integral inequalities constantin p. niculescu

8/12/2019 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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integral inequalities constantin p. niculescu

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