
arithmetic and geometric means of the xk .
(a) Then taking f (x) as the convex function − log x, Theorem 1.2 gives:
e≥G
e
A
(b) Taking f (x) as the function log 1−x
which is convex if 0 < x ≤ 12 , Theox
rem 1.2 gives
e
e
G(x)
A(x)
≥
e − x)
e − x)
A(1
G(1
provided that xk ∈ (0, 12 ] for all k.
The example (a) is a special case of a family of inequalities found by a different method in [1]. The example (b) is, of course, an analogue of Ky-Fan’s
Inequality [2].
A Variant of Jensen’s Inequality
A.McD. Mercer
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References
[1] A.McD. MERCER, A monotonicity property of power means, J. Ineq. Pure
and Appl. Math., 3(3) (2002), Article 40. [ONLINE: http://jipam.
vu.edu.au/v3n3/014_02.html].
[2] D.S. MITRINOVIĆ, J.E. PEČARIĆ AND A.M. FINK, Classical and New
Inequalities in Analysis, Kluwer Academic Publishers, Dordrecht, 1993.
A Variant of Jensen’s Inequality
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