Journal of Inequalities in Pure and
Applied Mathematics
ANDERSSON’S INEQUALITY AND BEST POSSIBLE
INEQUALITIES
volume 4, issue 3, article 54,
2003.
A.M. FINK
Mathematics Department
Iowa State University
Ames, IA 50011
E-Mail: fink@math.iastate.edu
Received 08 October, 2002;
accepted 01 May, 2003.
Communicated by: D. Hinton
Abstract
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2000
Victoria University
ISSN (electronic): 1443-5756
106-02
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Andersson [1] proved that if for each i, fi (0) = 0 and fi is convex and
increasing, then
n
1Y
Z
(1)
0
1
n
2n Y
fi (x)dx ≥
n+1 1
Z
1
fi (x)dx
0
with equality when each fi is linear.
Elsewhere [2] we have proved that if fi ∈ M = {f |f (0) = 0 and
increasing and bounded} and
Z t
Z
c = dσ xdσ(x) ≥ 0,
dσ ∈ M
0
f (x)
x
is
1
A.M. Fink
xdσ(x) ≥ 0
t
Z
for t ∈ [0, 1], and
1
xdσ(x) > 0
0
then
Z
(2)
0
n
1Y
1
R1
fi (x)dσ(x) ≥
Andersson’s Inequality and
Best Possible Inequalities
R0
1
0
n Z
xn dσ(x) Y 1
n
fi (x)dσ(x).
0
xdσ(x)
1
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One notices that if f is convex and increasing with f (0) = 0 then f ∈ M .
R1 0
For f (x)
=
f (xt)dt when f 0 exists. The question arises if in fact Andersson’s
x
0
inequality can be extended beyond (2).
Lemma 1 (Andersson). If fi (0) = R0, increasing
i =
R 1 ∗ and Rconvex,
R 1 1, 2 ∗and
1
1
∗
f2 = α2 x where α2 is chosen so that 0 f2 = 0 f2 then 0 f1 f2 ≥ 0 f1 f2 .
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J. Ineq. Pure and Appl. Math. 4(3) Art. 54, 2003
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We will examine whether Andersson’s Lemma is best possible. We now
discuss the notion of best possible.
An (integral) inequality I(f, dµ) ≥ 0 is best possible if the following situation holds. We consider both the functions and measures as ‘variables’. Let the
functions be in some universe U usually consisting of continuous functions and
b , usually regular Borel measures. Suppose we
the measures in some universe U
c⊂U
b so that I(f, dµ) ≥ 0 for all f ∈ M if and only
can find M ⊂ U and M
c (given that µ ∈ U
b ) and I(f, dµ) ≥ 0 for all µ ∈ M
c if and only if
if µ ∈ M
c) give us a best possible
f ∈ M (given that f ∈ U ). We then say the pair (M, M
inequality.
As an historical example, Chebyshev [3] in 1882 submitted a paper in which
he proved that
Z b
Z b
Z b
Z b
(3)
f (x)g(x)p(x)dx
p(x)dx ≥
f (x)p(x)dx
g(x)p(x)dx
a
a
a
a
provided that p ≥ 0 and f and g were monotone in the same sense. Even before
this paper appeared in 1883, it was shown to be not best possible since the pairs
f, g for which (3) holds can be expanded. Consider the identity
Z Z
1 b b
(4)
(f (x) − f (y)][g(x) − g(y)]p(x)p(y)dxdy
2 a a
Z b
Z b
Z b Z b
=
f gp
p−
fp
gp.
a
a
a
a
Andersson’s Inequality and
Best Possible Inequalities
A.M. Fink
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So (3) holds if f and g are similarly ordered, i.e.
J. Ineq. Pure and Appl. Math. 4(3) Art. 54, 2003
(5)
[f (x) − f (y)][g(x) − g(y)] ≥ 0, x, y ∈ [a, b].
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For example x2 and x4 are similarly ordered but not monotone.
Jodeit and Fink [4] invented the notion of ‘best possible’ in a manuscript
circulated in 1975 and published in parts in [3] and [4]. They showed that if we
b to be regular Borel measures
take U Rto be pairs of continuous functions and U
b
µ with a dµ > 0, then
Z b
Z b
Z b
Z b
(6)
f g dµ
dµ ≥
f dµ
g dµ
a
a
a
a
c1 = {µ|µ ≥
is a best possible inequality if M1 = {(f, g)| (5) holds} ⊂ U and M
0} i.e.
c1 , and
(6) holds for all pairs in M1 if and only if µ ∈ M
c1 if and only if (f, g) ∈ M1 .
(6) holds for all µ ∈ M
The sufficiency in both cases is the identity corresponding to (4). If dµ =
δx + δy where x and y ∈ [a, b], the inequality (6) gives (5), and if f = g =
xA , A ⊂ [a, b], then (6) is µ(A)µ(a, b) ≥ µ(A)2 which gives µ(A) ≥ 0. Strictly
speaking this pair is not in M1 , but can be approximated in L1 by continous
functions.
If we return to Chebyshev’s hypothesis that f and g are monotone in the
same sense, let us take U be the class of pairs of continuous functions, neither
b as above, M0 = {f, g ∈ U | f and g are simularly
of which is a constant and U
monotone} and
Z t
Z b
c
M0 = µ dµ ≥ 0,
dµ ≥ 0 for a ≤ t ≤ b .
a
Andersson’s Inequality and
Best Possible Inequalities
A.M. Fink
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t
J. Ineq. Pure and Appl. Math. 4(3) Art. 54, 2003
c0 .
Lemma 2. The inequality (6) holds for all (f, g) ∈ M0 if and only if µ ∈ M
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Rx
Proof.
There exist measures dτ and dλ such that f (x) = 0 dτ and g(x) =
Rx
dλ. We may assume f (0) = g(0) since adding a constant to a function does
0
not alter (6). Letting x0+ = 0 if x ≤ 0 and 1 if x > 0 we can rewrite (6) after an
interchange of order of integration as
Z
1
Z
1
Z
dλ(s)dτ (t)
(7)
0
1
1
Z
(x − t)0+ (x − s)0+ dµ(x)
0
Z 1
Z 1
0
0
−
(x − t)+ dµ(x)
(x − s)+ dµ(x) ≥ 0.
dµ
0
0
0
0
Since f, g are arbitrary increasing functions, dλ and dτ ≥ 0 so (6) holds if and
only if the [ ] ≥ 0 for each t and s. For example we may take both these
measures, dτ, dλ to be point atoms. The equivalent condition then is that
Z 1 Z 1
Z 1 Z 1
(8)
dµ
dµ ≥
dµ
dµ.
0
t∨s
t
s
Rs
R1
By symmetry we may assume that t ≥ s so that (8) may be written 0 dµ t dµ
c0 (6) holds and (6) holds for all f, g ∈ M0 only if
≥
if dµ ∈ M
R s 0. Consequently,
R1
dµ t dµ ≥ 0. But for s = t this is the product of two numbers whose sum
0
is positive so each factor must be non-negative, completing the proof.
Lemma 3. Suppose f and g are bounded integrable functions on [0, 1]. If (6)
c0 then f and g are both monotone in the same sense.
holds for all µ ∈ M
Proof. First let dµ = δx + δy where δx is an atom at x. Then (6) becomes
[f (x) − f (y)][g(x) − g(y)] ≥ 0, i.e. f and g are similarly ordered. If x < y < z,
Andersson’s Inequality and
Best Possible Inequalities
A.M. Fink
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J. Ineq. Pure and Appl. Math. 4(3) Art. 54, 2003
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take dτ = δx − δy + δz so that µ ∈ M0 . To ease the burden of notation let the
values of f at x, y, z be a, b, c and the corresponding values of g be A, B, C. By
(6) we have
(9)
aA − bB + cC ≥ (a − b + c)(A − B + C).
By similar ordering we have
(10) (a − b)(A − B) ≥ 0, (a − c)(A − C) ≥ 0, and (b − c)(B − C) ≥ 0;
and (9) may be rewritten as
(11)
A.M. Fink
(a − b)(C − B) + (c − b)(A − B) ≤ 0.
Now if one of the two terms in (10) is positive, the other is negative and all the
factors are non-zero. By (10) the two terms are the same sign. Thus
(12)
Andersson’s Inequality and
Best Possible Inequalities
(a − b)(C − B) ≤ 0 and (c − b)(A − B) ≤ 0.
Now (10) and (12) hold for any triple. We will show that if f is not monotone, then g is a constant.
We say that we have configuration I if a < b and c < b, and configuration II
if a > b and c > b.
We claim that for both configurations I and II we must have A = B = C.
Take configuration I. Now b − a > 0 implies that B − A ≥ 0 by (10) and
C − B ≥ 0 by (12). Also b − c > 0 yields (B − C) ≥ 0 by (10) and A − B ≥ 0
by (12). Combining these we have A = B = C. The proof for configuration II
is the same.
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Assume now that configuration I exists, so A = B = C. Let x < x0 < y. If
a0 < b (a0 = f (x0 )) then x0 , y, z form a configuration I and A0 = B. If a0 ≥ b,
then x, x0 , z form a configuration I and A0 = B. If x0 < x and a0 < b, then
again x0 , y, z form a configuration I and A0 = B. Finally if a0 ≥ b and x0 < x
then x0 , x, b for a configuration II and A0 = B. Thus for x < y g(0) ≡ g(y).
The proof for x > y is similar yielding that g is a constant.
If a configuration II exists, then the proof is similar, or alternately we can
apply the configuration I argument to the pair −f, −g.
Finally if f is not monotone on [0, 1] then either a configuration I or II must
exist and g is a constant. Consequently, if neither f nor g are constants, then
both are monotone and by similar ordering, monotone in the same sense.
Note that if one of f, g is a constant, then (6) is an identity for any measure.
Theorem 4.
Andersson’s Inequality and
Best Possible Inequalities
A.M. Fink
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i) Let M be defined as above and N = {g|g(0) = 0 and g is increasing and
bounded}. Then for F (x) ≡ f (x)
x
Z 1
(13)
f gdσ(x)
0
Z
≥
0
1
−1 Z
xdσ(x)
0
1
Z
F (x)xdσ
1
g(x)xdσ(x)
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0
c.
holds for all pairs (f, g) ∈ M × N if and only if dσ ∈ M
ii) Let f (0) = g(0) = 0 and fx and g be of bounded variation on [0, 1]. If (13)
c then either f or g is a constant (in which case (13)
holds for all dσ ∈ M
x
is an identity) or fx , g ∈ M × N .
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The proof starts with the observation that (13) is in fact a Chebyshev inequality
Z
1
Z
0
Z
dτ ≥
F g dτ
(14)
1
0
1
Z
F dτ
0
1
g dτ
0
where dτ = x dσ; and F, g are the functions. The theorem is a corollary of the
two lemmas.
Andersson’s inequality (2) now follows by induction, replacing one f by f ∗
at a time. Note that the case n = 2 of Andersson’s inequality (2) has the proof
Z 1
Z 1
Z 1
∗
f1 f2 ≥
f1 f2 ≥
f1∗ f2∗
0
0
Andersson’s Inequality and
Best Possible Inequalities
A.M. Fink
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and it is only the first one which is best possible! The inequality between the
extremes is perhaps ‘best possible’.
Remark 1. Of course x can be replaced by any function that is zero at zero
(x)
and positive elsewhere, i.e. f (x)
can be replaced by fp(x)
and the measure dτ =
x
p(x)dσ(x).
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References
[1] B.J. ANDERSSON, An inequality for convex functions, Nordisk Mat.
Tidsk, 6 (1958), 25–26.
[2] A.M. FINK, Andersson’s inequality, Math Ineq. and Applic., to appear.
[3] P.L. C̆EBYŠEV, O priblizennyh vyraz̆enijah odnih integralov c̆erez drugie.
Soobs̆c̆enija i Protokoly Zasedamiı̆ Matematis̆eskogo Obs̆estva pri Imperatorskom Har’kovskom Universite, 2 (1882), 93–98, Polnoe Sobranie
Soc̆inenii P.L. C̆ebys̆eva. Moskva, Leningrad 1948, pp. 128–131.
[4] A.M. FINK AND M. JODEIT, Jr., On Čhebyšhev’s other inequality, 1984.
Inequalities in Statistics and Probability (Lecture Notes IMS No. 5) Inst.
Math. Statist. Hayward Calif., 115–129.
[5] A.M. FINK, Toward a theory of best possible inequalities, Nieuw Archief
Voor Wiskunde, 12 (1994), 19–29.
Andersson’s Inequality and
Best Possible Inequalities
A.M. Fink
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