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.

Proof: Suppose ф ≠X is a complete metric space such that X is meager in itself. Then X=  k M k with each M k 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 B k =B(p k ; ε k ) ε k <2 -k Such that B k  M k = ф and B k+1  B(p k ; ½ ε k )  B k k=1,2,…

Since ε k m we have B n  B(p m ; ½ ε m ), so that d(p m,p)  d(p m,p n )+d(p n,p) < ½ ε m +d(p n,p)  ½ ε m As n . Hence p  B m for every m. Since B m  M m C,we now see that p  M m for every m, so that p  M m =X. this contradicts p  X. Hence X in not meager i.e. X is nonmeager in itself.

Proof: We are given that the sequence ( T n x ) is bounded.  For every x  X  a real number c x such that (1) T n x  c x n=1,2,… For every k  N, let A k  X be the set of all x such that T n x  k for all n. We claim that A k is closed. Let x  A k, then there is a sequence (x j ) in A k converging to x. This means that for every fixed n we have (2) T n x j  k

Taking limit j  in (2) lim T n x j  k lim T n x j  k (since norm is continuous) T n (lim x j )  k (since each T n is continuous) T n x  k So x  A k, and A k is closed. By (1) and the defination of A k we have, each x  X belongs to some A k. Hence X=  k A k k=1,2,3,… Since X is complete, Baire’s theorem implies that some A k contain an open ball, say, (3) B 0 =B(x 0 ;r)  A k0 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 A k0 we thus have T n z  k 0 for all n. Also T n x 0  k 0 since x 0  B 0. From (4) we obtain x=(z-x 0 )/ . This gives for all n T n x = T n (z-x 0 ) /   2 x ( T n z + T n x 0 )/r  2 x (k 0 +k 0 )/r =(4k 0 x )/r Hence for all n, T n = sup x =1 T n x  (4k 0 )/r Hence the sequence of norms T n is bounded.

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

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(B n ) contains an open ball V n about 0  Y, where B n =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  A AA (  =2)

For w  X by A+w we mean A+w=  x  X  x=a+w, a  A  A A+w a a+w

We consider the open ball B 1 =B(0; ½ )  X. Any fixed x  X is in kB 1 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 kB 1 )=  k kT(B 1 )=  k kT(B 1 ) Now since Y is complete, it is non meager in itself, by Baire’s category theorem.  at least one kT(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 u n, v n such that u n =Tw n  T(B 1 ) such that u n  y+y 0 v n =Tz n  T(B 1 ) such that v n  y 0. Since w n, z n  B 1 and B 1 has radius ½, it follows that w n –z n  w n + z n < ½+½=1 so w n -z n  B 0, so T(w n -z n ) =Tw n –Tz n =u n -v n  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 B n =B(0;2 -n )  X. Also since T is linear,  T(B n )=2 -n T(B 0 ) So from (4) we obtain (5) V n =B(0;  /2 n )  T(B n ) 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 n th step we can choose an x n  B n such that (6) y -  k Tx k <  /2 n+1 k=1,2,…,n; n=1,2,… Let z n =x 1 +…+x n. Since x k  B k, we have x k m

z n -z m   k=(m+1),…,n x <  k=(m+1),…,  1/2 k  0 As m  . Hence (z n ) is a cauchy sequence in X and X is complete so (z n ) is convergent.  x  X such that z n  x. Also x =  k=1,…,  x k   k=1,…,  x k   k=1,…,  1/2 k =1 So x  B 0. Since T is continuous, Tz n  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.

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 the theorem.

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

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 (z n ) be a cauchy sequence in X  Y.  z n =(x n,y n ). Then for every  >0, there is an N s.t. z n -z m = (x n,y n ) – (x m,y m ) = (x n -x m,y n -y m ) = x n -x m + y n -y m N Hence (x n ) & (y n ) are cauchy sequences in X & Y respectively.Also X & Y are complete.

 x  X & y  Y s.t. x n  x & y n  y.  z n =(x n,y n )  (x,y)=z(say)  Taking limit m   in (1) we get lim m z n -z m = z n – lim m z m = z n – z N Since (z n ) 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)=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.  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 -1 x  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.

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.