121 Views

Download Presentation
## Functional Analysis

- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -

**Defination**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.**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 defination, 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.**We may thus choose a point p1 in M1C and an open ball about**it, say, B1=B(p1;ε1) M1Cε1<½ By 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(p1; ½ ε1) ε2<½ ε1 By 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 BmMmC ,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.**Statement: Let (Tn) be a sequnce of bounded linear operators**Tn:XYfrom 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)**limTnxj k limTnxj k (since norm is continuous) Tn(limxj) k (since each Tn is continuous) Tnx k So 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)Ak0 Let 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 )/r Hence for all n, Tn = sup x =1Tnx (4k0)/r Hence the sequence of norms Tn is bounded.**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:**The closure of the image of the open ball B1=B(0;½) contains an open ball B*. T(Bn) contains an open ball Vn about 0Y, where Bn=B(0;2-n). T(B0) contains an open ball about 0Y.**a. Let AX we write A to mean**A=xX x= a, aA (=2) A A**For wX by A+w we mean**A+w=xXx=a+w, aA A a A+w 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)= kkT(B1)= kkT(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 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 • 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 0 As m . Hence (zn) is a cauchy sequence in X and X is complete so (zn) is convergent. xX such that zn x. Also x = k=1,…, xk k=1,…,xk k=1,…, 1/2k=1 So 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.**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=Tx is 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=Tx • is 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 >N • Hence (xn) & (yn) are cauchy sequences in X & Y respectively.Also X & Y are complete.** xX & yYs.t. xn x & yn y.**• zn=(xn,yn) (x,y)=z(say) • Taking limit m in (1) we get • limmzn-zm = zn – limmzm • = zn – z < n > N • Since (zn) was arbitrary cauchy sequence in XY & znz we get XY is complete hence a Banach space. • Now since T is closed • 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 scalars**• then • 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 X T is bounded. Hence the proof.**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.