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# LINEAR ANALYSIS MATH 554

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This 12 page Class Notes was uploaded by Addison Beer on Wednesday September 9, 2015. The Class Notes belongs to MATH 554 at University of Washington taught by Anne Greenbaum in Fall. Since its upload, it has received 27 views. For similar materials see /class/192052/math-554-university-of-washington in Mathematics (M) at University of Washington.

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Date Created: 09/09/15

arm S Norms A norm is a way of measuring the length of a vector Let V be a vector space A norm on V is a function V gt 0 00 satisfying i Va 6 V v 2 0 and v 0 if 0 ii Va 6 E V a and iii triangle inequality Vega 6 V l g l The pair V2 H is called a normed linear space or normed vector space Fact A norm on a vector space V induces a metric d on V by diam H39v va Exercise Show that d is a metric on V All topological properties eg open setsv closed setsv convergence of sequences continuity of functions compactnessv etc will refer to those of the metric space K d Examples 1 1 norm on 15 quot 1 g p g 30 a p 00 llwllx gigging 10 6 15 quot b 1 S p lt 00 llwllp El21 The triangle inequality 71 Z 39lquot 311 S izl is known as lvlinkowski s inequality It is a consequence of Holder s inequality Integral versions of these inequalities are proved in real analysis texts eg Folland or Royden The proofs for vectors in 16 are analogous to the proofs for integrals 3 y 12 1 S I lt 00 30 Related observation for 1 3 p lt gt0v the map so r gt soquot for x Z 0 is convex c 0 ltplt 1 2quot l 7 is not a norm on 16 quot If as 61 and y 62 26 Linear Algebra and Alatz ix Analysis so the triangle inequality does not hold 3190 0ltplt1 kw Related observation for 0 lt p lt 1 the map so r gt soquot for x Z 0 is not convex 2 EV norm on W subspace of IF 1 g p 3 gt0 a p 00 53 6 IE 30 supi21 lt 00 supi21 for L39 6 53 b 1 s p lt co x err 231 lt co user 21wvgt for L39 E 1 Exercise Show that the triangle inequality follows from the nite dimensional case 3 LP norm on CavbD 1 g p g 00 a p 00 Hfo supagmgb 1190 Since fL39 is a continuous real valued function on the compact set a b it takes on its maximum so the sup is actually a max here Hfo 1133 Sim l b 1 s p lt co pr 11an Use continuity of f to show that f 0 gt 90 E 0 on cub The triangle inequality L Sifts gipdxp s worm b gx 7 dx is lvlinkowski s inequalityg a consequence of Holder s inequality c 0 lt p lt 1 fz397 dwgtp is not a norm on 0mm Pseudo example Let a 0 b 1 H 103 H 00 dbl 0 ltwgl 311C19 1 01 Emma 01 Emmi l IOhI Then A T to l L V El 5 a v A ml l Q V A EH 3 v A s V Elia arms 27 so the triangle inequality fails Here f and g are not continuous Exercise Adjust these f and g to be continuous eg k to construct a legitimate counterexample to the triangle inequality Remark There is also a Minkowski inequality for integrals if 1 g p lt 00 and u E Cla b X 0 dl then b d P i d b i f Mailulu dz Sf ivaulipdw dz 1 C C 1 Continuous Linear Operators on Normed Linear Spaces Theorem Suppose V and W W are normed linear spaces and L V gt W is a linear transformation Then the following are equivalent a L is continuous 19 L is uniformly eontinuous e 3 C so that Vu E V LuHW g Proof a gt c Suppose L is continuous Then L is continuous at u 0 Let 6 1 Then 35 gt 0 so that if g 5 then LuHW g 1 as L0 0 For any u 0 i g 5 so HL g 1 ie LuHW g Let C 0 gt b 81199030 W 6 V HL39UHW S CHUva The WW2 6 V HL39Ul L39U2Hw Lu1 U2HW g Cu1 Hence L is uniformly continuous given 6 let 5 etc In fact L is uniformly Lipschitz continuous with Lipschitz constant C bgta is immediate De nition If L V gt W is a linear operator where V and W are normed linear spaces and supvelaw0 lt gt0 then L is called a bounded linear operator from V to W Remarks 1 Note that it is the norm ratio or stretching factor that is bounded not llL UllW t U E Vlv H ml w l M l v Exercise Show that if 3 K Vu E VLuw 3 K then L E 0 2 The theorem above says that if V and W are normed linear spaces and L V gt W is linear then L is continuous 4 L is uniformly continuous ltgt L is a bounded linear operator 28 Linear Algebra and Alatz ix Analysis De nition If V and W are normed linear spaces and L V gt W is a bounded linear operator de ne the operator norm of L to be LU w llLll sup H 176VU 0 Remarks 1 There are other equivalent de nitions of the operator norm L 3 HL UHW 176VH 17HV1 L 311 HL39UHW UEVZHVHVSl minC Va 6 V L UHW g ie is the smallest stretching factor upper bound C Exercise Show these are equivalent to the de nition above 2 The most common use of the operator norm is the obvious but powerful inequality Va 6 V L39UHW S 39 Equivalence of Norms Lemma If V is a normed linear space then V gt R is continuous Proof For v1v2 E V v2 v2 g and thus g 3981 U2H Similarly 3982 3981 S 3982 39le 3981 v2 30 ill39vlll HWHE S 3981 U2H Given 6 gt 0 let 5 6 etc De nition Two norms 1 and 2 both on the same vector space V are called equivalent norms on V if 3 constants C1C2 gt 0 for which Va 6 V g g C2v2 Remarks 1 Two norms 0 and 3 on V are equivalent i quot the identity map I V gt V 3 is bicontinuous vHs S C1H39UHQ gt I V 0 gt V 3 is continuous and lt gt I V gt V is continuous 2 Equivalence of norms denoted temporarily by is an equivalence relation on the set of all norms on a xed vector space V i 0 w 0 ii 0 w 3 i quot H 39 6 w H 39 Ha and iii if H 39 Ha w H 39 Us and H 39 6 w H 39 Hm then H 39 Ha w H 39 Her For nite dimensional vector spaces V all norms are equivalent arms 29 The Norm Equivalence Theorem If V is a nite dimensional vector space then any two norms on V are equivalent Proof Fix a basis for V and identify V with 16 quot E V lt gt L39 6 Iquot where v 2 royal Using this identi catiom we can restrict our attention to IFquot Let 1 231 wi 22 denote the euclidean norm reg 2 norm on 16 quot Because equivalence of norms is an equivalence relation it suf ces to show that any given norm on 16 quot is equivalent to the euclidean norm For as 6 IFquot g Ely1 6i 3 i i 21 wi 22 Ely1 2 by the Schwarz inequality in Rquot Thus S Moi 1W 2 211 6139 2 Thus the identity map I 16m gt 15 is continuous which is half of what we have to show Composing the map with 16m gt R which is continuous by the preceding Lemmav we conclude that 15 gt R is continuous Let S w 6 IFquot 1 Then S is compact in F3 g and thus takes on its minimum on S which must be gt 0 since 0 g 3 Let m minwzw 1 gt 0 Hence if where 6i 1 then 2 m For any 10 6 15 quot with L39 0 1 so 2 my ie g So and are equivalent Remarks 1 All norms on a xed nite dimensional vector space are equivalent Be careful thougln when studying problems eg in numerical PDE where there is a sequence of nite dimensional spaces of increasing dimensions the constants C1 and C2 in the equivalence can depend on the dimension eg g in 2 The Norm Equivalence Theorem is not true in in nite dimensional vector spaces 3 It can be shown that for a normed linear space V2 the closed unit ball v E V v 3 1 is compact i dim V lt 0 Examples 1 0111630 2 6 15m 3NVn Z N 10quot 0 for 1 g p lt q 3 03 the E norm and E norm are not equivalent We show the case p 1 q gt0 First note that g so I 1332 gt 1382 is continuous But if y1130303y2 131 33 1313103ietc3 then ynHoc 1Vn but lynHi n So there does not exist a constant C for which Vw 6 F30 le 3 CHLCHOC A M v On Cav 0 for 1 g p lt q 3 00 the LP and Lquot norms are not equivalent We will show the case p Li 00 here g b so I CavbDv gt Cav 0 is continuous Remark Since the integral 1a jjuwdw is clearly continuous on Cav 12 1 since Iu1 Iu2 g b ja gu1 L u2 gdw u1 u21v composition of these two continuous operators implies the standard result that if an gt u uniformly on a b then gt WLOG assume a 0 b 1 Let uquot be 30 Linear Algebra and Alatz ix Analysis Then unHl 1 but uon 71 So there does not exist a constant C for which Va 6 Claablll Hullx S Cllulllv 3 1er2 subspace of 16 with norm pr 239 the closed unit ball w 6 2 LL39Hp g 1 is not conipact The sequence 61362 is bounded el 3 1 and all are in the closed unit ball but no subsequence converges because ei e 2 2 v5 for 1j Exercise Does the sequence 61 e2 63 converge weakly in E2 A sequence is said to converge weakly to w in E2 if V31 6 2 gym 3 gt any where 3 is the inner product on 52 Norms induced by inner products Let V be a vector space and 1 be an inner product on V De ne v viv By the properties of an inner product HUN 2 0 with v 0 if v 0 and V or e IE Vv E V a To show that is actually a norm on V we need the triangle inequality We begin by rst showing the Cauchy Schwarz inequality LemmaThe CauchySchwarz inequality For all by w E V we have g Moreoven we have equality if v and w are linearly dependent This latter statement is sometimes called the converse of Cauchy Schwarz Proof Case i If v 0 or w 0 clear Case ii If 1 and Z 0 then 0 g 2Rev3w 21 so g 1 with equality iffy Case iii For any v 0 and w 0 choose or 6 IE with gag 1 and 04v 111 2 0 Let v1 my and ohmJ 7 v lt 1 H Ull39llwll lt 139 1 1111 H37 Then v1 w1 1 and39v1w1 2 0 so with equality if v1 Exercise In case iii of the above proofv show by w are linearly dependent iflquot v1 1111 arms 31 Now the triangle inequality follows 2Rev2w S H39b39ll2 2iltvawgt H wll2 S H39UH2 QH39UH 39 H wll H wll2 ll39vll Hull So V is a norm on V It is called the norm induced by the inner product An inner product induces a norm which induces a metric V2 lt gt V2 lt gt V2 d Examples 1 The Euclidean norm ie E2 norm on 16 quot is induced by the standard inner product lt90 31gt 231 llwll2 231 n 2 11 v 2 Let A 6 FMquot by Hermitian symmetric and positive de nite and let n n for 6 IFquot 11 31 Then is an inner product on 15 quot which induces the norm 11 31 Remark An alternate convention is to de ne so 31 to be 231 221 Eat3w y in which case VLL39HALL39 3 The E2 norm on E2 subspace of 16 is induced by the inner product 231 21 1wa V3321 21 l 4 The L2 norm uw 2dw2 on Clavbl is induced by the inner product ltui vgt lineman Closed unit balls v E V g 1 in nite dimensional normed linear spaces V Example For E norms in R2 1 g p g 00 7 p22 7 p21 De nition A subset C of a vector space V is called convex if VuweCXVtelU l tv1 tuEC 32 Linear Algebra and Alatz ix Analysis Remarks 1 This means that the line segment joining v and w is in C if v and w are in C 111 is on this line segment 2 The linear combination to 1 for t 6 l0 1 is often called a convex combination of v and Let B v E V v S 1 denote the closed unit ball in a nite dimensional normed linear space Facts 1 B is convex 2 B is compact 3 B is symmetric if v E B and a 6 IE with a 1 then ow E B 4 The origin is in the interior of B Lemma If dim V lt 0 and B C V satis es the four conditions in the statement of facts above then there is a unique norm on V for which B is the closed unit ball in infc gt 0 e B Remark The condition that 0 be in the interior of a set is independent of the norm by the norm equivalence theorenn all norms induce the same topology on V ie have the same collection of open sets Exercise Show that the object de ned in the lemma above does indeed de ne a nornn and that B is its closed unit ball The uniqueness of this norm follows from the fact that in any normed linear space v infc gt 0 e B where B is the closed unit ball B v v 3 1 Hence there is a one to one correspondence between norms on a nite dimensional vector space and subsets B satisfying the four conditions stated above arms 33 Completeness Completeness in a normed linear space V means completeness in the metric space V d where duw u wH every Cauchy sequence in V ie V6 gt 03NVnm Z N lt 6 has a limit in V ie Eu 6 gt 0 as n gt 00 Example Iquot endowed with the euclidean norm 27 up is complete Topological properties are those which depend only on the collection of open sets eg open closed compact whether a sequence converges etc Completeness is not a topological property Example Let f 1 00 gt 01 be given by f with the usual metric on R Then f is a homeomorphism bijective bicontinuous but 100 is complete while 01 is not complete Completeness is a uniform property Theorem If Xp uncl Yo ure metric spaces and 99Xp gt Yo is u uniform homeomorphism bijectiue bicontinuous and go and 99 1 are both uniformly continuous then Xp is complete i Yo is complete The key step in the proof of this theorem is to show that if go X gt Y is a uniform homeomorphism then 99 preserves Cauchy sequences ie a sequence is Cauchy in X p iflquot 99Xn is Cauchy in Y 0 Since bounded linear operators between normed linear spaces are automatically uniformly continuous several facts follow immediately Corollary If two norms 1 and 2 on a vector space V are equivalent then V 1 is complete iflquot V 2 is complete Corollary Every nite dimensional normed linear space is complete Proof If dim V n lt gt0 choose a basis of V and use it to identify V with 16 Since 16 is complete in the euclidean norm the corollary follows from the norm equivalence theorem But not every in nite dimensional normed linear space is complete De nition A complete normed linear space is called a Banach space An inner product space for which the induced norm is complete is called a Hilbert space Examples To show that a normed linear space is complete we must show that every Cauchy sequence converges in that space The basic strategy for showing that a space is complete is a three step process that can be described as follows given a Cauchy sequence i construct what you think is its limit ii show the limit is in the space V iii show the sequence converges to the limit in V 1 Let M be a metric space Let CM denote the vector space of continuous functions u M gt 16 Let CHM denote the subspace of CM consisting of all bounded continuous functions CAM u E CM 3 KVL e M uc 3 K On 0M de ne the sup norm supme M 34 Linear Algebra and Alatz ix Analysis Fact CM is complete Proof Let em C CM be Cauchy in Given 6 gt 0 EN so that Vnv m 2 N un um lt e For each x E M unw umw 3 un umv so for each x E M is a Cauchy sequence in E which has a limit in IE which we will call since IE is complete limH30 Given 6 gt 0 3NVn b Z NVL39 E M unw umL39 lt 6 Taking the limit for each xed w as m gt gt0v we get V71 2 NVL39 E M g 6 Thus an gt u uniformly so u is continuous since the uniform limit of continuous functions is continuous Clearly u is bounded choose N for 6 1 then Vac E M g 1 so u E CM And now we have un u gt 0 as n gt gt0v ie uquot gt u in CoMv H I W is complete for 1 g p 3 0 p 00 This is a special case of 1 where M N 132 3 Given 1 3 p lt gt0 Let be a Cauchy sequence in E write wk 2 aucbe e gt 0 3 KVk 2 K 90c L39 lt e For each m E N no i semi S ME Hack willa izl so for each m E N is a Cauchy sequence in 16 which has a limit let am limH30 Let x be the sequence x a13a2va3 so far we just know that L39 6 16 Given 6 gt 0 3 KV 2 K lt 6 Then for any N and for 1 1 2 K mi p lt 6 taking the limit as E gt 00 at p g l 6 then taking the limit as N gt 00 til7 g 6 Thus L39K L39 6 72 so also 10 xx 6 72 and we have th 2 K L39 g 6 Thus L39 gt 0 as k gt gt0v ie wk gt w in E I If M is a compact metric space then every continuous function u M gt IE is bounded so C M CM In particular CM is complete in the sup norm u supmeM uL39 special case of 1 For example ClavbJ is complete in the L30 norm For 1 g p lt 00 Cla b is not complete in the LP norm 1 Example On 031 let an be g 1 g 1 n Then an E ClU 1 Exercise Show that an is Cauchy in Up We must show that there does not exist a u E C031 for which a gt 0 Exercise Show that if u E C031 and u gt 0 then and E 1 for i lt L39 g 1 contradicting the continuity of u at L39 0for0 wlt l 2 Norms 35 5 133 6 IE 30 3NVn Z N 70quot 0 is not complete in any 1 norm 1 g p 3 gt0 This can be shewn using the sequences described below 1 3 p lt gt0 Cheese any x e PIE 3 and censider the truncated sequences 31 1010 32 x1x20331w2w370netc Exercise Show that is Cauchy in 138 p but that there is ne 3 6 133 for which LUMP gt 039 p gt0 Sanie idea cheese any x E l g for which linin 0 and censider the sequence of truncated sequences Completion of a Metric Space Fact Let X p be a nietricspace Then there exists a complete metric space X 5 and an inclusion map 1 X gt X for which i is injective i is an isenietry from X to ie Vang E X and iXis dense in X Moreover all such X43 are isenietrically isenierphic The metric space X 5 is called the completion of X 0 One way to censtruct such an X is to take equivalence classes of Cauchy sequences in X to be elenients ef X Representations of Completions In SOIIIC situatiens the cenipletien of a metric space can be identi ed with a larger vector space which actually includes X and whose elements are objects of a similar nature to the elements of X One example is R completion of the ratienals Q The completion ef Cab in the LP nerni fer 1 3 p lt gt0 can be represented as U a 12 the vector space of equivalence classes of Lebesgue measurable functiens u cub gt IF fer which a a a a ja guz39g dz lt gt0 With nernl L gux dxgtp Fact A subset of a complete metric space is complete iff it is closed Proposition Let V be a Banach space and W C V be a subspace The norm on V restricts to a norm on W We have W is complete iff W is closed Examples 1 ODORquot E ODORquot linlmnxum 0 2 CARquot E ODORquot 3K gt 0 9 Vac with Z 0 Remarks 36 Linear Algebra and Alatz ix Analysis If M is a metric space and u M gt IE is a function de ne the support of u to be the closurc of w e M uc 7E 0 The support of a function is automatically closed The complement of the support of a function is the interior of E M 0 A H v 2 Elements of CARquot are continuous functions with compact support 3 ODORquot is complete in the sup norm exercise This can either be shown directly or by showing that COOKquot is a closed subspace of ODORquot 4 CARquot is not complete In fact CARquot is dense in COOKquot So COOKquot is a represen tation of the completion of CARquot in the sup norm Series in normed linear spaces Let K H De nition We say the series convcrgcs in V if 3v 6 V 9 limNn30 Sg 0 where y V SA v 251 on is the JV partial sum We say this series convcrgcs obsolutcly if 221 lt 30 be a normed linear space Consider a series 21 on in V Caution Strictly speaking if a series converges absolutelyquot in a normed linear space it does not have to converge in that space Example The series 1 0 0 0 0 0 it converges absolutely in 16 but it doesn t converge in E30 Proposition A normed linear space L2 H is complete i quot every absolutely convergent series actually converges in L2 Proof Skctch gt Given an absolutely convergent seriesv show that the sequence of partial sums is Cauchy for m gt 7 HSquot Sn 23quotle Ho 4 Given a Cauchy sequence you choose nlvm lt inductively so that for k 132 3 Van Z nk g 2quot Then in particular Cn 1 g 2quot Show that the series rem Eggs acm is absolutely convergent Let x be its limit Show that LCquot gt

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