Analysis Review

8/10/2019 Analysis Review

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Analysis - Final Review of MATH527A/BBased on notes by Dr. H. Flascka

1 Metric Spaces

DefinitionA metric d on a space Msatisfies the fol-lowing:

d(x, y) = 0 iffx = y d(x, y) =d(y, x), x, yM d(x, z)d(x, y) + d(y, z)

withd : M M[0, ).Example: lp(R,N with||x||p = (

N

|xi|p)(1/p). Inparticular l1 l2 . . . l. Also for1 p < q ,lq lp.Convergence in metric space: Ifxn x in (M, d)then >0,N such that

n > Nd(xn, x)< Cauchy convergence A sequence xn is cauchy iff

the following holds: > 0 there existsN such thatn, mNd(xn, xm)<

In particular if all Cauchy sequences converge in thespace M then the space is complete. ALso all con-

verging sequences are Cauchy.CoarsenessIfxnx in (M, d) implies thatxn

xin (M, d)thend is coarser thand.

Equivalence of metrics: If two metrics are equiva-lent then

xnx in (M, d)xnx in (M, d)E.g. OnRn all lp norms are equivalent. An alternatecharacterization is the following: if given x and >0, >0 such that

d(x, y)< d(x, y)< and >0

d(x, y)< d(x, y)< .

Continuity in Metric spaceContinuity off :V Watv

0can be defined through the following:

>0

there exists >0 such that

||v v0||< ||f(v) f(v0)||<

this is implies by||f(v)||W C||v||V. Also can bedefined sequentially, i.e. ifxn x and iff(xn)f(x)thenf is continuous.

Holders inequality: For all sequences we havethe following:

N

|xixj| ||x||p||y||q

ifxlp andylq, and further1/p + 1/q= 1.Youngs Inequality

abap/p + b1/qor more generally:

ab a0

f(t)dt +

b0

f1(t)dt

These are used to prove for example the Minkowskiinequality for L1, L2 or the sequence space metricsetc.. See homework # 2.

2 Normed linear spaces = vectorspaces

Definition We define a vector space V on a field F(usually on R) with the following: First the proper-ties of the vector space

1. v+ w= w + v

2. u + (u + w) = (u + v) + w

3.0V, s.t.0 + u= u

4. vV, s.t. v+ v= 0and operations involving the field elementsa, bF:

1. a(v+ w) = aw + av

2. (a + b)v= av + bv

3. a(bv) = b(av)

4. 1 v= v for 1 being the identityThe norm on this space satisfies the following condi-tions:

||v|| 0 ||av||=|a| ||v||

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||v+ u|| ||v|| + ||u||

A metric is thus defined for d(x, y) = ||xy ||, soevery vector space is a metric space.

Continuity of linear maps A linear map f : (V, || ||V)(W, | | | |W)is defined with the following:

f(av+ bw) = af(v) + bf(w)

f(0v) = 0w

The following are equivalent for normed spacesW, V:

f is continuous

f is continuous at0.

C 0.

The mechanism we use is summing over many normsor taking a supremum.

3 Hilbert Spaces

Definition. Normed linear spaces that are complete.Norm must satisfy paralellogram law:

||x + y||2 + ||x y||2 2||x||2 + 2||y||2

Cauchy-Schwartz says

|< x, y >| ||x||H||y||HForMa closed linear space we can show that

M M =H

so we can decompose anyz H as z = u+ v foruM andvM.

Also (for convex subspace as well), we can finduniquewMsuch that

minvM

||z v||=||w z||, wM

4 Topological Spaces

DefinitionThe following 3 axioms define a topologyTon a setXwhich is a collection ofopen sets,

, X T Finite intersections are in space itself i.e. U, V

T U V T. Arbitrary unions are in space as well i.e.UT

then A U

T.

An open Uis defined such thatyU,V such thatyVU.

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The relative topology is the topology that is re-stricted to a subsetY GXsuch that

S={HY|H=G Y, G T }AHausdorfftopology is such that for all x, yX

there exists open setsUx, Vy such thatUV =.

A subsetVX is a neighborhoodofxX ifVis open andVx.

Convergence in topological sense is definedthrough the following: xn

x iff

V

N(x) there

existsNsuch that

nNxnVContinuity at a point is given by the following: forf :XY then forfto be continuous atxXthenW N(f(x)),V N(x)such thatf(V)W.

Continuityof functions in the topogical sense isimplied by the following: Let (X, T and (Y, S) aretopological spaces, thenf :XY is continuous ifff1(G) T for allG S. Note that sequential con-tinuity does not directly imply the above (example isco-countable topology on R).

AHomeomorphism is a function that is one-to-one and onto.

ANeighborhood or local base is defined throughthe following axioms:

V N(x)xV IfU, V N(x)thenW N(x)such thatW

V U. IfV N(x) there exists W N(x) such that

W V W

V

yW U N(y)and UV.We can define a topology using the local base i.e. Ois open isx O there exists V N(x) such thatVO.

Abaseis defined through the following axioms forB B:

BB B= X ForB1, B2 BandxB1B2then there exists

(B3x) Bsuch that B3B1 B2.Additionally we can defined the unique topology

TonXdefined through the base above throughO T yO, B Bsuch that(Bx)O

In particular we can thus find that in O T thenthereforeO =UB,UOU.

The weak topologyon X is defined as the topol-ogy that makes all linear functionals continuous, i.e.f L(V, R)is continuous.

Finally a countable local base makes the space firstcountable. A countable base or basis gives a secondcountable. In addition we prove that the space is sec-ond countable iff it is separable which implies thatthere is a countable set{xn} M such that givenanyx

Mand >0 then

xnsuch thatd(xn, x)< .

5 Size of sets

5.1 Small/largeness

Dense set A set A is dense in X ifU A= forallUopen X.

Interior pointsAX, a pointxA is an interiorpoint ifUopen x such thatUA. Therefore

Ao

={x|x interior ofA}= VopenA

V

Closure ForA X, x A if for all Vopen x,V A=or

A={x|x closure ofA}=

AF,Fclosed

F

It is a fact that Ao A A, and so we can givethe interpretation thatAo is open,A is closed.

Nowhere denseAXis nowhere dense ifA hasempty interior, i.e. (A)o =

. Also, A is nowhere

dense if (Ac)o = X or its complements interior isdense.

Meager or 1st category sets ifA is the countableunion of nowhere dense sets, thenAis meager or 1stcategory. For example{rn}is an enumeration of therationals and so since Q is a countable union of singleelement sets, then it is meager. A single element setis nowhere dense.

2nd category or residual setsSimply, the comple-ment of a meager set is residual.

Density (again) IfAXis dense in X thenA =X. Also, ifA

Xis dense inB, thenB

A. Also,A

is dense inXif and only if its complement has emptyinterior, i.e. (Ac)o =.

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very small nowhere densevery large dense interior

small meagerlarge residual

medium small empty interiormedium large dense

Measure zero A subset A of Rn has measurezero (or Lebesgue measure zero) if > 0, thereis a countable collection{Bn} of open balls Bn =B(xn, n) =

{x| ||

x

xn||2< n

}such that

A=n=1

Bn

and

n=1volume ofBn . Example: Q in R.The cantor middle thirds set has measure zero eventhough this set has measure zero. Any countableunion of elements have measure zero.

Generic sets Let (X, T) be a topological space. ApropertyP is generic (in the sense ofBairecategory)if

{x|P(x)is true}is a dense residual set

E.g. The property of being an irrational number isgeneric since the complement is meager (Q) and Qc

is dense. The set of differentiable fcns in C[0, 1] isnot generic. The property of quadratic equations tohave distinct roots is generic.

Zariski generic setsLet X= Rn, and equip it withthe Zariski topology. A set in this topology is consid-ered open if its complement is the solution to a sys-tem of polynomial equations. The closed sets havemeasure zero. Therefore A property is generic if theset ofx that have propertyPis a Zariski generic set.Equivalently, the set not having the property is givenby the solution set of finitely many algebraic equa-tions.

E.g. A quadratic equation x2 +bx + c has non-distinct roots whenb2 4c= 0, therefore it is Zariskiclosed, and thus the property ofnothaving this prop-erty is Zariski generic.

Baire category theorem?

5.2 Measure theory

A Measurable space is a set Xtogether with a col-lection of subsetsBofX satisfying:

X B, BifA B Ac B

ifA1, A2 B n=1

An B

B is called a -algebra. Also closed under countable

intersections.

AMeasure satisfies the following relations on ameasurable space(X, B),

:B [0, ]such that() = 0

and

n=1

An

=

n=1

(An)

whenever the set{An}are mutually disjoint and be-long toB.

E.g. we can prove the following: LetA B thenB= (B A) Aand (A) = (B A) + (A). Alsoif (A) (B). More importantly, if An B andA1A2 . . . then

n=1

An

= lim

n(An)

also letBk B, then

n=1

Bn

n=1

(Bn)

The Borel -algebra is the smallest such -algebracontaining all the open sets in a topologyT.

disjointness is important in proving the resultsconcerning the measure. Therefore if given a generalset{Ein} B, then define

F1= E1, F2= E2 FC1 , orFj =Ejj1n=1

Fn

6 Convergence and compactness

6.1 Convergence and completeness

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For f :M R continuous, and so xsuch thatfor allxM then

a= f(x)f(x)f+(x) = A

There is also a parallel definition of compactnessfor a topological space (X, T) , i.e. consider{U} apossibly uncountable open covering of a set AX,i.e.

A

U

then ifA is compact there exists afinitesubcovering,i.e.

AN

n=1

Un

for all open coverings. The definition for a space itselfis directly analogous.

A related property is the so-calledfinite intersec-tion property, i.e. if{F} is a collection of closedsubsets of a topological space(X, T)then the collec-tion has this property if a finite intersection of suchsets is non-empty:

Ni=1

Fi=

If a topological space is compact then for any col-lection of closed sets that has the finite intersectionproperty, then

F=

Related facts: A closed subset of a compact topo-logical space is compact. ALso, a compact subset of aHausdorff space is closed.

6.4 Uniformity

Definition ofuniform continuity: Let f : (M, d)(M, d) (metric spaces) then f is uniformly continu-ous if >0, >0, such that

d(x, y)< d(f(x), f(y))<

Note depends only on , and not x which refers topointwise convergence.

Fact: A continuous function from a compact metricspace inherits some structure from the domain, andthusf is uniformly continuous.

Now we define pointwise convergence of func-tions: Let(M, d) and (M, d) be metric spaces, then{fn} are a sequence of functions from M to M andsuppose thatf :MM, then if

limn

fn(x) =f(x)xM

thenfn approachesfpointwise!Now for uniform convergence. Then if >

0, Nsuch thatn > N

d(f

n(x), f(x))< , x

M

The Weierstrass M-test: Suppose fn are a se-quence of functions in (M, d)and suppose that

||fn||an,with||an||10there is an R0 depending on but not t,such that for all R > R0,

t

[a, b],

RR

f(t, x)dx F(t) <

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Differentiating parametrized integrals, can bedone under the hypothesis that

R

f(t, x)dx= F(t),

R

f

tdx = G(t)converge uniformly

thenF(t) = G(t).

7 Lebesgue Integration

The motivation for developing the idea of Lebesgueintegration comes from consideration of the comple-tion ofC[0, 1] under the L1 metric for example. Itturns out these are measurable functions and themetric is Lebesgue integration (?).

Define U(f, ) =n

i=0maxx(ti,ti+1) f(t)(ti+1ti) and L(f, ) =

ni=0minx(ti,ti+1) f(t)(ti+1 ti).

These are approximations to the integral/area underthe curve off. Therefore the limiting process of thecan be defined:

U(f) = inf

U(f, ), L(f) = sup

L(f, )

In particular iffis Riemann integrable then U(f) =L(f) and so for any continuous function this condi-tion is met since they will be Riemann integrable.Can also that a general function on [0, 1]is Riemannintegrable if and only if the set of points of disconti-nuity off is of Lebesgue measure zero.

We definemeasurable functions, take(X, B)and(Y, C)be measurable spaces, then a functionf :XYis measurable iff1(A) Bfor allA C.

This definition can be extended to the case that re-ally interest us, which means that Y = R. Thereforefirst define the extended real numbers:

Rex = R , = [, ]

with the following rules : areal,a =, ifa >0, thena =,0 = 0, isundefined.

So consider (X, B) a measurable space. Thenf :X Rex is measurable if the set

{t|f(t)< } B, R

This property is inherited to a sequence of suchfunctions, i.e. if{fn}are measurable then iffnfpointwise, thenf is measurable.

7.1 Lebesgue integral

From here on we expect (X, B, ) is the domain.Firstly lets define the characteristic function, i.e.A(x)such that

A(x) =

1, xA0, x /A

Anonnegative simple function Let A1 . . . AnXbe pairwise disjoint measurable sets, a function

(x) =n

j=1

jAk

then is a simple function (step function). Withinthe Lebesgue theory we expect that

=n

j=1

j(Aj)

The idea is to approximate the integral of a gen-eral function f with successive simple functions ofincreasingn.

So let f : X

[0,

] and measurable, then its

Lebesgue integral is given by f= sup

0f, simple

Now consider a general function that can take neg-ative values, then we can integrate each part sepa-rately, i.e. f = f+ f. Work out which one ispositive/negative! hah!

There is a analogy to sequence where the measureis the number of points in a set. The function is thesequence. See page 355.

For non-negative fcns the Lebesgue integral retain

linearity and positive definiteness, and the the inte-gral is zero ifffis zero almost everywhere = the set

where the function is not zero is measure zero.

7.2 Convergence theorems

Fatous Lemma We introduce first the idea oflimitinferior and limit superior. In the sense of real num-bers

liminfan = least limit point of the sequence

or the least point such that for any open set contain-ing it, there are only finite number of points to theleft. Similarly forlim sup an,

lim sup an = greatest limit point of the sequence

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If the series converges, then these two things areequal. For Fatous lemma we use the notion thatthings cant escape to, i.e,

X

lim inffndlim inf

fnd

Monotone convergence theoremstates that if wehave a sequence of non-negative measurable func-tions{fn} such that fn+1(x) fn(x) for each x (anon-decreasing set) andlim fn = f(pointwise) then

lim

fndx=

lim fndx

Note: power series can be integrated term by termin compact subsets of their interval of convergence.

Dominated convergence theorem states that fora sequence{fn}that can be dominated|fn| g andlim fn = fa.e. andg is integrable, then

lim

fndx=

lim fndx

7.3 Lebesgue spacesLet(X, B, )be a measure space. LP(x)(1p 0, such that|f(x)| M, for a.e.xX

The equivalence class is as before,fg if they differonly on a set of measure zero. The norm is actuallycalled the essential supremum, i.e.

ess sup f= infM

Cauchy sequence in L: Let{fn} be a Cauchysequence in L(X) , then there is a set E X of measure zero with the property that >0,Nsuch that

|fn(x) fm(x)|> , xX Ewhenevern,m > N

Another important aspect is the Riesz-FischerTheorem, for1p , Lp() is a complete met-ric space. Also,Lp()are normed linear spaces.

Chebyshevs Inequality Let f be a measurable,extended real-valued function on a measure space(X, B, ). For every >0,

({x| |f(x)|> })1

|f|d

RegularLet(X, T)be a topological space and sup-pose that X is locally compact: everyx X has aneighborhood whose closure is compact. Let be ameasure defined onB (Borel -algebra). Then isregularif for every measurable setE,

(E) = inf{(O)|OE, Oopen}= sup{(K)|KE , Kcompact}

It means that the measure of a set Ecan be approx-imated by a compact set enclosing it, or open setscontained in E. An alternate definition (or conse-quence of the above): A measure is Borel regular ifEmeasurable and >0,Oopen, OE, Kcompact B such that(OE)< , (BK)

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