week2-metric and normed spaces

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

1 Lecture 2.2: Distance function

1.1 Definitions and examples

Definition 1 Let E be a set and d : E E R be a function. d is a distance on E if:1. (x, y) E E, d(x, y) 0;2. (x, y) E E, d(x, y) = 0 if and only if x = 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 Set E with a given distance defined on it, i.e. the pair (E, d), is a metric space. Ifd is a pseudodistance on E, then (E, d) is a pseudometric space. If d is a quasidistance on E,then (E, d) is a quasimetric space.

Example 1 Setting

d(x, y) =

{0 if x = y,1 if x 6= y,

where x and y are elements of an arbitrary set E, we obtain a metric space (E, d) (a discrete spaceor space of isolated points). Indeed, by definition of d, the first three points of Definition are satisfied.For the last point we have

1. If x = y the triangle inequality becomes:

0 2 if z 6= x or 0 0 if z = x.

2. If x 6= y the triangle inequality becomes:1 2 if z 6= x and z 6= y or 1 1 if z = x or z = y.

Therefore the triangle inequality is satisfied.

Example 2 Consider the space C([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 on C([0, 1]):

1MAS, ECP

2 Week 2: Metrics and normed spaces

1. As the modulus is a positive function on R:

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 Consider E = Rn with n N and p [1,[. Lets define for x = (x1, . . . , xn) Eand y = (y1, . . . , yn) E

dp(x, y) =

(ni=1

|xi yi|p) 1

p

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

|xi yi|.

1. We can easily see that d satisfies the assertions of Definition 1 and hence, d is a metric inR

n .

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

Let x, y, z be three points in Rn and let A = x z, B = z y. Then x 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 for p = 1. Suppose that p > 1. To prove Minkowskis inequality (1),we use Hlders inequality (see Lemma 2 for the proof):

ni=1

|AiBi| (

ni=1

|Ai|p) 1

p(

ni=1

|Bi|p) 1

p

, (2)

where 1p+ 1

p= 1, and the following indentity for any a and b in R or C :

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

Week 2: Metrics and normed spaces 3

Thus, we can write

ni=1

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

|Ai|(|Ai|+ |Bi|)p1 +ni=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:ni=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

ni=1

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

ni=1

|Bi|p) 1

p(

ni=1

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

p

,

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 that 1 1

p= 1

pwe

finally obtain that

(ni=1

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

p

[

ni=1

|Ai|p] 1p

+

[ni=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. Consider E = Rn with the metric

d2(x, y) =

(ni=1

(xi yi)2) 1

2

.

This metric is the Euclidean distance function.

4. Consider E = { functions from R to R defined in 0}. For f and g in E, we defined(f, g) = |g(0) f(0)|.

Let us prove that d is a pseudodistance. For f(x) = x2 and g(x) = x3 (f 6= g), by definition ofd, d(f, g) = 0, so point 2 of Definition 1 does not hold. Points 1, 3 and 4 are true thanks tothe properties of the modulus. Hence we conclude that d 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 function 1 and then for all positive a and b it holds

ab a0(x)dx+

b01(y)dy. (3)

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


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 If A = supx[0,[ (x) A we have b0

1(y)dy = whatdoes not contradict (3).

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

p= 1. For all x = (x1, . . . , xn) Rn

and y = (y1, . . . , yn) Rn it holdsni=1

|xiyi| (

ni=1

|xi|p) 1

p(

ni=1

|yi|p) 1

p

. (4)

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


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

ab ap

p+bp

p, (5)

where 1p+ 1

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

p1 .

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 Lemma 1. Lets take (x) = xp1. Since p > 1, (0) = 0and is a continuous and strictly increasing function. Therefore, 1(y) = y

1p1 and from (7) we

obtain

ab a0xp1dx+

b0y

1p1dy =

ap

p+bp

p. (6)

The equality in (6) takes place if and only if b = 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 0 it holds

ab a0(x)dx

b

1(y)dy. (7)

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

2. Prove the inequality

ab ap

p+bp

p,

where a 0, b > 0, p ]0, 1[ and p = pp1 < 0.

3. Consider dp for p ]0, 1[. Is it a metric on Rn?Definition 3 Let (E, d) be a metric space. Let A E. The distance between a set A and apoint x X is defined by d(A, x) = infaA d(a, x).Problem 2 Prove that

1. x A implies d(A, x) = 0, but not conversely;2. d(A, x) is a continuous function of x (for fixed A) (see Week 1 and Subsection 2.2 for the

definition of the continuous function);


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

4. A = A F , where F is the set of all points x such that d(A, x) = 0.Problem 3 Let there be two subsets of a metric space (E, d).Then the number

z(A,B) = inf(a,b)AB

d(a, b)

is called the distance between A and B. Show that z(A,B) = 0 if A B 6= ?, but not conversely.Hence, z(A,B) is not a distance on P(E), the set of all subsets of E. (A / E but A ( E, thusA P(E)).Show that for A and B 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 in E (dH is a pseudodistance in E). Note that dH(A,B)is called Hausdorff 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. Given x in E, define the open ball around x with radiusr > 0 by

Br(x) = {y E| d(x, y) < r}.

Then, we define a topology on E by (see Week 1)

T = {O E| x O r > 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 x X has an openneighborhood Ox disjoint from Y , and hence is at a positive distance rx from Y (recall Problem 2).Similarly, every point y Y is at a positive distance ry from X. Consider the open sets

U = xXB rx2(x), V = yYB ry

2(y).

We have X U , Y V . 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 : K1dX(x1, x2) dY (f(x1), f(x2)) K2dX(x1, x2) x1, x2 X,then f is isomorphism of (X, dX) on (Y, dY ). If, in addition, K2 = K1 = 1, then f is an isometry.

Example 6 Let us consider two different metrics on X: d1 and d2. Thus we have two metric spaces(X, d1) and (X, d2). The mapping f : (X, d1) (X, d2), f(x) = 3x is bi-Lipschitz, then two metricsd1 and d2 are equivalent.

2.2 Convergence

We know the definition of the convergence introduced for topological spaces and thus it is the samein the metric spaces:

A sequence of points (xn) in a metric space (E, d) is said to converge to a point l E if every openneighborhood B(l) of l contains all points xn starting from a certain index:

> 0 N N , n N d(xn, l) < .

Clearly, in a metric space


Proposition 1 1. xn converges to l in a metric space (E, d) if and only if limn d(xn, l) = 0.

2. a function f : (X, dX) (Y, dY ) is continuous, if from xn x for n in X follows thatf(xn) f(x) for n in Y .

It is an immediate consequence of the definition of a limit that

1. No sequence can have two distinct limits;

2. If a sequence (xn) converges to a point l, then so does every subsequence of (xn).

In addition to Week 1, in metric spaces we have

Theorem 2 A necessary and sufficient condition for a point l to be a limit point of a set F in ametric space is that there exist a sequence (xn) of distinct points of F converging to l.

Proof. The condition is necessary, since if l is a limit point of F , then every open neighborhoodB 1

n(l) contains at least one point xn F B 1

n(l) different of l. We note that

B1(l) ) B 12(l) ) . . . ) B 1

n(l) ) . . .

Therefore, these points xn can be chosen different and these points form a sequence (xn) convergingto l. The sufficiency is obvious.

Problem 5 Let (X, dX) and (Y, dy) be two metric spaces and f : X Y . Prove that1. If f is an isometry, then f is continuous.

2. If f is a Lipschitz continous function, then f is continuous.

3. If f is a contraction, then f is continuous.

2.3 Complete metric spaces

Definition 10 In a metric space (E, d), we call Cauchy sequence, a sequence (un) such that

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

Definition 11 A metric space (E, d) is called complete if all Cauchy sequences of elements of Econverge in E.

Example 7 1. R is complete. Q isnt.

2. (C([a, b]), d) is complete. (C([a, b]), d2) isnt. See Example 3 to definitions of d and d2.

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 for k + in (E, d), then xn x for n + in (E, d).

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


xx

xn

xmxnk

xk

> > < < 2

< 2>

Figure 4 As distances d(xn, x) < and d(x, xm) < , then by the triangle inequality d(xn, xm) < 2 (theleft-hand schema). As d(xnk , xk) < and d(xnk , x) < , then by the triangle inequality d(xk, x) < 2(the right-hand schema).

2.4 Completion of a metric space

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

of E if E G and its closure EG = G, i.e., if E is a dense subset of G.Example 8 The space of all real numbers R is the completion of the space of all rational numbersQ .

Theorem 3 Every metric space (E, d) has a completion. This completion is unique in the followingsense: if there are two completions E1 and E2, then they are isometric.

For the proof see A.N. Kolmogorov, S.V. Fomin Introductory Real Analysis.

2.5 Separable spaces

Definition 13 A metric space is said to be separable if it has a countable dense subset.

Example 9 Rn for n N contains the countable dense set of all points x = (x1, . . . , xn) withrational coordinates. C[a, b] are separable spaces.

The space of sequences 2 = {x = (x1, . . .)| d2(x, y)2 = i1 |xi yi|2 < } contains thecountable dense set of all points x = (x1, . . .) with only finit number of nonzero coordiantes,which are rational.

The space C[a, b] of all continuous functions on [a, b] with a metricd(g, f) = max

x[a,b]|g(x) f(x)|

contains the countable dense set of polynomials with rational coefficients.

Let = {bounded sequences x = (x1, . . .)| d(x, y) = sup

k

|xk yk|}. is an example of a nonseparable space. Let us show that contains an uncountable denseset.

In fact, consider the set F of all sequences consisting exclusively of zeros and ones. In thiscase, F has the power of the continuum, since there is a bijection between F and the set of allsubsets of the set of natural numbers N : for all A ( N we associate xA = xn such that

xn =

{1, if n A,0, if n / A.


We note that the distance between any two points of F equals 1:

d(xA, xB) = 1 if A 6= B.

Suppose we surround each point of F by an open sphere of radius 12, thereby obtaining an

uncountably infinite family of pairwise disjoint spheres. Then if some set M is dense in ,there must be at least one point of M in each of the spheres. It follows that M cannot becountable and hence that cannot be separable.

2.6 Compactness in metric spaces

Since metric spaces are topological spaces, all results and definitions of the compactness in thetopological 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 subset M and > 0. A set A E is saidto be an -net for the set M if,

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

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

2-net of R2 .

Definition 15 In a metric space (E, d) a subset M is called totally bounded if for all > 0 thereexists a finit -net of M .

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 sphere S in the space 2

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

x2n = 1}

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

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

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

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

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

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

where the nth coordinate of en is one and the others are all zero. The distance between any two pointsen and em (n 6= m) is

2. Hence S cannot have a finite -net with 0, we choose n such that

1

2n1 1 is apositive homogeneous function, which is convex and strictly positive if x 6= 0, what implies that pis a norm.

We can see that in the same linear vector space E we can define different norms. In the next Sectionwe will answere the question: if there are two different norms 1 and 2 on E, what about theproperties of (E, 1) and (E, 2)?

3.2 Converging sequences and continuous applications

Let us reformulate in terms of the norm the definitions of converging sequences and continuousapplications from Week 1, given in terms of topology (see Lecture 2.5 for relations between a metric,a norm and a topology):

1. (E, ) be a normed linear vector space and (xn) be a sequence of elements of E. We say that(xn) converges to l E if

d(xn, l) = xn l 0 for n +.

2. Let f be a mapping of a normed space (X, ) to a normed space (Y, ). The mapping fis continuous if from the convergence of xn to l in X for n + follows that the sequence{f(xn)}nN converge to f(l) in Y :

xn lX 0 for n + f(xn) f(l)Y 0 for n + (10)Example 18 Let (X, X) be a normed linear vector space. The norm X is a continuousfunction on X:

if xn lX 0 for n + xnX lX for n +.To prove it is sufficient to notice that for all x and y in X it holds the following inequality:

| xX yX| x yX .Therefore, we have that

| lX xnX | l xnXand, since l xnX 0 for n +, it implies that lX xnX 0 for n +.Remark 4 Any linear continuous mapping f : (X, X) (Y, Y ) is Lipschitz continuous:

K 0 f(x) f(y)Y Kx yX x, y X.If in addition, K < 1, f is a contraction.

4 Lecture 2.5: Underlying metric and topology to a normed

space

We can recognize in Example 17 the formulas of the metrics given in Example 3 written for a distancebetween x and 0.


4.1 Metric and a norm

Proposition 4 Let (E, ) be a normed space. Then the function

(x, y) = x y for (x, y) E E (11)

is a metric in E. Moreover, the metric is absolutely homogenous, i.e.

(x, y) = ||(x, y) for (x, y, ) E E C , (12)

and invariant by translation (see Fig. 6)

(x+ z, y + z) = (x, y). (13)

x

x

y

y

z

0

Figure 6 Translation of two vectors x and y by a vector z.

Proposition 5 A metric d in a space E is defined by a norm in E by formula (11) if and only if dsatisfies (12) and (13).

Problem 7 If d satisfies (12) and (13), prove that N(x) = d(x, 0) is a norm.

4.2 Equivalent norms

Definition 19 Let 1 and 2 be two norms in a vector space E. The norm 2 is calledstronger than the norm 1 if there exists a positive constant C > 0 such that

x E x1 Cx2.

Example 19 Let us consider the space C([a, b]) of continuous functions on [a, b] with two followingnorms:

f = maxaxb

|f(x)|, f2 =( b

a|f(x)|2dx

) 12

We find that is stronger than 2:

f2 ( b

a1dx

) 12

maxaxb

|f(x)| = b af.


A stronger norm will provide a stronger topology. If there are two norms 1 and 2 on E and 2 is stronger than 1, it means that

1. the convergence in (E, 2) implies the convergence in (E, 1) (but not converse!)2. if a set F is dense in (E, 2) then F is also dense in (E, 1) (but not converse!)

Definition 20 Let 1 and 2 be two norms in a vector space E. The norms are called equivalentif there exist two positive constants c > 0 and C > 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 10 If E is a finite-dimensional vector space dim(E) < , then all norms in E areequivalent.

Proof. Since E is a finite-dimensional vector space, there exists a basis {ui, 1 i n} such thatfor all x E there exist unique i (i = 1, . . . , n) such that

x =ni=1

iui.

Let us denote by 2 the usual Euclidean norm:

x2 = ni=1

iui2 =(

ni=1

|i|2) 1

2

.

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

x = ni=1

iui ni=1

|i|ui.

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

x ni=1

|i|ui (

ni=1

|i|2) 1

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 of C > 0 such that x2 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 for x2 = 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 1mxy2 = 1

m

yy2 y2 = 1my

y2y2 =

1

my.

Now we choose C = 1/m.

Corollary 1 Let E be a finite-dimensional vector space. There is only one topology induced by thenorms.

5 Lecture 2.6: An example of a normed space Lp

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 |R

n+m = measure |R

n measure |R

m

allows to have the Lebesgue integral in Rn .

5.1.1 Definition of a measure

We denote bythe disjoint union.

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

A1 T contains ?.A2 T is stable by the complementarity:

if A T then \ A T .

A3 T is stable by countable unions:

if n N An T then nN An T .


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

T contains

and the axiom (A3) by (A3):

T is stable by countable intersections.

Example 20 Let us consider a set of three elements = {a, b, c}. We can construct the followingdifferent -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:

Proposition 6 If T 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 \:

if A T and B T then B \ 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 and T be a -algebraon it. The application A T 7 (A) R is called a measure on T , if it satisfies the followingconditions:

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

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

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

B1 B2 . . . , B = nBn, then B 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 and T be a -algebra on it. A mapping : T R+ is a measure onT iff it satisfied two following conditions:

(?) = 0, (nNAn) = n=0 (An), if An T for all n N and Ai Aj = ? for i 6= 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.

2. If R 6= ?, then m(R) = (b a)(d c).More generally, we say that

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

nk=1Rk (Ri Rk = ? for i 6= k) then

m(R) =n

k=1

m(Rk).

We now construct the measure for all subsets of R2 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 11 All set operations:, , \, and

of two basic sets give a basic set (here AB = (A \B) (B \A), see Fig 8). If B nBn (the sum is not necessary finite), where B and Bn are basic sets, then

m(B) n

m(Bn).

Let Bn be disjoint basic sets for n N , and let B be a basic set, which can be presentedby the disjoint union of Bn:

B =n=1

Bn.

Then

m(B) =n=1

m(Bn).

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 set A in R2 by a basic set using

(A) = infAB=

nBn

m(B),

called the exterior measure and defined as an infimum of the measure of all cover of A 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

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

Definition 23 The set A R0 is measurable by Lebesgue if > 0 there exists a basic set B such that (AB) < .

Definition 24 MR0 = {measurable (by Lebesgue) subsets of R0}.


Theorem 12 1. MR0 is a -algebra.2. The exterior measure considered onMR0 is a measure. It is called the Lebesgue measure

and noted .

There are properties of Lebesgues measure:

Theorem 13 1. Lebesgue measure is -additive:if {An}nN (MR0 and A =

n=1An, then

A MR0 and (A) =

n=1 (An).

2. If A1 A2 . . . are measurable and A = 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

R

2 = n,mZRn,m,

and we say that A R2 is measurable if An,m = ARn,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, Theorem 13 is still true, but for the property2) we need to add the condition that (A1) < +.Example 21 All open and closed sets on Rn are measurable.

5.1.3 Measurable functions

Definition 25 Fuction f : B Rn R is measurable on B if1. B is measurable,

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

Proposition 9 Function f : 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 if x Q ,0 if x R \ Q is measurable.

3. Let A be a measurable set. Then 1A(x) =

{1 if x A,0 if x / A is measurable (the converse is also

true).

There are main properties of measurable functions:


Proposition 10 1. If f is measurable then k R kf is also measurable.2. If f is measurable then c R, f + c is also measurable.3. If f and g are measurable then the sum f + g, the product f g and the composition f g are

also measurable.

4. If f 6= 0 is measurable then 1fis also measurable.

5. If n N fn is measurable then sup fn, inf fn, lim fn (if it exists) and nN fn (if it exists) arealso measurable.

We note as a collorary of Proposition 10, that

If f is measurable, then f 2 and |f | are measurable. (15)Definition 26 (Equivalent functions) Let f and g be two measurable functions. We say that fis equivalent to g (f g) if they take different values on a set which Lebesgue measure is zero:

({x : f(x) 6= g(x)}) = 0.In this case, f = g almost everywhere (a.e.).

We notice that the relation gives classes of equivalentness: if f g then g f , and moreover, ifalso g h, it implies that f h.Example 23 The function 1

Q

(x) and the trivial function f 0 belong to the same class of equiva-lentness.

We finish the introduction to the measurable functions by Lusins Theorem (see Fig 9 for an example):

Theorem 14 (Lusin) Function f is measurable on [a, b] if and only if

> 0 f C([a, b]) such that ({x : f(x) 6= f(x)}) .

f

f

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

5.1.4 Integral of Lebesgue

Definition 27 A measurable function f : Rm R is called simple if f can be presented as a linearcombination of indicator functions of measurable by Lebesgue sets:

f(x) =n

k=1

ak1Ak(x),


where n N (n ), ak R for all k and Ak are measurable by Lebesgue sets.Or equivalently, a measurable function f is simple if f takes no bigger than countable number ofvalues.

Example 24 1. A continuous function f : [a, b] R is not a simple function.2. Let denote by [x] the integer part of x. The function f(x) = [x]1[a,b] is a simple function.

The notion of simple functions allows to precise the property to be measurable:

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

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

Definition 28 Let f be a simple function and be the Lebesgue measure. The function f is calledintegrable (by Lebesgue) on a measurable set A, if the sum

Af(x)d

def=n

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

is absolutely convergent.

Example 25 Let f(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 function f is called Lebesgue-integrable on a measurable by Lebesgueset A if there exists a sequence of simple functions (fn)nN which are integrable and fn f :

Af(x)d

def= lim

n

Afn(x)d,

where is 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 theLebesgue integral.

We also notice that if there exists the Riemann integral of |f |, it implies the existance of the Lebesgueintegral of f . In this case, if there also exists the Riemann integral of f , it is equal to the Lebesgueintegral of f .

This is an example of the function which is Lebesgue-integrable, but not Riemann integrable:

Example 26 Let f be the Dirichlet function:

f(x) = 1Q[0,1](x) =

{1, x Q [0, 1]0, x (R \ Q) [0, 1] .

The Dirichlet function is Lebesgue-integrable (it is a simple function):[0,1]

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

24 Week 2: Metrics and normed spacesreplacements

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 integral ba 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=0 i({x| |f(x) i| < y}) with i [f(a), f(b)].

where we have used that Q 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 15 (Beppo-Levi)

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

limfn =

lim fn. (16)

If (fn) is an increasing or decreasing sequence of Lebesgue-integrable functions, then it holds (16).Theorem 16 (Dominated Convergence) Lets consider a measurable and 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 function f on . There exists an Lebesgue-integrable function g, g(x)d


Definition 30 The set of Lebesgue-integrable functions from to R is noted L1() or simply L1

when no confusion is possible. For f L1() we define

fL1() =|f |d.

Each element of L1 is a classe of equivalent functions, which are equal almost everywhere.

Definition 31 Let 0 < p < , and f : R be measurable functions. The function f belongs toLp() if |f |p L1(). For f Lp() we define

fLp() =(

|f |pd

) 1p

.

Remark 6 For p = 1 we obtain from Definition 31 that f L1() implies that1. f is measurable,

2. |f | L1(),which exactly means that f is Lebesgue-integrable. Hence, Definition 31 for p = 1 is equivalent toDefinition 30. 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 of C() by the norm L1.Definition 32 Let f be a function from to R. We define essential supremum of f as a number

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

[sup

x\Af(x)

]= inf

B(: (\B)=0

[supxB

f(x)

]. (17)

Example 27 Let f(x) = 1Q[0,1](x). By definition of the essential supremum, ess supx[0,1] f(x) = 0,

but supx[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 and f : R be a continuous function. Then it holds

ess supx f(x) = supx

f(x).

Definition 33 We note L() the set of measurable functions from to R for which there exists areal number C such that for almost every x in , |f(x)| C, i.e.

ess supx |f(x)|


Problem 8 Prove using Remark 8 1) that

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

Proposition 11 Let f Lp() for 0 < p . Then

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

Proof. For 0 < p 0 (a+ b)p 2p(ap + bp).For example, if a b, then

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

Definition 34 Let p [1,].A function f belongs to Lp,loc() when f1K belongs to L

p() for every compact K , where 1K isthe characteristic function of K: 1K(x) = 1 if x K and 0 otherwise.Definition 35 Let p [1,].We call Hlder conjugate (or dual index) of p, the number p = 1+ 1

p1 so that1p+ 1

p= 1 (see

Section 1.1) (if p = 1 then p = and p = then p = 1).Note that the Hlder conjugate of 2 is 2.

Proposition 12 (Hlders Inequality) Let p [1,] and p be its Hlder conjugate. Let f Lp() and g Lp(). Then fg L1 and

fgL1() fLp()gLp().

Problem 11 Using results of Section 1.1, prove Hlders inequality.

If we summarize all the results, we have

Corollary 2 Let p [1,]. is a norm on Lp().


5.3 Applications of Hlders inequality

Proposition 13 Let 0 < p q , ( Rn , ()


We now apply Hlders inequality with q and q:

fpLp() =|f |pd =

|f |p |f |(1)pd

(

(|f |p) p1pd

)pp1

((|f |(1)p)

p2(1)pd

) (1)pp2

=(

|f |p1d

)pp1

(|f |p2d

) (1)pp2

,

from where

fLp() =(

|f |pd

) 1p

(|f |p1d

) p1

(|f |p2d

) 1p2

= fLp1()f1Lp2(),which gives (19).

Problem 13 Prove (19) for p2 =.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 with fi Lpi() and 1p = 1

pi 1. Then

fi Lp() and fiLp() fiLpi()

Corollary 3 If f Lp() Lq() then f Lr() for any r such that p r q.

Lecture 2.2: Distance functionDefinitions and examples

Lecture 2.3: Underlying topology to a metric space. CompletenessTopology in a metric spaceConvergenceComplete metric spacesCompletion of a metric spaceSeparable spacesCompactness in metric spaces

Lecture 2.4: Normed vector spacesDefinition and examplesConverging sequences and continuous applications

Lecture 2.5: Underlying metric and topology to a normed spaceMetric and a normEquivalent norms

Lecture 2.6: An example of a normed space LpA brief summary of the theory of measure and of Lebesgue's integralDefinition of a measureMeasure of Lebesgue in R2Measurable functionsIntegral of Lebesgue

Lp spacesApplications of Hlder's inequality

week2-metric and normed spaces

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