1 / 33

360 likes | 587 Views

Functional Analysis. Baire’s Category theorem. 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,

Download Presentation
## Functional Analysis

**An Image/Link below is provided (as is) to download presentation**
Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.
Content is provided to you AS IS for your information and personal use only.
Download presentation by click this link.
While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.
During download, if you can't get a presentation, the file might be deleted by the publisher.

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.

More Related