Functional Analysis - PowerPoint Presentation

Slide1

Functional AnalysisSlide2

Baire’s

Category theoremSlide3

Definition

Let X be a metric space. A subset M of X is said to be

Rare(Nowhere Dense) in X if its closure M has no interior points,

Meager(of First Category) in X if M is the union of countably many sets each of which is rare in X,Nonmeager(of Second category) in X if M is not meager in X.Slide4

Statement:

If a metric space X≠

ф

is complete, it is nonmeager in itself.Proof: Suppose ф≠X is a complete metric space such that X is meager in itself. Then

X=



k Mk with each Mk rare in X. Now M1 is rare in X, so that, by definition, M1 does not contain a nonempty open set. But X does(as X is open). This implies M1≠X. Hence M1C=X-M1 of M1 is not empty and open. Slide5

We may thus choose a point p

1

in M

1C and an open ball about it, say, B1=B(p1;ε1) 

M

1

C ε1<½By assumption, M2 is rare in X, so that M2 does not contain a nonempty open set. Hence it does not contain the open ball B(p1; ½ ε1). This implies that M2C B(p1; ½ ε1) is non empty and open, so that we may choose an open ball in this set, say, B2

=B(p2;ε2)

 M2C B(p

1; ½ ε1)

ε2<½ ε1

By induction we thus obtain a sequence of balls

Bk=B(pk;εk) εk<2-kSuch that Bk  Mk= ф and Bk+1  B(pk; ½ εk)  Bk k=1,2,… Slide6

Since

ε

k

<2-k, the sequence (pk) of the centers is Cauchy and converges, say, pk pX

because X is complete by assumption. Also, for every m and n>m we have

B

nB(pm; ½ εm), so that d(pm,p)d(pm,pn)+d(pn,p) <

½ εm +d(p

n,p)  ½ ε

m As n . Hence pBm

for every m. Since BmM

m

C ,we now see that pMm for every m, so that pMm=X. this contradicts pX. Hence X in not meager i.e. X is nonmeager in itself.Slide7

Uniform

Boundedness

TheoremSlide8

Statement: Let (

T

n

) be a sequnce of bounded linear operators Tn:X Yfrom a Banach space X into a normed space Y such that (

T

n

x ) is bounded for every xX, then the sequence of norms Tn is bounded. Slide9

Proof: We are given that the sequence (

T

n

x ) is bounded.  For every xX  a real number c

x

such that

(1) Tnx cx n=1,2,…For every kN, let AkX be the set of all x such that Tnx k for all n.We claim that A

k is closed.Let xAk

, then there is a sequence (xj) in

Ak converging to x. This means that for every fixed n we have (2)

Tnxj k

Slide10

Taking limit j

 in (2)

lim Tnxj k

lim

Tnxj k (since norm is continuous) Tn(lim xj)  k (since each Tn is continuous) Tnx  kSo xA

k, and Ak is closed.

By (1) and the definition of Ak

we have, each xX belongs to some A

k. Hence X=kA

k

k=1,2,3,…Since X is complete, Baire’s theorem implies that some Ak contain an open ball, say, (3) B0=B(x0;r)Ak0Let xX be arbitrary, not zero. Slide11

We set z=x

0

+

x =r/2 x (4)Then z-x0 <r, so that zB0. By (3) and from the definition

of A

k0

we thus have Tnz k0 for all n. Also Tnx0 k0 since x0B0. From (4) we obtain x=(z-x0)/.This gives for all n Tn

x = Tn(z-x

0) /  2 x ( T

nz + Tnx

0 )/r  2 x (k0

+k

0)/r =(4k0 x )/rHence for all n, Tn = sup x =1 Tnx (4k0)/rHence the sequence of norms Tn is bounded. Slide12

Open Mapping TheoremSlide13

Definition(Open

mapping)

Let X and Y be metric spaces. Then T: D(T)

Y with domain D(T)X is called an open mapping if for every open set in D(T) the image is an open set in YSlide14

Lemma (Open unit ball)

Statement: A bounded linear operator T from a

Banach space X onto a Banach space Y has the property that the image T(B0) of the open unit ball B0=B(0;1)X contains an open ball about 0

YSlide15

Proof: We prove the result in following steps:

The closure of the image of the open ball B

1

=B(0;½) contains an open ball B*. T(Bn

) contains an open ball

V

n about 0Y, where Bn=B(0;2-n). T(B0) contains an open ball about 0Y.Slide16

a. Let A



X we write

A to mean A=xX x= a, aA

A

A

(=2)Slide17

For

w



X by A+w we mean A+w=xXx=a+w

,

aA

AA+w

a

a+wSlide18

We consider the open ball B

1

=B(0;

½)X. Any fixed xX is in kB1 with real k sufficiently large (k>2 x ). Hence X=

k

kB

1 k=1,2,…Since T is surjective and linear,(1) Y=T(X)=T(k kB1)= k kT(B1)= k kT(B1) Now since Y is complete, it is non meager in itself, by Baire’s category theorem. at least one

kT(B1) must contain an open ball.This

implies that T(B1) also contains an open ball, say, B*=B(y

0;) T(B1). It follows that

B*-y0=B(0; )  T(B

1

) –y0. This completes with the proof of (a.) Slide19

b. We prove that B*-y

0



T(B0), where B0 is given in the statement. For this we claim that(3) T(B1) –y0 T(B

0

).

Let yT(B1) –y0. Then y+y0 T(B1).Also we have y0T(B1)sequences un, vn such that un

=Twn  T(B1

) such that un  y+y0

vn=Tz

n  T(B1) such that v

n

 y0. Since wn, znB1 and B1 has radius ½, it follows that wn –zn  wn + zn < ½+½=1 so wn-znB0,

so T(wn-z

n) =Twn

–Tzn =un

-vn  y+y

0

-y

0

=ySlide20

Thus

yT

(B

0). Since yT(B1) –y0 was

arbitrary, this proves (3). So from (2) we have

B*-y

0=B(0;)T(B0) Let Bn=B(0;2-n)  X. Also since T is linear, T(Bn)=2-nT(B0)So from (4) we obtain (5) V

n=B(0;/2n) T(B

n)This completes the proof of (b.)

c. We finally prove that V1

=B(0; /2)  T(B0)by showing that every yV

1

is in T(B0). So let yV1.From (5) with n=1 we have V1T(B1)Slide21

Hence y

 T(B

1

). So by definition  vT(B1) such that y-v </4.Now

vT

(B

1) so v=Tx1 for some x1B1.Hence y-Tx1 < /4.From this and (5) with n=2 we get that y-Tx1V2 T(B2). As before there is an x2B2 such that (y-Tx

1)-Tx2 < /8.Hence y-Tx

1-Tx2V

3T(B3), and so on. In the nth

step we can choose an xnB

n

such that (6) y -kTxk < /2n+1 k=1,2,…,n; n=1,2,… Let zn=x1+…+xn. Since xk Bk, we have xk <1/2k. This yields for n>mSlide22

z

n

-zm   k=(m+1),…,n x < 

k=(m+1),…, 

1/2

k  0As m  . Hence (zn) is a cauchy sequence in X and X is complete so (zn) is convergent.  xXsuch that zn  x. Also x = k=1,…, xk  k=1,…,

xk  

k=1,…, 1/2k=1

So xB0 . Since T is continuous,

Tzn  Tx

, and

Tx=y (by (6))Hence yT(B0)Since yV1 was chosen arbitrarily so V1T(B0)This completes the proof of (c.) hence the lemma.Slide23

Statement(Open mapping theorem) (Bounded inverse theorem): A bounded linear operator T from a

Banach

space X onto a

Banach space Y is an open mapping. Hence if T is bijective, T-1 is continuous thus bounded.Slide24

Proof: We prove that for every open set A in X has the image set T(A) open in Y. For this we prove that T(A) is union of open balls and for this we prove that for every y



T(A) there is an open ball about y.

Now let y=TxT(A). Since A is open, therefore it has an open ball with center at x.  A-x contains an open ball with center at 0. Let r be the radius of the open ball & k=1/r. Then k(A-x) contains open unit ball B(0;1). Now apply lemma to k(A-x) we get

T(k(A-x))= k(T(A)-

Tx

) contains an open ball about 0 & so does T(A)-Tx.Hence T(A) contains an open ball about Tx=y.Slide25

Since

y

T

(A) was arbitrary, we get T(A) is open. Finally if T is bijective, i.e. T-1 exists then it is continuous as it is proved to be open. Also T is linear and bounded so T-1 is also linear.

Now since T

-1

is continuous and linear hence it is bounded. Hence the proof of the theorem.Slide26

Closed Graph TheoremSlide27

Definition

(Closed linear operator)

Let X and Y be

normed spaces and T: D(T) Y a linear operator with domain D(T)X. Then T is called closed linear operator if its graph G(T)=(

x,y

)

xD(T), y=Txis closed in the normed space XY, where the two algebriac operations of a vector space in XY are defined by (x1,y1)+(x2+y2)=(x1+x2,y1

+y2) (x,y

)=(x,y) ( a scalar)And the norm on XY is defined by

(x,y) = x + y Slide28

Statement (Closed Graph theorem) : Let X & Y be

Banach

spaces and

T: D(T)  Y a closed linear operator, where D(T)X. Then if D(T) is cloed in X, the operator T is bounded.Slide29

Proof: Now T: D(T)



Y is a closed linear operator.

 by defination G(T)=(x,y)xD(T), y=Tx



is closed in

normed space XY.We show that XY is a Banach space .Let (zn) be a cauchy sequence in XY.zn=(xn,yn). Then for every >0, there is an N s.t. z

n-zm = (x

n,yn) – (

xm,ym)

= (xn-x

m

,yn-ym) = xn-xm + yn-ym <   m,n >NHence (xn) & (yn) are cauchy sequences in X & Y respectively.Also X & Y are complete.Slide30



xX

&

yY s.t. xn  x & y

n

y.

zn=(xn,yn)  (x,y)=z(say)Taking limit m   in (1) we get limm zn-zm = zn – limm z

m = zn – z <

  n > NSince (zn

) was arbitrary cauchy sequence in XY & z

nz  we get XY is complete hence a Banach space.

Now since T is closed

 G(T) is closed in XY & also D(T) is closed in XY. Hence G(T) & D(T) are complete.Now consider the map P: G(T)  D(T) defined by P(x,Tx)=xSlide31

Let (x

1

,Tx

1), (x2,Tx2) G(T) and ,  be scalarsthen P((x1,Tx

1

)+(x

2,Tx2))=P((x1, Tx1)+ (x2,Tx2)) =P((x1,T(x1))+(x2, Tx2)) =P(x1+x

2, T(x1)+T(x

2)) =P(x1+x

2, T(x1+x

2)) = x1+x

2

= P(x1,Tx1)+P(x2,Tx2) P is linear.Now P(x,Tx) = x  x + Tx = (x,Tx)P is bounded.Now clearly by definition P is bijective.P-1 exists & P-1: D(T)  G(T) is given by P-1(x) =(

x,Tx)Slide32

Since G(T) & D(T) are complete we apply open mapping theorem, we say that P

-1

is bounded

i.e.  a real number b>0 s.t. P-1x  b x  x D(T) (

x,Tx

)  b x  x D(T)

So we have Tx  Tx + x = (x,Tx)  b x  x XT is bounded. Hence the proof.Slide33

TEST

Do any two.

State & prove

Baire’s

category theorem.

State & prove Uniform

boundedness theorem.State & prove Open mapping theorem.State & prove Closed graph theorem.