lecture 02 functional analysis week02 v3

8/13/2019 Lecture 02 Functional Analysis Week02 V3

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Week 2: Metrics and normed spacesDocument prepared by Anna Rozanova-Pierrat1

1 Lecture 2.2: Distance function

1.1 Definitions and examples

Definition 1 LetEbe a set andd: E E R be a function. d is adistance onE if:1.(x, y) E E, d(x, y) 0;2.(x, y) E E, d(x, y) = 0 if and only ifx= y;3.(x, y) E E, d(x, y) =d(y, x);4.(x,y,z) E E E, d(x, y) d(x, z) + d(z, y) (triangular inequality).

If point 2 does not hold, d is called a pseudodistance. If point 3 does not hold, d is called aquasidistance.

Definition 2 SetEwith a given distance defined on it, i.e. the pair(E, d), is ametric space. Ifd is a pseudodistance onE, then (E, d) is apseudometric space. Ifd is a quasidistance onE,then(E, d) is aquasimetric space.

Example 1 Setting

d(x, y) =

0 ifx= y,1 ifx =y,

wherex andy are elements of an arbitrary setE, we obtain a metric space(E, d) (a discrete spaceor space of isolated points). Indeed, by definition ofd, the first three points of Definition are satisfied.For the last point we have

1. Ifx= y the triangle inequality becomes:

0 2 ifz=x or 0 0 ifz= x.

2. Ifx =y the triangle inequality becomes:1 2 ifz=x andz=y or 1 1 ifz= x orz=y.

Therefore the triangle inequality is satisfied.

Example 2 Consider the spaceC([0, 1]) of all continuous functions on [0, 1]. Let us verify that

d(f, g) = maxx[0,1] |f(x) g(x)|is a distance onC([0, 1]):

1MAS, ECP


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2 Week 2: Metrics and normed spaces

1. As the modulus is a positive function onR:

x R |x| 0,

we have that (f, g) C([0, 1]) C([0, 1]), d(f, g) 0.

2. We have for all(f, g) C([0, 1]) C([0, 1])d(f, g) = 0 max

x[0,1]|f(x) g(x)| = 0 x [0, 1] 0 |f(x) g(x)| 0

x [0, 1] |f(x) g(x)| = 0 x [0, 1] f(x) =g(x).

3. For all(f, g) C([0, 1]) C([0, 1]) we haved(f, g) = max

x[0

,1] |

f(x)

g(x)

|= max

x[0

,1] |

[f(x)

g(x)]

|= max

x[0

,1] |

g(x)

f(x)

|=d(g, f).

4. For all(f , g , h) C([0, 1]) C([0, 1]) C([0, 1]) we haved(f, g) = max

x[0,1]|f(x) g(x)| = max

x[0,1]|f(x) h(x) + h(x) g(x)|

maxx[0,1]

(|f(x) h(x)| + |h(x) g(x)|) maxx[0,1]

|f(x) h(x)| + maxx[0,1]

|h(x) g(x)|=d(f, h) + d(h, g).

We conclude that(C([0, 1]), d) is a metric space.

Example 3 ConsiderE = Rn withnN andp

[1,

[. Lets define forx = (x

1, . . . , xn)

E

andy= (y1, . . . , yn) E

dp(x, y) =

ni=1

|xi yi|p 1

p

andd(x, y) = maxi[1,...,n]

|xi yi|.

1. We can easily see thatd satisfies the assertions of Definition1and hence, d is a metric inRn.

2. Let us prove thatdp is a metric. It is obvious that points 1-3 are true fordp for allp [1, [.We need to prove the triangle inequality 4.

Letx,y,z be three points inRn

and letA= x z, B = z y. Thenx y =A+B and thetriangle inequality 4 takes the form of Minkowskis inequality

ni=1

|Ai+ Bi|p 1

p

ni=1

|Ai|p 1

p

+

ni=1

|Bi|p 1

p

. (1)

The inequality is obvious forp= 1. Suppose thatp >1. To prove Minkowskis inequality (1),we use Hlders inequality (see Lemma2for the proof):

n

i=1|AiBi|

n

i=1|Ai|p

1p

n

i=1|Bi|p

1p

, (2)

where 1p

+ 1p

= 1, and the following indentity for anya andb inR orC:

(|a| + |b|)p =|a|(|a| + |b|)p1 + |b|(|a| + |b|)p1.


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Week 2: Metrics and normed spaces 3

Thus, we can write

n

i=1(|Ai| + |Bi|)p =

n

i=1|Ai|(|Ai| + |Bi|)p1 +

n

i=1|Bi|(|Ai| + |Bi|)p1.

We apply Hlders inequality to each sum in the right-hand part of the equality, using the factthat(p 1)p =p:n

i=1

|Ai|(|Ai|+|Bi|)p1

ni=1

|Ai|p 1

p

ni=1

(|Ai| + |Bi|)(p1)p 1

p

=

ni=1

|Ai|p 1

p

ni=1

(|Ai| + |Bi|)p 1

p

,

from where with

n

i=1|Bi|(|Ai| + |Bi|)p1

n

i=1|Bi|p

1p

n

i=1(|Ai| + |Bi|)p

1p

,

we find that

ni=1

(|Ai| + |Bi|)p

ni=1

(|Ai| + |Bi|)p 1

p

n

i=1

|Ai|p 1p

+

ni=1

|Bi|p 1p

.

Dividing both sides of this inequality by(n

i=1(|Ai| + |Bi|)p)1p , and noticing that1 1

p = 1

p we

finally obtain that

n

i=1

(

|Ai

|+

|Bi

|)p

1p

n

i=1 |

Ai|p

1p

+ n

i=1 |

Bi|p

1p

.

To finish the proof, we use the fact that|Ai+ Bi| |Ai| + |Bi| for all i and consequently wehave (1).

3. ConsiderE= Rn with the metric

d2(x, y) =

ni=1

(xi yi)2 1

2

.

This metric is the Euclidean distance function.

4. ConsiderE= { functions fromR to R defined in0}. Forf andg inE, we defined(f, g) =|g(0) f(0)|.

Let us prove thatd is a pseudodistance. Forf(x) =x2 andg(x) =x3 (f=g), by definition ofd, d(f, g) = 0, so point 2 of Definition1 does not hold. Points 1, 3 and 4 are true thanks tothe properties of the modulus. Hence we conclude thatd is a pseudodistance.

Let us prove the inequality of Hlder. Firstly we prove the following technical Lemma:

Lemma 1 Let: [0, [ [0, [ be a continuous strictly increasing function. Then there exists theinverse function1 and then for all positivea andb it holds

ab a0

(x)dx+ b0

1(y)dy. (3)

The equality in (3) takes place if and only ifb= (a).


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00 aa

b

b

y= (x)y= (x)

S1S1

S2S2

Figure 1 For the left-hand figure b < (a)and for the right-hand figure b > (a). Clearly S < S1+S2.

0 a

b

y=(x)

S1

S2

Figure 2 b= (a). Clearly S= S1+S2.

0

Ab

y = (x)x= 1(y)

Figure 3 An example of b0

1(y)dy=for b > A.

Proof. The proof is based on the geometric sense of the integral as a area of the subgraph and onthe three following figures, where S1 =

a0 (x)dx, S2 =

b0

1(y)dy, and S= ab is the area of therectangle.

Remark 1 IfA= supx[0,[ (x) A we haveb

01(y)dy=what

does not contradict (3).

Lemma 2 (inequality of Hlder) Letp]1, [ and 1p

+ 1p

= 1. For allx = (x1, . . . , xn) Rnandy= (y1, . . . , yn) Rn it holds

ni=1

|xiyi| ni=1

|xi|p1p n

i=1

|yi|p1p

. (4)

Proof. The proof of Hlders inequality is based on the following fact:


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for all p [1, [ and any positive constants aand b, it holds

ab ap

p +

bp

p, (5)

where 1p

+ 1p

= 1, i.e.,p = pp1 .

For p= p = 2 inequality (5) is obvious: since (a b)2 0 and thus a2 2ab+b2 0, from whereab a2

2 + b

2

2.

Let us prove (5) in the general case. We use Lemma1. Lets take(x) =xp1. Sincep >1, (0) = 0and is a continuous and strictly increasing function. Therefore, 1(y) = y

1p1 and from (7) we

obtain

ab

a

0

xp1dx+ b

0

y 1p1 dy=

ap

p

+bp

p. (6)

The equality in (6) takes place if and only ifb= ap1 which is equivalent to bp

=ap(p1) =ap.

Take a= xidp(x,0)

and b= yidp (y,0)

. Thanks to (5) we find

|xiyi|dp(x, 0)dp(y, 0)

|xi|p

p[dp(x, 0)]p+

|yi|pp[dp(x, 0)]p

,

from where, by taking the sum over all i from 1 to n,

ni=1

|xiyi

|dp(x, 0)dp(y, 0) ni=1

|xi

|p

p[dp(x, 0)]p+

ni=1

|yi

|p

p[dp(x, 0)]p =

1

p+

1

p = 1.

Therefore, we obtain (4) by the multiplication of the last inequality by dp(x, 0)dp(y, 0).

Problem 1 1. Prove (geometrically) that if: [0, [ [0, [is a continuous strictly decreasingfunction with

10 (x)dx < , then there exists the inverse function1 and then for alla0

andb >0 it holds

ab a0

(x)dx b

1(y)dy. (7)

The equality in (7) takes place if and only ifb= (a).

2. Prove the inequality

ab ap

p +

bp

p,

wherea 0, b >0, p]0, 1[ andp = pp1


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3. d(A, x) = 0 if and only ifx is a contact point ofA;

4. A= A F, whereF is the set of all pointsx such thatd(A, x) = 0.Problem 3

Let there be two subsets of a metric space(E, d).Then the numberz(A, B) = inf(a,b)ABd(a, b)is called the distance betweenA andB. Show thatz(A, B) = 0 ifA B= , but not conversely.Hence, z(A, B) is not a distance onP(E), the set of all subsets of E. (A / E but A E, thusA P(E)).Show that forA andB two non-empty closed subsets of a metric space(E, d), the following function

dH(A, B) = max{ supaA

infbB

d(a, b), supbB

infaA

d(a, b) },

is a distance on the set of all closed subsets inE (dH is a pseudodistance inE). Note thatdH(A, B)is calledHausdorff distance.

2 Lecture 2.3: Underlying topology to a metric space. Com-

pleteness

2.1 Topology in a metric space

Definition 4 Let(E, d) be a metric space. Givenx inE, define the open ball aroundx with radiusr >0 by

Br(x) ={y E|d(x, y)< r}.Then, we define a topology on Eby (see Week 1)

T = {O E| x Or >0, Br(x) O}. (8)

Theorem 1 Every metric space(E, d) is a normal space, and thus, a Hausdorff space, and thus, aT1-space.

Proof. Let X and Y be any two disjoint closed subsets of (E, d). Every point xXhas an openneighborhoodOx disjoint from Y, and hence is at a positive distance rx from Y(recall Problem2).

Similarly, every point y Y is at a positive distance ry fromX. Consider the open setsU=xXB rx

2(x), V =yYB ry

2(y).

We have X U, YV. Moreover, U and V are disjoint. In fact, suppose the contrary that thereis a point z U V. Then there are points x0 X,y0 Y such that

d(x0, z)0,N >0, m, n > Nd(um, un)< .

Definition 11 A metric space(E, d) is calledcomplete if all Cauchy sequences of elements ofEconverge inE.

Example 7 1. R is complete. Q isnt.

2. (C([a, b]), d) is complete. (C([a, b]), d2) isnt. See Example3to definitions ofd andd2.

Proposition 2 1. Every convergent sequence(xn) in(E, d) is a Cauchy sequence in(E, d).

2. If (xn) is the Cauchy sequence in (E, d) and if there exists a subsequence{xnk} such thatxnk x fork+ in(E, d), thenxnx forn + in(E, d).

Proof. Ideas for the proof are given in Figure 4.

2.4 Completion of a metric space

Definition 12 Let(E, d)be a metric space. A complete metric space(G, d)is called acompletion

ofE ifE G and its closureEG =G, i.e., ifE is a dense subset ofG.Example 8 The space of all real numbersR is the completion of the space of all rational numbersQ.


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xx

xnxm

xnk

xk

> > <


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2.6 Compactness in metric spaces

Since metric spaces are topological spaces, all results and definitions of the compactness in the

topological spaces hold for metric spaces as well. Let us just detail the specific properties of thecompactness in metric spaces.

Definition 14 Let(E, d) be a metric space containing a subsetM and >0. A setAE is saidto be an-net for the setM if,

x M there is at least one point a A such that d(x, a) .

It is possible that A M= , but ifAis an -net forM, it is possible to construct 2-setB M.Example 10 The set of all points with integer coordinates is a 1

2-net ofR2.

Definition 15 In a metric space(E, d) a subsetM is calledtotally boundedif for all >0 thereexists a finit-net ofM.

We notice that:

1. If a set M is totally bounded, then its closure M is also totally bounded.

2. Every subset of a totally bounded set is itself totally bounded.

Every totally bounded set is bounded, being the union of a finite number of bounded sets. Theconverse is not true, as shown in the following example:

Example 11 The unit sphereS in the space2

S= {x= (x1, . . . , xn, . . .) 2|d2(x, 0) =n=1

x2n = 1}

is bounded but not totally bounded. In fact, let us consider inS the points

e1 = (1, 0, 0, . . . , 0, 0, . . .),

e2 = (0, 1, 0, . . . , 0, 0, . . .),

. . . . . . . . . . . . . . . . . . . . . . . .

en = (0, 0, 0, . . . , 1, 0, . . .),

. . . . . . . . . . . . . . . . . . . . . . . . ,

where thenth coordinate ofen is one and the others are all zero. The distance between any two pointsen andem (n=m) is

2. HenceScannot have a finite-net with 0, we choosen such that

1

2n1 0 such that

x E cx2 x1 Cx2.

We notice that norms are equivalent iff associated balls can be included in one another (after apossible homothetic transformation).

Theorem 9 IfEis a finite-dimensional vector spacedim(E)


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Let us denote by 2 the usual Euclidean norm:

x2=

n

i=1

iui2=

n

i=1 |

i|2

12

.

We want to prove that any norm inEis equivalent to the Euclidean norm 2. We start withthe proof of the existence ofc >0 such that cx x2. Using the triangular inequality, we havethat

x=n

i=1

iui n

i=1

|i|ui.

As for alli ui are positive numbers, we can apply Lemma2for p= 2, which gives Cauchy-Schwartzinequality in Rn,

x n

i=1

|i|ui

ni=1

|i|21

2

ni=1

ui2 1

2

=x2

ni=1

ui2 1

2

.

We set c= (n

i=1 ui2)12 and finally obtain that

cx x2.

Let us justify the existance ofC > 0 such thatx2 Cx. The inequality cx x2 impliesthat if a sequence (xn) converges with respect to 2, then (xn) converges with respect to too.Consequently, the norm is continuous in (E, 2) (as application from E to R+) and attainsits minimum m >0 on the unit sphere

S1(0) = {x E| x2 = 1},

(which is compact by the Heine-Borel theorem):

minx2=1

x= m >0.

We can thus write that forx2= 1

x2m= 1 m x. (14)

Suppose now that x= yy2 , i. e.x2= 1, but y is not necessary in S1(0). Consequently, from (14)we find using the linear property of the norm that

y2 1m

xy2= 1m

y

y2

y2= 1

myy2y2 =

1

my.

Now we choose C= 1/m.

Corollary 2 LetEbe a finite-dimensional vector space. There is only one topology induced by thenorms.


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5 Lecture 2.6: An example of a normed spaceLp

5.1 A brief summary of the theory of measure and of Lebesgues integral

We give some basic ideas of the constuction of Lebesgues measure, for sets in R2 following A.N.Kolmogorov, S.V. Fomin Introductory Real Analysis.

Theorem of Fubini for the direct product of the measures

measure |Rn+m = measure |Rn measure |Rmallows to have the Lebesgue integral in Rn.

5.1.1 Definition of a measure

We denote by

the disjoint union.

Definition 21 (-algebra)A -algebraTon a set is a nonempty family of subsets of satisfyingthe following axioms :

A1T contains.A2T is stable by the complementarity:

ifA T then \ A T.

A3T is stable by countable unions:ifn N An T thennNAn T.

In this definition, it is of course possible to change the axiom (A1) by (A1):

T contains and the axiom (A3) by (A3):

Tis stable by countable intersections.Example 20 Let us consider a set of three elements ={a,b,c}. We can construct the fol lowingdifferent-algebras:

{, } {, {a}, {b, c}, } {, {b}, {c, a}, } {, {c}, {a, b}, }

{, {a}, {b}, {c}, {a, b}, {b, c}, {c, a}, } =P()We can verify that there are no other tribes on this set.

It is easy to show that:


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Proposition 6 IfT is a-algebra on a set then : T contains and.

T is stable by complementarity. T is stable by countable intersections and unions. T is stable by differences\:

ifA T andB T thenB \ A T.

Remark 5 Note that the definition of a-algebra does not coincide with the definition of a topology.

Definition 22 (Abstract measure defined on a-algebra)Let be a set andT be a-algebraon it. The applicationA T (A) R is called a measure onT, if it satisfies the followingconditions:

1. (A) 0A T,2. (A

B) =(A) + (B)A, B T, A B = ,

3. if(Bn)nN a sequence of elements ofT such that

B1 B2 . . . , B = nBn, thenB T and(B) = limn(Bn) = supnN

(Bn).

From the definition, it follows the two properties of the measure:

Proposition 7 (Properties of measure)

1. () = 0,

2. (A B) =(A) + (B) (A B).We also note that

Proposition 8 Let be a set andT be a-algebra on it. A mapping:T R+ is a measure onT iff it satisfied two following conditions:

() = 0,

(nNAn) = n=0 (An), ifAn T for alln N andAi Aj = fori=j .5.1.2 Measure of Lebesgue in R2

Let us consider the set of all rectangles R in R2 (closed, open and semiclosed or semiopen):

R={(x, y) R2| (a,b,c,d) R4 a d), then m() = 0.


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a) b) c)

Figure 7 Examples of rectangles in R2: a) a closed rectangle [a, b] [c, d]; b) an open rectangle ]a, b[]c, d[;c) examples of semiclosed or semiopen rectangles: form the top to the bottom ]a, b[]c, d],[a, b]]c, d]and ]a, b[[c, d].

2. IfR= , then m(R) = (b a)(d c).

More generally, we say that

1. For allR the measure m(R) 0 and m(R) R+.2. The measure m(R) is additive: ifR=

nk=1 Rk (Ri Rk = for i=k) then

m(R) =n

k=1

m(Rk).

We now construct the measure for all subsets ofR2 by two steps:

1. Construction of the measure, denoted m, of basic sets B which can be presented as a finiteunion of disjoint rectangles:

B =n

k=1

Rk m(B) =n

k=1

m(Rk).

Moreover, for basic sets it holds the following theorem

Theorem 10 All set operations:

, , \, and

of two basic sets give a basic set (hereAB = (A \ B) (B\ A), see Fig8). IfB nBn (the sum is not necessary finite), whereB andBn are basic sets, then

m(B) n

m(Bn).

LetBn be disjoint basic sets forn N, and letB be a basic set, which can be presentedby the disjoint union ofBn:

B =n=1

Bn.

Then

m(B) =n=1

m(Bn).


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A

B

Figure 8 Let A and B be sets with a non empty inersection. The set AB = (A \ B) (B \ A)is presentedby the blue region.

2. Approximation of a setAin R2 by a basic set using

(A) = inf AB=

nBn

m(B),

called the exterior measure and defined as an infimum of the measure of all cover ofA byfinite or infinite union of disjoint rectangles.

Lets firstly consider all sets in R0={0 x 1; 0 y 1} of finite measure. For all sets A R0we define the exterior measure (A) = infAB=

nBn

m(B). We notice that

IfAis a basic set, then (A) =m(A). IfA nAn R0 for n J N, then (A) n (An).

Definition 23 The setA R0 is measurable by Lebesgue if

>0 there exists a basic setB such that(AB)< .

Definition 24MR0 = {measurable (by Lebesgue) subsets ofR0}.Theorem 11 1.MR0 is a-algebra.

2. The exterior measureconsidered onMR0 is a measure. It is calledthe Lebesgue measureand noted.

There are properties of Lebesgues measure:

Theorem 12 1. Lebesgue measure is-additive:if{An}nN MR0 andA=

n=1 An, then

A MR0 and(A) =

n=1 (An).

2. IfA1 A2 . . . are measurable andA=nAn, thenA MR0 and(A) = limn (An).

We can now generalize the definition of measure in all R2: we present R2 as an union of rectanglesRn,m={n < x n+ 1, m < y m+ 1} of finite measure

R2 = n,mZRn,m,


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and we say thatA R2 is measurable ifAn,m= A Rn,m is measurable for all n and m in Z. Hence,we define

(A) =

n,m(An,m).

This time (A) can take the value +. In this case, Theorem12is still true, but for the property2) we need to add the condition that (A1)< +.Example 21 All open and closed sets onRn are measurable.

5.1.3 Measurable functions

Definition 25 Fuctionf :B Rn R is measurableonB if

1. B is measurable,

2. for alla R f1(]a, +[) ={x B :f(x)> a} is measurable.We can reformulate the definition of the measurable scalar function by

Proposition 9 Functionf : R R is measurable if and only if

c R the set{x|f(x)< c} is measurable.

There are several examples of measurable functions:

Example 22 1. Continous functions are measurable, as the inverse image of any open set is openand any open set is measurable.

2. 1Q(x) =

1 ifx Q,0 ifx R \Q is measurable.

3. LetA be a measurable set. Then1A(x) =

1 ifx A,0 ifx / A is measurable (the converse is also

true).

There are main properties of measurable functions:

Proposition 10 1. Iff is measurable thenk R kf is also measurable.2. Iff is measurable thenc R, f+ c is also measurable.3. Iff andg are measurable then the sumf+g, the productf g and the compositionf g are

also measurable.

4. Iff= 0 is measurable then 1f

is also measurable.

5. Ifn N fn is measurable thensup fn, inffn,lim fn (if it exists) andnN fn (if it exists) arealso measurable.

We note as a collorary of Proposition10, that

Iff is measurable, then f2 and|f| are measurable. (15)


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Definition 26 (Equivalent functions) Letf andg be two measurable functions. We say thatfisequivalent to g (f g) if they take different values on a set which Lebesgue measure is zero:

({x: f(x) =g(x)}) = 0.In this case, f=g almost everywhere (a.e.).

We notice that the relation gives classes of equivalentness: iff g then gf, and moreover, ifalso g h, it implies that f h.Example 23 The function1Q(x) and the trivial functionf 0 belong to the same class of equiva-lentness.

We finish the introduction to the measurable functions by Lusins Theorem (see Fig9for an example):

Theorem 13 (Lusin) Functionf is measurable on[a, b] if and only if

>0 f C([a, b]) such that({x: f(x) =f(x)}) .

ff

Figure 9 An example off(the blue line) and f (the red line) for Lusins Theorem.

5.1.4 Integral of Lebesgue

Definition 27 A measurable functionf : Rm R is calledsimpleiffcan be presented as a linearcombination of indicator functions of measurable by Lebesgue sets:

f(x) =n

k=1

ak1Ak(x),

wheren N (n ), ak R for allk andAk are measurable by Lebesgue sets.The notion of simple functions allows to precise the property to be measurable:

Lemma 3 Functionfis measurable if and only if there exists a sequence(fn)nN of simple functionssuch thatfn converge uniformly to f: fn f.

The notion of simple functions also allows to introduce the Lebesgue integrals.


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Definition 28 Letf be a simple function and be the Lebesgue measure. The functionf is calledintegrable (by Lebesgue) on a measurable setA, if the sum

A f(x)d

def

=n yn(An), whereAn={x|x A, f(x) =yn}

is absolutely convergent.

Example 24 Letf(x) = 1 on [a, b] which is a simple function. Then, by definition

[a,b]

1d= 1 ([a, b]) =b a.

Consequently, we define the Lebesgue integral as a limit of the integrals for simple functions:

Definition 29 A measurable functionfis calledLebesgue-integrableon a measurable by Lebesgue

setA if there exists a sequence of simple functions(fn)nN which are integrable andfn f:A

f(x)ddef= lim

n

A

fn(x)d,

whereis Lebesgue measure.

We dont give the properties of Lebesgues integral, which can be easily found in the literature,but we would like to notice the difference of the constuctions of Riemann and Lebesgue integrals(see Fig.10). Disretization on x for the Riemann integral is raplaced by discretization by y in the

aa bb

f(a)

f(b)

xi xi+1i

a= x0 < . . . < xi < . . . < xn=b

f(i) iy

Figure 10 Left-hand figure: construction of the Riemann integralba f(x)dx = limn

ni=0(xi+1

xi)f(i). Right-hand figure: construction of the Lebesgue integral[a,b]f(x)dx =

limy0,nn

i=0i({x| |f(x) i|


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The Dirichlet function is Lebesgue-integrable:[0,1]

f(x)d= 1 (Q [0, 1]) + 0 ((R \Q) [0, 1]) = 1 0 + 0 1 = 0,

where we have used thatQ is a countable set and therefore, its Lebesgue measure(Q) = 0. But theDirichlet function is not Riemann integrable: f is discontinous in every point of [0, 1], of a positivemeasure equal to 1.

These are especially two theorems which we will use in Week 3:

Theorem 14 (Beppo-Levi)

If(fn) is an increasing sequence of measurable positive functions taking values inR+, then

lim

fn =

lim fn. (16)

If(fn)is an increasing or decreasing sequence of Lebesgue-integrable functions, then it holds (16).Theorem 15 (Dominated Convergence)Lets consider a measurableand defined on it Lebesguemeasure. If(fn) is a sequence of functions which satisfies:

n N fn is Lebesgue measurable on. Sequence(fn) converge almost everywhere to a measurable functionf on. There exists an Lebesgue-integrable functiong, g(x)d


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Remark 6 Forp= 1 we obtain from Definition31thatf L1() implies that1. f is measurable,

2.|f| L1

(),which exactly means thatf is Lebesgue-integrable. Hence, Definition31 forp = 1 is equivalent toDefinition30. Functions that are equal almost everywhere are identified.

Remark 7 The space(C(), L1()) is not complete. By the way,

L1() =C()

L1 ,

i.e. L1() is the completion ofC() by the norm L1.Definition 32 Letfbe a function from to R. We define essential supremum offas a number

ess supx f(x) = inf (A)=0

supx\A

f(x)

= inf

B:(\B)=0

supxB

f(x)

. (17)

Example 26 Letf(x) = 1Q[0,1](x). By definition of the essential supremum,ess supx[0,1] f(x) = 0,butsupx[0,1] f(x) = 1.

Remark 8 1. It holds

ess supx f(x) = inf{M R : f(x) M a. e. in}.

2. Let Rn

be an open set andf : R be a continuous function. Then it holdsess supx f(x) = sup

xf(x).

Problem 8 Prove using Remark81) that

|f(x)| fL() a. e. in.

Definition 33 We noteL() the set of measurable functions fromto R for which there exists areal numberCsuch that for almost everyx in,|f(x)| C, i.e.

ess supx f(x)< .

Functions equal almost everywhere are identified. We note

fL()= ess supx f(x).

Proposition 11 Letf Lp() for0< p . Then

fLp()= 0 f= 0a.e. on.

Proof. For 0 < p


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Problem 9 Prove it forp=:

fL()= 0 f= 0a.e. on.

There are some indications:

1. Use Remark81).

2. Show forA = fL() thatA+1n

= {x :f(x) A + 1n}has Lebesgue measure(A+ 1

n) =

() forn= 1, 2, . . ..

3. ConsiderA=n=1A+1n

which is a full Lebesgue measure: (A) =().

Problem 10 Show thatLp() for0 < p is a linear vector space. Indication: use the numericalinequality

a, b

0 andp >0 (a + b)p

2p(ap + bp).

For example, ifa b, then

(a+ b)p (2a)p = 2pap 2p(ap + bp).

Definition 34 Letp [1, ].A functionfbelongs to Lp,loc() whenf1Kbelongs to L

p() for every compactK , where1K isthe characteristic function ofK: 1K(x) = 1 ifx K and0 otherwise.Definition 35 Letp [1, ].We callHlder conjugate (or dual index) ofp, the numberp = 1 + 1

p1 so that 1

p

+ 1

p = 1 (see

Section ) (ifp= 1 thenp =andp= thenp = 1).Note that the Hlder conjugate of2 is2.

Proposition 12 (Hlders Inequality) Let p [1, ] and p be its Hlder conjugate. Let fLp() andg Lp(). Thenf g L1 and

f gL1() fLp()gLp().

Problem 11 Using results of Section35, prove Hlders inequality.

If we summarize all the results, we have

Corollary 3 Letp [1, ]. is a norm onLp().

5.3 Applications of Hlders inequality

Proposition 13 Let0< p q , Rn, ()< . Then1. Lq() Lp(),2.fLp()()

1p 1q fLq().

Proof. We denote byQ= qp 1 and calculate the dual index ofQ:

Q = Q

Q 1= q

qp.


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We now apply Hlders inequality with Q and Q:

fpLp()=

|f|pd=

|f|p 1d

|f|pLQ 1LQ =

(|f|p)Qd 1Q

1d

1Q =

|f|qdp

q () qpq ,

from where

fLp()=

|f|pd

1p

|f|qd 1q

()qp

pq ,

which gives 2).

Iff Lq, thanks to 2),fLp()< , what implies that f Lp, i.e. 1) holds. Let us notice that the condition ()< is very important. For instance,Problem 12 For = R+ give an example of a functionf such that for0< p < q

f Lq(R+), butf / Lp(R+).

Proposition 14 Let0< p1 < p < p2 . Lets define by the equality1

p=

p1+

1 p2

(18)

Note that forp= p1 = 1 and forp= p2 = 0. Such]0, 1[ exists and it is unique. Then fromf Lp1() Lp2() follows thatf Lp() and

fLp() fLp1()f1Lp2(). (19)

Proof. Let us consider p2 < . We denote by q= p1p and calculate the dual index ofqusing (18):

1 =p

p1+

(1 )pp2

pp1

1

q = q

q 1 1

q = 1 1

q = 1 p

p1=

(1 )pp2

,

i.e. q = p2

(1 )p.

We now apply Hlders inequality with qand q:

fpLp()=

|f|pd=

|f|p |f|(1)pd

(|f|p) p1pd

pp1

(|f|(1)p) p2(1)pd (1)p

p2=

|f|p1d

pp1

|f|p2d (1)p

p2,

from where

fLp()=

|f|pd

1p

|f|p1d p1

|f|p2d1

p2

=fLp1()f

1

Lp2(),which gives (19).

Problem 13 Prove (19) forp2= .


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We finish with the interpolation inequality (see H. Brezis Functional Analysis, Sobolev Spaces andPartial Differential Equations):

Proposition 15 Let

{fi, i

I

}be a family of functions withfi

Lpi() and 1

p = 1pi

1. Thenfi Lp() and

fiLp() fiLpi()Corollary 4 Iff Lp() Lq() thenf Lr() for anyr such thatp r q.

lecture 02 functional analysis week02 v3

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