Journal of Inequalities in Pure and
Applied Mathematics
http://jipam.vu.edu.au/
Volume 4, Issue 4, Article 73, 2003
A VARIANT OF JENSEN’S INEQUALITY
A.McD. MERCER
D EPARTMENT OF M ATHEMATICS AND S TATISTICS ,
U NIVERSITY OF G UELPH ,
G UELPH , O NTARIO N1G 2W1,
C ANADA .
amercer@reach.net
Received 14 August, 2003; accepted 27 August, 2003
Communicated by P.S. Bullen
A BSTRACT. 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 ).
Key words and phrases: Jensen’s inequality, Convex functions.
2000 Mathematics Subject Classification. 26D15.
1. M AIN T HEOREM
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 ).
P
P
Note: Here and, in all that follows,
means n1 .
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 ),
ISSN (electronic): 1443-5756
c 2003 Victoria University. All rights reserved.
116-03
(1 ≤ k ≤ n).
2
A.M C D. M ERCER
2. T HE P ROOFS
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 )
and this concludes the proof.
3. T WO E XAMPLES
e = x1 + xn − A and G
e = x1 xn , where A and G denote the usual arithmetic and
Let us write A
G
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 , Theorem 1.2 gives
x
e
A(x)
e − x)
A(1
≥
e
G(x)
e − x)
G(1
provided that xk ∈ (0, 21 ] 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].
R EFERENCES
[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.
J. Inequal. Pure and Appl. Math., 4(4) Art. 73, 2003
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