Integral inequalities Constantin P. Niculescu
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Transcript of Integral inequalities Constantin P. Niculescu
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Integral inequalities
Constantin P. Niculescu
Basic remark: Iff : [a; b]! R is (Riemann) integrableand nonnegative, thenZ b
af(t)dt0:
Equality occurs if and only iff = 0 almost everywhere
(a.e.)
When f is continuous, f = 0 a.e. if and only iff = 0
everywhere.
Important Consequence: Monotony of integral,
fg implies Z ba
f(t)dt Z ba
g(t)dt:
Lecture presented on December 16, 2008, at the Abdus SalamSchool of Mathematical Sciences, Lahore.
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In Probability Theory, integrable functions are random
variables. Most important inequalities refer to:
M(f) = 1
b aZ b
af(t)dt (mean value off)
V ar(f) =M
(f M(f))2
(variance off)
=
1
b a Z b
a f
2
(x)dx 1
b a Z b
a f(x) dx!2
:
Theorem 1 Chebyshevs inequality: Iff; g : [a; b]! Rhave the same monotony, then
1
b a Z b
a f(t)g(t)dt 1
b a Z b
a f(t)dt! 1
b a Z b
a g(t)d
iff; ghave opposite monotony, then the inequality should
be reversed.
Application: Let f : [a; b]! R be a dierentiable func-tion having bounded derivative. Then
V ar(f) (b a)2
12 sup
axb
f0(x)2 :
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Theorem 2 (The Mean Value Theorem). Letf : [a; b]!R be a continuous function andg: [a; b]! R be a non-negative integrable function. Then there is c2 [a; b]such thatZ b
af(x)g(x) dx=f(c)
Z ba
g(x) dx:
Theorem 3 (Boundedness). If f : [a; b]! R is inte-grable, then f is bounded,jfj is integrable and
1
b aZ b
af(t)dt
1b a
Z ba
jf(t)j dt
supatb jf(t)j :
Remark 4 Iff0 is integrable, then
0 1
b a Z b
ajf(x)j dx
1
b a Z b
af(x)dx
b a3
supaxb
f0(x) :
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Remark 5 Suppose thatfis continuously dierentiable
on [a; b] andf(a) =f(b) = 0: Then
supatb
jf(t)j b a2
Z ba
f0(t) dt:
Theorem 6 (Cauchy-Schwarz inequality).
Iff; g: [a; b]! R are integrable, then
Z ba
f(t)g(t)dt
Z b
af2(t)dt
!1=2 Z ba
g2(t)dt
!1=2
with equality if andg are proportional a.e.
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Special Inequalities
Youngs inequality. Let f : [0; a] ! [0; f(a)] be astrictly increasing continuous function such that f(0) =
0:Using the denition of derivative show that
F(x) = Z x0 f(t) dt + Z f(x)
0 f1(t) dt xf(x)
is dierentiable on [0; a] and F0(x) = 0 for all x2[a; b]: Find from here that
xyZ x
0f(t) dt +
Z y0
f1(t) dt:
for all 0xa and 0yf(a).
Figure 1: The geometric meaning of Youngs inequality.
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Special case (corresponding for f(x) =xp1 and f1(x) =xq1) : For all a; b 0p; q2 (1; 1) and 1=p+1=q= 1; then
ab ap
p +
bq
q if p; q2(1; 1) and 1
p+
1
q = 1;
ab
ap
p
+bq
q
if p
2(
1; 1)
nf0
gand
1
p
+1
q
= 1
The equality holds if (and only if) ap =bq:
Theorems 7 and 8 below refer to arbitrary measure spaces
(X; ; ):
Theorem 7 (The Rogers-Hlder inequality for p > 1).
Let p; q2 (1; 1) with 1=p+ 1=q = 1; and let f2Lp() and g 2 Lq(): Then f g is in L1() and wehave
ZX f gd ZX jf gj d (1)and Z
Xjf gj d kfkLp kgkLq: (2)
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Thus ZX
f g d kfkLp kgkLq: (3)
The above result extends in a straightforward manner
for the pairs p = 1; q = 1 and p = 1; q = 1:In the complementary domain, p2 (1; 1)nf0g and1=p + 1=q= 1; the inequality sign should be reversed.
Forp=q = 2;we retrieve theCauchy-Schwarz inequal-
ity.
Proof. If f or g is zero -almost everywhere, then the
second inequality is trivial. Otherwise, using the Younginequality, we have
jf(x)jkfkLp
jg(x)jkgkLq 1
pjf(x)j
p
kfkpLp+
1
qjg(x)j
q
kgkqLqfor all x in X, such that f g2L1(). Thus
1
kfkLp kgkLqZ
Xjf gj d1
and this proves (2). The inequality (3) is immediate.
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Remark 8 (Conditions for equality). The basic observa-
tion is the fact that
f 0 andZ
Xf d= 0 implyf = 0 -almost everywhere.
Consequently we have equality in (1) if, and only if,
f(x)g(x) =ei jf(x)g(x)jfor some real constant and for -almost every x:
Suppose that p; q2 (1; 1): In order to get equality in(2) it is necessary and sucient to have
jf(x)jkfkLp
jg(x)jkgkLq=
1
pjf(x)j
p
kfkpLp+
1
qjg(x)j
q
kgkqLqalmost everywhere. The equality case in Youngs inequal-
ity shows that this is equivalent tojf(x)jp = kfkpLp =
jg(x)
jq =
kg
kqLq almost everywhere, that is,
A jf(x)jp =B jg(x)jq almost everywherefor some nonnegative numbers A and B.
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Ifp= 1andq =1;we have equality in (2) if, and onlyif, there is a constant0such thatjg(x)j almosteverywhere, andjg(x)j=for almost every point wheref(x)6= 0:
Theorem 9 (Minkowskis inequality). For 1 p
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According to the Rogers-Holder inequality,
jjf+ gjjpLp =Z
Xjf+ gjp d
Z
Xjf+ gjp1 jfj d +
ZX
jf+ gjp1 jgj d
Z
Xjfjp d
1=p ZX
jf+ gj(p1)q d1=q
+
+Z
Xjgjp d1=p Z
Xjf+ gj(p1)q d1=q
= (jjfjjLp+ jjgjjLp) jjf+ gjjp=qLp ;
where 1=p+ 1=q = 1; and it remains to observe thatp p=q= 1:
Remark 10 Ifp = 1; we obtain equality in (4) if, andonly if, there is a positive measurable function ' suchthat
f(x)'(x) =g(x)
almost everywhere on the set
fx: f(x)g(x)
6= 0
g:
If p2 (1; 1) and f is not 0 almost everywhere, thenwe have equality in (4) if, and only if, g = f almosteverywhere, for some0:
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Landaus inequality. Let f : [0; 1)! R be a twicedierentiable function. PutMk = supx0 f(k)(x)fork = 0; 1; 2: Iff and f00 are bounded, then f0 isalso bounded and
M12q
M0M2:
Proof. Notice that
f(x) =f(x0)+Z x
x0
f0(t) f0(x0)
dt+f0(x0)(xx0)
The case of functions on the entire real line.
Extension to the case of functions with Lipschitz deriva-
tive.
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Inequalities involving convex functions
Hermite-Hadamard inequality: Let f : [a; b]! R be aconvex function. Then
f
a + b
2
1
b
a
Z ba
f(x) dx f(a) + f(b)2
(HH)
with equality only for ane functions.
The geometric meaning.
The case of arbitrary probability measures. See [2].
Jensens inequality: If' : [; ]![a; b]is an integrablefunction andf : [a; b]! R is a continuous convex func-tion, then
f 1
Z
'(x) dx!
1
Z
f('(x)) dx:
(J)
The case of arbitrary probability measures.
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An application of the Jensen inequality:
Hardys inequality: Suppose thatf2 Lp(0; 1); f 0;where p2(1; 1): Put
F(x) =1
x
Z x0
f(t) dt; x >0:
Then
kFkLp p
p 1 kfkLp
with equality if, and only if, f = 0 almost everywhere.
The above inequality yields the norm of the averaging
operatorf! F; from Lp(0; 1) into Lp(0; 1):
The constant p=(p 1) is best possible (though un-tainted). The optimality can be easily checked by consid-
ering the sequence of functionsfn(t) =t1=p (0;n](t):
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A more general result (also known as Hardys inequality):
If fis a nonnegative locally integrable function on(0; 1)andp; r >1; thenZ1
0xprFp(x)dx
p
r 1p Z1
0tpr fp(t) dt:
(5)
Moreover, if the right hand side is nite, so is the left
hand side.
This can be deduced (via rescaling) from the following
lemma (applied tou=xp; p >1; and h=f):
Lemma. Suppose that u : (0; 1)! R is convex andincreasing andhis a nonnegative locally integrable func-
tion. ThenZ10
u
1
x
Z x0
h(t) dt
dx
x
Z10
u(h(x))dx
x:
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Proof. In fact, by Jensens inequality,Z10
u
1
x
Z x0
h(t)dt
dx
x
Z10
1
x
Z x0
u(h(t)) dt
dx
x
=Z1
0
1
x2
Z10
u(h(t))[0;x](t) dt
dx
=
Z10
u(h(t))
Z1t
1
x2dx
dt
=Z1
0u(h(t)) dt
t:
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Exercises
1. Prove the inequalities
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