http://jipam.vu.edu.au/
Volume 6, Issue 2, Article 36, 2005
ON A F. QI INTEGRAL INEQUALITY
ALFRED WITKOWSKI MIELCZARSKIEGO4/29 85-796 BYDGOSZCZ, POLAND
Received 16 August, 2004; accepted 06 February, 2005 Communicated by F. Qi
ABSTRACT. Necessary and sufficient conditions under which the Qi integral inequality Z b
a
ft(x) dx≥ Z b
a
f(x) dx t−1
or its reverse hold for allt≥1are given.
Key words and phrases: Convexity, Qi type integral inequality.
2000 Mathematics Subject Classification. 26D15.
1. INTRODUCTION
In [5] Feng Qi formulated the following problem: Characterize a positive function f such that the inequality
(1.1)
Z b
a
ft(x) dx≥ Z b
a
f(x) dx t−1
holds fort >1.
In [1, 2, 3, 4] and the references therein, several sufficient conditions and generalizations are given. In all the cited papers the authors look for the solution of the Qi inequality with restricted t. This paper is another contribution to this subject. We shall try to establish conditions under which the inequality holds for allt >1.
Let (X, µ) be a finite measure space and f be a positive measurable function. Define for t∈R
(1.2) H(t) = H(t, f) = ln
Z
ftdµ−(t−1) ln Z
fdµ
.
It is clear that inequality (1.1) is equivalent to H(t) ≥ 0 for t > 1. We will say that for the functionf the Qi Inequality (QI) holds ifH(t, f)is nonnegative for allt≥1. We will also say
ISSN (electronic): 1443-5756
c 2005 Victoria University. All rights reserved.
152-04
that for the functionf the Reverse Qi Inequality (RQI) holds ifH(t, f)is non-positive for all t≥1.
By the Cauchy-Schwarz integral inequality, we have forp, q ∈R (1.3)
Z
fp+q2 dµ 2
≤ Z
fpdµ Z
fqdµ
which means that the functionH(t)is convex, that is
(1.4) H
t1+t2 2
≤ H(t1) +H(t2) 2 holds fort1, t2 ∈R, so its derivative
(1.5) H0(t) =
R ftlnfdµ R ftdµ −ln
Z fdµ
is increasing int∈R. Let
M = ess supx∈Xf(x) and µM =µ({x:f(x) =M}).
The following lemmas will be useful.
Note that from now on we will use the convention thatln∞=∞andln 0 =−∞.
Lemma 1.1. The following formula holds:
(1.6) lim
t→∞
H(t)
t = ln M R fdµ. Proof. Forε >0letmε=µ(x:f(x)> M −ε). Then
(M −ε)tmε ≤ Z
ftdµ≤Mtµ(X), so
tln(M −ε) + lnmε
≤H(t) + (t−1) ln Z
fdµ
≤tlnM + lnµ(X).
(1.7)
Dividing byton both sides of (1.7) yields lnM −ε
R fdµ ≤lim inf
t→∞
H(t)
t ≤lim sup
t→∞
H(t)
t ≤ln M R fdµ.
In caseM =∞,M −εstands for an arbitrary large number. This completes the proof.
Lemma 1.2. IfM <∞then
(1.8) lim
t→∞
H(t)−tln M R fdµ
= ln
µM Z
fdµ
.
Proof. Direct computation yields
t→∞lim
H(t)−tln M R fdµ
= lim
t→∞ln Z
f M
t
dµ+ ln Z
fdµ
= ln
µM Z
fdµ (1.9)
as(f /M)ttends monotonically to the characteristic function of{x:f(x) = M}.
2. FENGQIINTEGRAL INEQUALITY
In this section we consider the problem: Characterize positive functionsf that satisfy (QI).
Theorem 2.1. A constant functionM satisfies (QI) if and only ifµ(X)≤1andM ≥1/µ(X).
Proof. H(t) ≥ 0 is equivalent to M ≥ µ(X)t−2. This can be valid for allt > 1 only if the
conditions of the theorem are fulfilled.
From now on we assume that f is not constant, in which case the function H is strictly convex.
It is clear that the necessary condition for (QI) isH(1) ≥0or equivalentlyR
fdµ≥1.
Theorem 2.2. (a) IfH(1) = 0then (QI) holds if and only ifH0(1)≥0.
(b) IfH(1)>0then
(b1) ifH0(1) ≥0then (QI) holds;
(b2) ifH0(1) <0andM <R
fdµthen (QI) fails for larget;
(b3) ifH0(1) <0andM =R
fdµthen (QI) holds if and only ifµMM ≥1;
(b4) if H0(1) < 0 and M > R
fdµ then there exists an unique point t0 such that H0(t0) = 0and (QI) holds if and only ifH(t0)≥0.
Proof. (a) and (b1) follow immediately from convexity ofH.
From Lemma 1.1 we see thatHbecomes negative for larget, which proves (b2).
(b3) follows from Lemma 1.2 and from the fact that being convex the graph ofHlies above its horizontal asymptote.
Finally (b4) follows from the fact that H0 is strictly increasing andH0(t0) = 0for somet0, thenHattains its minimum att0. Observe that in this caseHmay be infinite for some finitet∞
and consequently for allt > t∞.
From the above theorem we obtain the following, surprising Corollary 2.3. Ifµ(X)<1then (QI) holds if and only ifH(1) ≥0.
Proof. We will show that ifµ(X)<1thenH0(1)≥0for allf, so the condition (b1) is satisfied.
Applying the integral Jensen Inequality to the convex functionxlnxwe obtain 1
µ(X) Z
flnfdµ≥ 1
µ(X) Z
fdµ
ln 1
µ(X) Z
fdµ
≥ 1 µ(X)
Z fdµ
ln
Z fdµ
which is equivalent toH0(1)≥0.
In the case (b4), solving the equation H0(t) = 0may not be an easy task, but the following corollaries may be helpful:
Corollary 2.4. Let
tL=
R flnfdµ−R
fdµlnR fdµ R flnfdµ . IfH0(tL)≥0then (QI) holds.
Proof. tLis the point where the supporting line drawn att = 1meets the OX-axis. The graph of H lies above it. In particularH(tL) ≥ 0. As H0(t) is nonnegative fort ≥ tLthe proof is
completed.
Corollary 2.5. If0< µM, M <∞let
tR =−ln(µMR fdµ) ln(M/R
fdµ). IfH0(tR)≤0ortR≤tLthen (QI) holds.
Proof. tRis the point where the supporting line drawn at∞(it exists by Lemma 1.2) meets the OX-axis. IftR≤tLthe two supporting lines meet above the OX-axis.
IfH0(tR)≤0we use an argument similar to that in the proof of the previous corollary.
3. REVERSEDFENGQIINEQUALITY
In this section we give sufficient and necessary conditions for the reversed problem: Charac- terize positive functionsf that satisfy (RQI).
Theorem 3.1. A constant function satisfies (RQI) if and only ifµ(X)≥1andM ≤1/µ(X).
The proof is similar to that of Theorem 2.1.
Theorem 3.2. For a non constant function f (RQI) holds if and only ifH(1) ≤ 0 andM ≤ R fdµ.
Proof. AsµMM < R
fdµ = exp(H(1)) ≤ 1 it follows from Lemma 1.1 and 1.2 that H is negative for larget. Being convex and non-positive att= 1,it must be decreasing.
On the other hand ifM >R
fdµthenHis positive for largetby Lemma 1.1.
4. FINAL REMARK
Finally we prove the following
Theorem 4.1. For every positive functionf there exists a constantc >0such thatcf satisfies (QI) or (RQI).
Proof. One can easily see that
H(t, cf) = lnc+H(t, f),
so cf satisfies (QI) for certain cif and only if H(t, f) is bounded from below. Similarly cf satisfies (RQI) only ifH(t, f)is bounded from above.
It follows immediately from Lemma 1.1 and Lemma 1.2 that the functionH(t)is bounded from below if and only ifM > R
fdµorM = R
fdµandµM > 0and is bounded from above if and only ifM ≤R
fdµ.
This completes the proof of our theorem.
Note that in caseM = R
fdµandµM > 0we can find constantsc1 and c2 such that (QI) holds forc1f and (RQI) holds forc2f.
REFERENCES
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sid=269].
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