FUNCTIONAL ANALYSIS Baires Category theorem Defination Let X

FUNCTIONAL ANALYSIS

Baire’s Category theorem

Defination Let X be a metric space. A subset M of X is said to be v Rare(Nowhere Dense) in X if its closure M has no interior points, v Meager(of First Category) in X if M is the union of countably many sets each of which is rare in X, v 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. Then X= k Mk with each Mk rare in X. Now M 1 is rare in X, so that, by defination, M 1 does not contain a nonempty open set. But X does(as X is open). This implies M 1≠X. Hence M 1 C=X-M 1 of M 1 is not empty and open.

We may thus choose a point p 1 in M 1 C and an open ball about it, say, B 1=B(p 1; ε 1) M 1 C ε 1<½ By assumption, M 2 is rare in X, so that M 2 does not contain a nonempty open set. Hence it does not contain the open ball B(p 1; ½ ε 1). This implies that M 2 C B(p 1; ½ ε 1) is non empty and open, so that we may choose an open ball in this set, say, B 2=B(p 2; ε 2) M 2 C B(p 1; ½ ε 1) ε 2<½ ε 1 By induction we thus obtain a sequence of balls Bk=B(pk; εk) εk<2 -k Such that Bk Mk= ф and

Since εk<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 Bn B(pm; ½ εm), so that d(pm, p) d(pm, pn)+d(pn, p) < ½ εm +d(pn, p) ½ εm As n . Hence p Bm for every m. Since Bm Mm. 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.

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 x X, then the sequence of norms Tn is bounded.

Proof: We are given that the sequence ( Tnx ) is bounded. For every x X a real number cx 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 Ak 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

Taking limit j in (2) lim Tnxj k (since norm is continuous) Tn(lim xj) k (since each Tn is continuous) T nx k So x Ak, and Ak is closed. By (1) and the defination of Ak we have, each x X belongs to some Ak. Hence X= k. Ak k=1, 2, 3, … Since X is complete, Baire’s theorem implies that some Ak contain an open ball, say, (3) B 0=B(x 0; r) Ak 0 Let x X be arbitrary, not zero.

We set z=x 0+ x =r/2 x (4) Then z-x 0 <r, so that z B 0. By (3) and from the defination of Ak 0 we thus have Tnz k 0 for all n. Also Tnx 0 k 0 since x 0 B 0. From (4) we obtain x=(z-x 0)/. This gives for all n Tnx = Tn(z-x 0) / 2 x ( Tnz + Tnx 0 )/r 2 x (k 0+k 0)/r =(4 k 0 x )/r Hence for all n, Tn = sup x =1 Tnx (4 k 0)/r Hence 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(B 0) of the open unit ball B 0=B(0; 1) X contains an open ball about 0 Y

Proof: We prove the result in following steps: a. The closure of the image of the open ball B 1=B(0; ½) contains an open ball B*. b. T(Bn) contains an open ball Vn about 0 Y, where Bn=B(0; 2 -n). c. T(B 0) contains an open ball about 0 Y.

a. Let A X we write A to mean A= x X x= a, a A ( =2) A A

For w X by A+w we mean A+w= x X x=a+w, a A A a A+w a+w

We consider the open ball B 1=B(0; ½) X. Any fixed x X is in k. B 1 with real k sufficiently large (k>2 x ). Hence X= k k. B 1 k=1, 2, … Since T is surjective and linear, (1) Y=T(X)=T( k k. B 1)= k k. T(B 1) Now since Y is complete, it is non meager in itself, by Baire’s category theorem. at least one k. T(B 1) must contain an open ball. This implies that T(B 1) also contains an open ball, say, B*=B(y 0; ) T(B 1). It follows that (2)B*-y 0=B(0; ) T(B 1) –y 0. This completes with the proof of (a. )

b. We prove that B*-y 0 T(B 0), where B 0 is given in the statement. For this we claim that (3) T(B 1) –y 0 T(B 0). Let y T(B 1) –y 0. Then y+y 0 T(B 1). Also we have y 0 T(B 1) sequences un, vn such that un=Twn T(B 1) such that un y+y 0 vn=Tzn T(B 1) such that v n y 0. Since wn, zn B 1 and B 1 has radius ½, it follows that wn –zn wn + zn < ½+½=1 so wn-zn B 0, so T(wn-zn) =Twn –Tzn =un-vn y+y 0 -y 0=y

Thus y T(B 0). Since y T(B 1) –y 0 was arbitrary, this proves (3). So from (2) we have (4) B*-y 0=B(0; ) T(B 0) Let Bn=B(0; 2 -n) X. Also since T is linear, T(Bn)=2 -n. T(B 0) So from (4) we obtain (5) Vn=B(0; /2 n) T(Bn) This completes the proof of (b. ) c. We finally prove that V 1=B(0; /2) T(B 0) by showing that every y V 1 is in T(B 0). So let y V 1. From (5) with n=1 we have V 1 T(B 1)

Hence y T(B 1). So by defination v T(B 1) such that y-v < /4. Now v T(B 1) so v=Tx 1 for some x 1 B 1. Hence y-Tx 1 < /4. From this and (5) with n=2 we get that y-Tx 1 V 2 T(B 2). As before there is an x 2 B 2 such that (y-Tx 1)-Tx 2 < /8. Hence y-Tx 1 -Tx 2 V 3 T(B 3), and so on. In the nth step we can choose an xn Bn such that (6) y - k. Txk < /2 n+1 k=1, 2, …, n; n=1, 2, … Let zn=x 1+…+xn. Since xk Bk, we have x <1/2 k. This yields for n>m

zn-zm k=(m+1), …, n x < k=(m+1), …, 1/2 k 0 As m . Hence (zn) is a cauchy sequence in X and X is complete so (zn) is convergent. x X such that zn x. Also x = k=1, …, xk k=1, …, 1/2 k=1 So x B 0. Since T is continuous, Tzn Tx, and Tx=y (by (6)) Hence y T(B 0) Since y V 1 was chosen arbitrarily so V 1 T(B 0) 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=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.

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 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) 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 (x 1, y 1)+(x 2+y 2)=(x 1+x 2, y 1+y 2) (x, y)=( x, y) ( a scalar) And the norm on X Y is defined by (x, y) = x + y

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.

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. zn-zm = (xn, yn) – (xm, ym) = (xn-xm, yn-ym) = xn-xm + yn-ym < m, n >N Hence (xn) & (yn) are cauchy sequences in X & Y respectively. Also X & Y are complete.

x X & y Y s. t. xn x & yn y. zn=(xn, yn) (x, y)=z(say) Taking limit m in (1) we get limm zn-zm = zn – limm zm = zn – z < n>N Since (zn) was arbitrary cauchy sequence in X Y & zn 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)=x

Let (x 1, Tx 1), (x 2, Tx 2) G(T) and , be scalars then P( (x 1, Tx 1)+(x 2, Tx 2))=P(( x 1, Tx 1)+ (x 2, Tx 2)) =P(( x 1, T( x 1))+( x 2, Tx 2)) =P( x 1+ x 2, T( x 1)+T( x 2)) =P( x 1+ x 2, T( x 1+ x 2)) = x 1+ x 2 = P(x 1, Tx 1)+ P(x 2, Tx 2) P is linear. Now P(x, Tx) = x x + Tx = (x, Tx) P is bounded. Now clearly by defination P is bijective.

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-1 x b x x D(T) (x, Tx) b x x D(T) So we have Tx + x = (x, Tx) b x x X T is bounded. Hence the proof.

TEST Do 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.