FUNCTIONAL ANALYSIS

Baire’s Category theorem

DefinationLet 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.

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. ThenX=k Mk

with each Mk rare in X.Now M1 is rare in X, so that, by defination, M1 does not contain a nonempty open set. But X does(as X is open). This implies M1≠X. Hence M1

C=X-M1 of M1 is not empty and open.

We may thus choose a point p1 in M1C and

an open ball about it, say, B1=B(p1;ε1) M1

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 M2

C 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) M2

C B(p1; ½ ε1) ε2<½ ε1By induction we thus obtain a sequence of balls Bk=B(pk;εk) εk<2-k

Such that Bk Mk= ф and Bk+1 B(pk; ½ εk) Bk k=1,2,…

Since εk<2-k, the sequence (pk) of the centers is Cauchy and converges, say, pk pX because X is complete by assumption. Also, for every m and n>m we have BnB(pm; ½ εm), so that d(pm,p)d(pm,pn)+d(pn,p) < ½ εm +d(pn,p) ½ εm As n . Hence pBm for every m. Since BmMm

C ,we now see that pMm for every m, so that pMm=X. this contradicts pX. Hence X in not meager i.e. X is nonmeager in itself.

Uniform Boundedness Theorem

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

bounded for every xX, then the sequence of norms Tn is bounded.

Proof: We are given that the sequence ( Tnx ) is bounded. For every xX a real number cx such that(1) Tnx cx n=1,2,…For every kN, let AkX be the set of all x such that Tnx k for all n.We claim that Ak is closed.Let xAk, then there is a sequence (xj) in Ak converging to x. This means that for every fixed n we have (2) Tnxj k

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 xAk, and Ak is closed.By (1) and the defination of Ak we have, each xX belongs to some Ak. Hence X=kAk 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 xX be arbitrary, not zero.

We set z=x0+x =r/2 x (4)Then z-x0 <r, so that zB0. By (3) and from the defination of Ak0 we thus have Tnz k0 for all n. Also Tnx0 k0 since x0B0. From (4) we obtain x=(z-x0)/.This gives for all n Tnx = Tn(z-x0) / 2 x ( Tnz + Tnx0 )/r 2 x (k0+k0)/r =(4k0 x )/rHence for all n, Tn = sup x =1 Tnx (4k0)/rHence the sequence of norms Tn is bounded.

Open Mapping Theorem

Defination(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 Y

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 0Y

Proof: We prove the result in following steps:

a. The closure of the image of the open ball B1=B(0;½) contains an open ball B*.

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

c. T(B0) contains an open ball about 0Y.

a. Let AX we write A to mean

A=xX x= a, aA

AA

(=2)

For wX by A+w we mean

A+w=xXx=a+w, aA

A

A+w

a

a+w

We consider the open ball B1=B(0; ½)X. Any fixed xX is in kB1 with real k sufficiently large (k>2 x ). Hence X=k kB1 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(y0;) T(B1). It follows that(2)B*-y0=B(0; ) T(B1) –y0. This completes

with the proof of (a.)

b. We prove that B*-y0T(B0), where B0 is given in the statement. For this we claim that(3) T(B1) –y0 T(B0).Let yT(B1) –y0. Then y+y0 T(B1).Also we have y0T(B1)sequences un, vn such that un=Twn T(B1) such that un y+y0 vn=Tzn T(B1) such that vn y0. Since wn, znB1 and B1 has radius ½, it follows that wn –zn wn + zn < ½+½=1 so wn-znB0, so T(wn-zn) =Twn –Tzn =un-vn y+y0-y0=y

Thus yT(B0). Since yT(B1) –y0 was arbitrary, this proves (3). So from (2) we have(4) B*-y0=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) Vn=B(0;/2n) T(Bn)This completes the proof of (b.)

c. We finally prove that V1=B(0; /2) T(B0)by showing that every yV1 is in T(B0). So

let yV1.From (5) with n=1 we have V1T(B1)

Hence y T(B1). So by defination vT(B1) such that y-v </4.Now vT(B1) so v=Tx1 for some x1B1.Hence y-Tx1 < /4.From this and (5) with n=2 we get that y-Tx1V2 T(B2). As before there is an x2B2 such that (y-Tx1)-Tx2 < /8.Hence y-Tx1-Tx2V3T(B3), and so on. In the nth step we can choose an xnBn 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>m

zn-zm k=(m+1),…,n x < k=(m+1),…, 1/2k 0As m . Hence (zn) is a cauchy sequence in X and X is complete so (zn) is convergent. xXsuch that zn x. Also x = k=1,…, xk k=1,…, xk k=1,…, 1/2k=1So xB0 . Since T is continuous, Tzn Tx, and Tx=y (by (6))Hence yT(B0)Since yV1 was chosen arbitrarily so V1T(B0)This completes the proof of (c.) hence the lemma.

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.

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=TxT(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.

Since yT(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.

Closed Graph Theorem

Defination (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)xD(T), y=Txis closed in the normed space XY, where the two algebriac operations of a vector space in XY are defined by (x1,y1)+(x2+y2)=(x1+x2,y1+y2) (x,y)=(x,y) ( a scalar)And the norm on XY is defined by (x,y) = x + y

Statement (Closed Graph theorem) : Let X & Y be Banach spaces andT: D(T) Y a closed linear operator, where D(T)X. Then if D(T) is

cloed in X, the operator T is bounded.

Proof: Now T: D(T) Y is a closed linear operator. by defination G(T)=(x,y)xD(T), y=Txis closed in normed space XY.We show that XY is a Banach space .Let (zn) be a cauchy sequence in XY.zn=(xn,yn). Then for every >0, there is an N s.t. zn-zm = (xn,yn) – (xm,ym) = (xn-xm,yn-ym) = xn-xm + yn-ym < m,n >NHence (xn) & (yn) are cauchy sequences in X & Y respectively.Also X & Y are complete.

xX & yY s.t. xn x & yn y.Þzn=(xn,yn) (x,y)=z(say)ÞTaking limit m in (1) we get limm zn-zm = zn – limm zm = zn – z < n > NSince (zn) was arbitrary cauchy sequence in XY & znz we get XY is complete hence a Banach space.Now since T is closedÞ G(T) is closed in XY & also D(T) is closed in XY. Hence G(T) & D(T) are complete.Now consider the map P: G(T) D(T) defined by P(x,Tx)=x

Let (x1,Tx1), (x2,Tx2) G(T) and , be scalarsthen P((x1,Tx1)+(x2,Tx2))=P((x1, Tx1)+ (x2,Tx2)) =P((x1,T(x1))+(x2, Tx2)) =P(x1+x2, T(x1)+T(x2)) =P(x1+x2, T(x1+x2)) = x1+x2 = P(x1,Tx1)+P(x2,Tx2) P is linear.Now P(x,Tx) = x x + Tx = (x,Tx)P is bounded.Now clearly by defination P is bijective.P-1 exists & P-1: D(T) G(T) is given by P-1(x) =(x,Tx)

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 XÞT is bounded. Hence the proof.

TESTDo any two.

1. State & prove Baire’s category theorem.

2. State & prove Uniform boundedness theorem.

3. State & prove Open mapping theorem.

4. State & prove Closed graph theorem.