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# Linear Analysis MATH 675

Mason

GPA 3.85

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This 5 page Class Notes was uploaded by Michale Kuhlman on Monday September 28, 2015. The Class Notes belongs to MATH 675 at George Mason University taught by Staff in Fall. Since its upload, it has received 31 views. For similar materials see /class/215007/math-675-george-mason-university in Mathematics (M) at George Mason University.

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Date Created: 09/28/15

Lecture 7 The HahnBanach Theorem Hyperplanes in Linear Spaces Recall from linear algebra the de nition of a plane in R3 Such a plane was determined by a point in the plane 950 yo 20 and a vector normal to the plane Then an arbitrary point x y 2 in the plane must satisfy lt13 7 10377 0 or equivalently lt13 3 10377 where f ltxy2gt and p 950310213 Of course this de nition can be extended to R where the same formula is used but all the vectors are nvectors The question is How do we extend the notion of a hyperplane to in nite dimensional linear spaces where we have no notion of orthogonality or of inner product77 We make the following observations about the R case 1 lt93 4 de nes a linear functional on R 2 The set WW1 931 lt93 77 0 51103 0 is a linear subspace of R 3 The hyperplane P through p orthogonal to 77 can be written P pi nullf 55155 pa 3 g nullf 4 Since all linear functionals on R can be written as lt55 4 for some 77 E R7 all hyperplanes in R can be written as P p nullf for some linear functional f and some xed vector p De nition 59 A subspace L of a real linear space L has codiinension 1 there exists a vector we 6 L we y 0 such that every x E L can be written uniquely aszazoyf0rsomea R andyEL Remark 60 Let f be a linear functional on a real linear space L Then by a previous theorem nullf has codiinension 1 De nition 61 Let L be a codimension 1 subspace of the linear space L A set M Q L is a hyperplane parallel to L there exists an element 10 E L such that MLzo z1zoy yELl note the typo on p 127 of Kohnogorov Theorem 62 31 a Let f lie a nonzero linear functional on a linear space L Then the set M zltzgt1 is a hyperplane parallel to L nullf b Let M L we lie any hyperplane parallel to L not containing 0 that is such that we E U Then there exists a unique nonzero linear functional f on L such that M 1 Proof a We must show that Mf L x0 for some 950 E L Let 950 be any vector such that fx0 1 such an 950 always exists since f is not the identically zero functional We will show that this 950 is the one we want Let x E Mf Any vector of the form x x0 y where y E L is in Mf since x mo y me y mo 1 Therefore L we g Mf Let x E Mf Q Since L is the nullspace of a linear functional x can be written at axe y for some oz E R and y E L But since at E Mf 1 x are y 060 y a so that x E L 950 as required b We would like to show rst that every x E L can be written uniquely as x axe y for some oz E R an y E L In order to show this note rst that since H has codirnension 1 there exists yo E L such that every x E L can be written uniquely as x yo y where E R and y E L Since 950 E L we have that for some H E R and y E L 950 yo y and y 0 since 950 E L Therefore yo xo E y and any x E L can be written as x yo y 960 ye 32 axoy where oz E R and y E L It is easy to show that this representation is unique De ne the linear functional f on L as follows Given x E L write x axe y with y E L and let a It is easy to verify that f is linear Also at E M if and only if x x0 y for some y E L if and only if 1 a The functional f is unique for suppose that g were another such func tional Then since 950 E M and g is identically 1 on M 9950 1 If y E D then y has the unique representation as y axe y where oz 0 Thus gy 0 Thus for any x E L 995 90050 y oz El Convex Functionals and The HahnBanach Theorem 32 De nition 63 Let L lie a real vector space The functionalp de ned on L is a convex functional ifp satis es Maw By S 04101 51021 for all z y E L and all real numlzers 0 g a B g 1 with a B 1 Clearly any linear functional is also a convex functional Compare this de nition With Kolinogorov p 130 De nition 3 Note that if L R then a convex functional on L is just a function that is concave upward at every x E Rt Theorem 64 HahnBanach Theoremi Suppose thatf is a linear functional de ned on a sulzspace L of L which satis es g pz Vz E Li Then there is a linear functional F de ned on L such that 1 for all z E L that is F is an extension of f and 2 g pz for all z E L Proof N Claim 1 Given 2 E L L we can nd an extension f of f from L to the subspace a2 y y E L Proof of Claim 1 Let 2 E L L5 The extension f Will be completely determined by the value of because az y a z y a z m We must only ensure that S px for all x E 12 y y E D That is we must have for all oz gt 0 and y E L fa2y a 2fy pyaz iaztry ia2fy 10yazl Rearranging this becomes any 7m 7 a2 m s 5pm a2 7 y forallozgt0andyELh Now for any 1 gt 0 and y1 y E L fy1 afy2 fwyi 0m oz a f lta y1a y2gt 37 Lecture 9 Banach Spaces De nition 71 A normed linear space is complete it is complete as a metric space in the metric induced by the norm A complete normed linear space is called a Banach space Theorem 72 Let T V1 a V2 lie a lzounded linear transfor7 mation and suppose that V2 is complete Then T can lie extended uniquely to a lzounded linear transformation Tfrom the completion of V1 71 to V2 in such a way that extends T means that for all 1 E V1 Tv Proof Let 71 be the completion of V1 CLAIM 1 71 is a normed linear space With the obvious de nition of addition and scalar multiplication and With the norm de ned by lim anlll Where 27 is a Cauchy sequence in V1 converging to v E 71 We Will use H H1 also to denote the norm on V1 We must de ne T Let x E V1 and let be a Cauchy sequence in V1 converging to x We must show that the sequence TQM is Cauchy in V2 To see this let 6 gt 0 Then there is an N so that if n m 2 N then Horn 7 acmHl lt Then if n m 2 N HTW Txml2 HTW xml2 S HTHHM xmlll lt 6 Therefore is Cauchy in V2 and hence converges in De ne To limrm CLAIM 2 The de nition of T is independent of the choice of the approxi mating sequence Proof of Claim 2 Let x E 71 and suppose that and are Cauchy sequences in V1 converging to at It is easy to see that the sequence xniyn is Cauchy in V1 and converges to 0 E V1 Since T is bounded Txn7yn A 0 in V2 Now suppose that 21 limTxn and 22 lim Then H21 22H2 S H21 TInl2 HTW Tynl2 TU7 22 Since each term on the right side of the inequality can be made as small as desired H21 7 22 0 or 21 22 CLAIM 3 T is linear CLAIM 4 T is bounded withNHTH HT Proof of Claim 4 Let x E V1 and let be a Cauchy sequence in V1 converging to 95 Then H x b 1imlTxnll2 S limHTHHxnlh HTHHIHL 38 This shows also that S To see that in fact equality holds note that N T T T T sup H IH2 2 sup H IH2 sup H IH2 HTH EE Z HxH1 zevl HxH1 zevl HxH1 Thus HTH 2 HTH SO HTH HTH Claim 5 The extension T is unique N N Proof of Claim 5 Suppose that there were two extensions T1 and T21 Let x E V1 and let be a Cauchy sequence in V1 converging to 95 Since T1 and T2 are both extensions of T Em Tm Tam and since limits in metric spaces are unique T1 at limrm mm The notion of the extension of an operator from a dense subset of a space to the whole space is very important in concrete situationsi Often an operator can be de ned easily only on a dense subspace of the space you are interested in If you can get the proper bound on the operator then it extends to a continuous operator in the whole space ie Corollary 73 Let V1 1 V2 lie normed linear spaces with V2 complete Let B Q V1 lie a dense sulzspace and suppose that there is a linear operator de ned on B such that there is a constant C gt 0 such that for all 1 E B Tv2 g CUH1 Then T can lie extended uniquely to a lzounded linear operator from the completion ofV1 to V2 Proof The proof follows from Theorem 72 and the observation that the completion of B is the same as the completion of V1 Example 74 a Let C0 1 be equipped with the norm HfH1 WW We have seen that this space is not complete The space L1 0 1 is de ned to be the completion of C0 1 in this normi The question remains What to objects in L1 look like We have de ned them as Cauchy sequencesi ls there any sense in which these objects are actually functions with point values We will see the answer to this when we discuss the Lebesgue measure and Lebesgue integrali

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