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Symbols","Pid":"PA391","PgNum":"391","Order":410},{"Title":"Index","Pid":"PA395","PgNum":"395","Order":414}]}},"max_resolution_image_width":1280,"max_resolution_image_height":2083,"quality_info":"We know this book has perfect quality.","volume_id":"ix4P1e6AkeIC","permission_info":"Pages displayed by permission of \u003ca class=link_aux href=\"https://books.google.com.sg/url?id=ix4P1e6AkeIC\u0026pg=PA355\u0026q=http://www.springer.com/shop\u0026clientid=ca-print-springer-kluwer_academic\u0026linkid=1\u0026usg=AOvVaw0EjTHoZhglhEeuCSkMAgXm\u0026source=gbs_pub_info_r\"\u003eSpringer Science \u0026amp; Business Media\u003c/a\u003e","is_ebook":false,"volumeresult":{"has_flowing_text":false,"has_scanned_text":true,"can_download_pdf":false,"can_download_epub":false,"is_pdf_drm_enabled":false,"is_epub_drm_enabled":false},"publisher":"Springer Science \u0026 Business Media","publication_date":"1994.01.25","subject":"Mathematics","num_pages":400,"sample_url":"https://play.google.com/books/reader?id=ix4P1e6AkeIC\u0026source=gbs_vpt_hover","synposis":"Functional analysis has become a sufficiently large area of mathematics that it is possible to find two research mathematicians, both of whom call themselves functional analysts, who have great difficulty understanding the work of the other. The common thread is the existence of a linear space with a topology or two (or more). Here the paths diverge in the choice of how that topology is defined and in whether to study the geometry of the linear space, or the linear operators on the space, or both. In this book I have tried to follow the common thread rather than any special topic. I have included some topics that a few years ago might have been thought of as specialized but which impress me as interesting and basic. Near the end of this work I gave into my natural temptation and included some operator theory that, though basic for operator theory, might be considered specialized by some functional analysts.","my_library_url":"https://www.google.com/accounts/Login?service=print\u0026continue=https://books.google.com.sg/books%3Fop%3Dlibrary\u0026hl=en","is_magazine":false,"is_public_domain":false},{"enableUserFeedbackUI":true,"pseudocontinuous":true,"is_cobrand":false,"sign_in_url":"https://www.google.com/accounts/Login?service=print\u0026continue=https://books.google.com.sg/books%3Fid%3Dix4P1e6AkeIC%26q%3Dassume%26source%3Dgbs_word_cloud_r%26hl%3Den\u0026hl=en","isEntityPageViewport":false,"showViewportOnboarding":false,"showViewportPlainTextOnboarding":false},{"page":[{"pid":"PA355","highlights":[{"X":75,"Y":546,"W":47,"H":11},{"X":75,"Y":647,"W":48,"H":12},{"X":268,"Y":681,"W":43,"H":14},{"X":309,"Y":815,"W":43,"H":14}],"flags":8,"order":374,"vq":"assume"}]},null,{"number_of_results":83,"search_results":[{"page_id":"PA8","page_number":"8","snippet_text":"... \u003cb\u003eassume\u003c/b\u003e that h = 0. ( Verify ! ) So we want to show that there is a unique vector ko in K such that || ko || = dist ( 0 , K ) = inf { || k || : keK } . Let d = dist ( 0 , K ) . By definition , there is a sequence { k } in K such that k\u0026nbsp;..."},{"page_id":"PA13","page_number":"13","snippet_text":"... \u003cb\u003eassume\u003c/b\u003e that M ‡ H , M1 ‡ ( 0 ) . Hence there is a vector fo in such that L ( fo ) 1. Now if he and a = L ( h ) , then L ( h - afo ) = L ( h ) — a = 0 ; so h − L ( h ) foEM . Thus - - 0 = ( h - L ( h ) fo , fo \u0026gt; = - = \u0026lt; h , f 。\u0026gt; — L\u0026nbsp;..."},{"page_id":"PA19","page_number":"19","snippet_text":"... V is an isometry if and only if \u0026lt; Vh , Vg ) = \u0026lt; h , g ) for all h , g in H. PROOF . \u003cb\u003eAssume\u003c/b\u003e ( Vh , Vg ) = \u0026lt; §5 . Isomorphic Hilbert Spaces and the Fourier Transform 19 §5. Isomorphic Hilbert Spaces and the Fourier Transform for the Circle."},{"page_id":"PA20","page_number":"20","snippet_text":"John B. Conway. PROOF . \u003cb\u003eAssume\u003c/b\u003e ( Vh , Vg ) = \u0026lt; h , g ) for all h , g in . Then || Vh || 2 = \u0026lt; Vh , Vh ) = \u0026lt; h , h ) = || h || 2 and V is an isometry . Now \u003cb\u003eassume\u003c/b\u003e that V is an isometry . If h , ge and EF , then || h + λg || 2 = || Vh +\u0026nbsp;..."},{"page_id":"PA21","page_number":"21","snippet_text":"... \u003cb\u003eassume\u003c/b\u003e this for the moment . Let D = { zeC : │z│ \u0026lt; 1 } . 5.6 . Theorem . If f : OD → C is a continuous function , then there is a sequence { P ( z , z ) } of polynomials in z and z such that p1 ( z , ž ) → f ( z ) uniformly on dD\u0026nbsp;..."},{"page_id":"PA28","page_number":"28","snippet_text":"... \u003cb\u003eassume\u003c/b\u003e that is a bounded measurable function and | 4 ( x ) | ≤ || 6 || ∞ for all x . So if ƒ eL2 ( u ) , then S1 of 12 dμ \u0026lt; | 6 || 2 Sff du . That is , MEBL2 ( μ ) ) and || M || ≤ || || ... If ε \u0026gt; 0 , the o - finiteness of the\u0026nbsp;..."},{"page_id":"PA33","page_number":"33","snippet_text":"... \u003cb\u003eassume\u003c/b\u003e Ah , h ) is real for every h in H. If αC and h , gЄH , then \u0026lt; A ( h + ag ) , h + ag ) = ( Ah , h ) + ã \u0026lt; Ah ... \u003cb\u003eassumed\u003c/b\u003e that is an R - Hilbert space . For example , if A = - [ _ ] on R on R2 , then \u0026lt; Ah , h ) = 0 for 0 all h\u0026nbsp;..."},{"page_id":"PA48","page_number":"48","snippet_text":"... \u003cb\u003eassume\u003c/b\u003e that \u0026lt; Th ,, h , \u0026gt; → 2 , where | 2 | = || T || . It will be shown that eo , ( T ) . Since | 2 | = || T || , 0 ≤ || ( T − 2 ) h , || 2 = || Th , || 2 - 2λ \u0026lt; Th „ , h „ \u0026gt; + λ2 ≤ 222 - 22 Th „ , h , \u0026gt; → 0 . Hence || ( T2 ) h\u0026nbsp;..."},{"page_id":"PA51","page_number":"51","snippet_text":"... \u003cb\u003eAssuming\u003c/b\u003e ( 6.4 ) , W ( a ) 0 and so h , and h , are linearly independent . PROOF . If W ( a ) = 0 , then linear ... \u003cb\u003eAssume\u003c/b\u003e ( 6.4 ) . If g is the Green function for L defined in ( 6.7 ) and G : L2 [ a , b ] → L2 [ a , b ] is the\u0026nbsp;..."},{"page_id":"PA52","page_number":"52","snippet_text":"... \u003cb\u003eAssume\u003c/b\u003e ( 6.4 ) . If heD , λEC \\ { 0 } , and Lh = λh , then Gh = λ - 1h . If he L2 [ a , b ] and Gh = λ - 1h , then he and Lh = 2h . PROOF . This is immediate from the theorem . 6.11 . Lemma . \u003cb\u003eAssume\u003c/b\u003e ( 6.4 ) . If aЄop ( G ) and a 0\u0026nbsp;..."},{"page_id":"PA54","page_number":"54","snippet_text":"... \u003cb\u003eassumed\u003c/b\u003e that the scalars a , in ( 7.3 ) are distinct . There is no loss in generality in \u003cb\u003eassuming\u003c/b\u003e this , however . In fact , if a ; = a ;, then we can replace P ; and P ; with P¡ + Pj . 7.4 . Proposition . An operator A on H is\u0026nbsp;..."},{"page_id":"PA55","page_number":"55","snippet_text":"... \u003cb\u003eassume\u003c/b\u003e that B is reduced by each ran P. Thus BP ; = P¡B for all i . If he , then Aha , Ph . Hence BAh = a ; BP ; h = Σ ; a ; P ; Bh = ABh . ( Why is the first equality valid ? ) i Using the notation of the preceding theorem , if AB = BA\u0026nbsp;..."},{"page_id":"PA57","page_number":"57","snippet_text":"... \u003cb\u003eassume\u003c/b\u003e that BA = AB whenever ATTA . Thus , B must commute with T itself . By ( 7.5 ) , B is reduced by each PH = Hn , n \u0026gt; 0 ; put B1 = B | H „ . Fix n \u0026gt; 0 for the moment and let A , be any bounded operator in B ( H ) . Define Ah Ah if\u0026nbsp;..."},{"page_id":"PA58","page_number":"58","snippet_text":"... \u003cb\u003eassume\u003c/b\u003e each 2,0 . If hЄH , h = ho + Σa = 1 h2 , where hoeker T and h „ P „ H for n \u0026gt; 1. Then Th = Σλnhn Hence ( Th , h ) = Σ 12hnho + Σ = 1hm ) = = 1 n n 12 h 20 since \u0026lt; h , hm \u0026gt; = 0 when nm . 00 n = 1 m = 0 \u0026#39; ( hnhm ) = 7.15 . Theorem\u0026nbsp;..."},{"page_id":"PA64","page_number":"64","snippet_text":"... \u003cb\u003eAssume\u003c/b\u003e that ( d ) holds and put q ( x ) = a . If ɛ \u0026gt; 0 , then q ( ( a + ε ) ̄ 1x ) = ( a + ε ) 1a \u0026lt; 1 . By ( d ) , 1 \u0026gt; ... \u003cb\u003eassume\u003c/b\u003e that the two norms are equivalent . Hence { x : || x || 1 \u0026lt; 1 } is an open neighborhood of 0 in the\u0026nbsp;..."},{"page_id":"PA69","page_number":"69","snippet_text":"... \u003cb\u003eAssume\u003c/b\u003e that A : → is an additive mapping ( that is , A ( x1 + x2 ) = A ( x1 ) + A ( x2 ) for all x , and x2 in X ) and show that conditions ( b ) , ( c ) , and ( d ) in Proposition 2.1 are equivalent to the continuity of A. §3 . Finite\u0026nbsp;..."},{"page_id":"PA70","page_number":"70","snippet_text":"... \u003cb\u003eassume\u003c/b\u003e that || Σ = 15 , e , || = max { | ¿ ; | : 1 \u0026lt; j \u0026lt; d } , where { e ; } is a Hamel basis for . Thus , for x = Σšje ,, || Tx || = || Σ ; š¡Te ; || ≤ Σ ||| Te , | C || x || , where C = Σ , || Te , || . By ( 2.1 ) , T is continuous\u0026nbsp;..."},{"page_id":"PA71","page_number":"71","snippet_text":"... \u003cb\u003eassume\u003c/b\u003e that WX / M and Q - 1 ( W ) is open in . Let r \u0026gt; 0 and put B , = { xeX : || x || \u0026lt; r } . It will be shown that Q ( B , ) = { x + M : || x + M || \u0026lt; r } . In fact , if | x || \u0026lt; r , then || x + M || ≤ || x || \u0026lt; r . On the other\u0026nbsp;..."},{"page_id":"PA74","page_number":"74","snippet_text":"... \u003cb\u003eAssume\u003c/b\u003e now that ker ƒ is closed and let Q : X → X / ker ƒ be the natural map . By ( 4.2 ) , Q is continuous . Let T : X / ker f → F be an isomorphism ; by ( 3.4 ) , Tis continuous . Thus , if g = ToQ : → F , g is continuous and ker f\u0026nbsp;..."},{"page_id":"PA78","page_number":"78","snippet_text":"... \u003cb\u003eassume\u003c/b\u003e negative values and that ( b ) in the definition only holds for a \u0026gt; 0 . 6.2 . The Hahn - Banach Theorem . Let X be a vector space over R and let q be a sublinear functional on X. If M is a linear manifold in X and f : M → R is a\u0026nbsp;..."},{"page_id":"PA80","page_number":"80","snippet_text":"... \u003cb\u003eassume\u003c/b\u003e that the extension F : X → R of f exists with F≤q . Let\u0026#39;s see what F must look like . Put α = F ( x ) . If t \u0026gt; 0 and y1EM , then F ( txo + y1 ) = tao + f ( y1 ) ≤ q ( txo + y1 ) . Hence αo - t ̄1ƒ ( y1 ) + t ̄1q ( txo + y1 )\u0026nbsp;..."},{"page_id":"PA82","page_number":"82","snippet_text":"... \u003cb\u003eassume\u003c/b\u003e F = R ; that is , 1 ° = 1. If xel \u0026quot; , let x \u0026#39; denote the element of 1 ° defined in part ( d ) above . Put M = { x - x : xel } . Note that ( x + xy ) \u0026#39; = x + ay \u0026#39; for any x , y in 1 ° and a in R ; hence M is a linear manifold in 1\u0026nbsp;..."},{"page_id":"PA83","page_number":"83","snippet_text":"... \u003cb\u003eassume\u003c/b\u003e that F = C. Let L1 be the functional obtained on lg . If xela , then x = x1 + ix , where x1 , x2 El . Define L ( x ) = L1 ( x1 ) + iL1 ( x2 ) . It is left as an exercise to show that L is C - linear . It\u0026#39;s clear that ( b ) , ( c )\u0026nbsp;..."},{"page_id":"PA85","page_number":"85","snippet_text":"... \u003cb\u003eassume\u003c/b\u003e μЄM ( K ) and sgdμ = 0 for every rational function g with poles in E. Let U be a component of C \\ K , and let woeEU . If w 。.∞0 , then the hypothesis and ( 8.3 ) imply that each derivative of û at wo vanishes . Hence û = 0 on U\u0026nbsp;..."},{"page_id":"PA87","page_number":"87","snippet_text":"... \u003cb\u003eassume\u003c/b\u003e that X = Y + P – P. 9.9 . Claim . X = Y + P = Y - P . - 1 Let xex ; so x = y + P1 - P2 , y in Y , P1 , P2 in P. Since is confinal there is a y1 in such that y1 \u0026gt; P1 . Hence p1 = Y1 ( Y1 - P1 ) - P. Thus x = y - P2 + P1Є ( Y - P )\u0026nbsp;..."},{"page_id":"PA92","page_number":"92","snippet_text":"... \u003cb\u003eassumed\u003c/b\u003e to be complete . The next example ( due to Alp Eden ) shows that the range must also be \u003cb\u003eassumed\u003c/b\u003e to be ... \u003cb\u003eassume\u003c/b\u003e that { Ax , } is convergent . It is possible to give a measure - theoretic solution to Exercise 2.3 , but\u0026nbsp;..."},{"page_id":"PA100","page_number":"100","snippet_text":"... \u003cb\u003eassuming\u003c/b\u003e that P is closed under the formation of finite sums and supremums of bounded families [ as in ( 1.4 ) ] . Sometimes it is convenient to \u003cb\u003eassume\u003c/b\u003e that consists of all continuous 100 IV . Locally Convex Spaces."},{"page_id":"PA101","page_number":"101","snippet_text":"John B. Conway. it is convenient to \u003cb\u003eassume\u003c/b\u003e that consists of all continuous seminorms . In either case the resulting topology on remains unchanged . 1.5 . Example . Let X be completely regular and let C ( X ) = all continuous functions\u0026nbsp;..."},{"page_id":"PA104","page_number":"104","snippet_text":"... \u003cb\u003eassume\u003c/b\u003e that is Hausdorff . ( a ) Show that is Hausdorff if and only if the singleton set { 0 } is closed . ( b ) If is Hausdorff , show that is a regular topological space . 4. Let be a TVS . Show : ( a ) if xe , the map x + x + xo is a\u0026nbsp;..."},{"page_id":"PA106","page_number":"106","snippet_text":"... \u003cb\u003eassume\u003c/b\u003e that X is metrizable and its metric is p . Let U1 = { x : p ( x , 0 ) \u0026lt; 1 / n } . Because is locally convex , there are continuous seminorms q1 , ... , qk and positive numbers \u0026amp; 1 , ... , \u0026amp; such that ~ } = 1 { x : q ; ( x ) \u0026lt; ε\u0026nbsp;..."},{"page_id":"PA107","page_number":"107","snippet_text":"... \u003cb\u003eassume\u003c/b\u003e that X is a LCS that has a bounded open set U. It must be shown that there is norm on that defines the same topology . By translation , it may be \u003cb\u003eassumed\u003c/b\u003e that OEU ( see Exercise 4i ) . By local convexity , there is a continuous\u0026nbsp;..."},{"page_id":"PA110","page_number":"110","snippet_text":"... a contradiction . The condition that K be compact in the preceding lemma is necessary ; is not enough to \u003cb\u003eassume\u003c/b\u003e that K is closed . ( What is a counterexample ? ) 3.9 . Theorem . Let X be a real LCS 110 IV . Locally Convex Spaces."},{"page_id":"PA113","page_number":"113","snippet_text":"... \u003cb\u003eassumes\u003c/b\u003e the value r at t = 1. Let 0 \u0026lt; t \u0026lt; 1 such that s \u0026#39; % | f ( x ) | \u0026quot; dx = r / 2 . Define g , h : ( 0 , 1 ) → R ... \u003cb\u003eassumed\u003c/b\u003e that OEG . Thus there is an R \u0026gt; 0 with B ( 0 ; R ) ≤ G. By the preceding paragraph , B ( 0 ; 2n ( 1\u0026nbsp;..."},{"page_id":"PA114","page_number":"114","snippet_text":"... \u003cb\u003eassume\u003c/b\u003e LEC ( X ) * . There are compact sets K1 , ... , K , and positive numbers α1 , ... ,, such that L ( f ) \u0026lt; = 1PK , ( f ) ( 3.1f ) . Let K = = U11K ; and = max { α ;: 1 \u0026lt; j \u0026lt; n } . Then \\ L ( ƒ ) | ≤αpx ( ƒ ) . Hence if feC ( X )\u0026nbsp;..."},{"page_id":"PA118","page_number":"118","snippet_text":"... \u003cb\u003eassume\u003c/b\u003e that each restriction is continuous . If p is a continuous seminorm on y , then po TX ; is a T - continuous seminorm for every i . By ( 5.6c ) , poT is continuous on . By Exercise 1.23 , T is continuous . It may have occurred to\u0026nbsp;..."},{"page_id":"PA127","page_number":"127","snippet_text":"... \u003cb\u003eassume\u003c/b\u003e that x。ЄX \\ A1 . A1 is a closed convex balanced set so by ( IV.3.13 ) there is an x * in X * , an a in R , and an ɛ \u0026gt; 0 such that Rea1 , x * \u0026gt; \u0026lt; a \u0026lt; a + ε \u0026lt; Re \u0026lt; xo , x * \u0026gt; -1 for all a1 in A1 . Since OEA1 , 0 = \u0026lt; 0 , x * \u0026gt; \u0026lt; a\u0026nbsp;..."},{"page_id":"PA133","page_number":"133","snippet_text":"... \u003cb\u003eassume\u003c/b\u003e that there is an xo in such that || xo || = 1 and \u0026lt; X0 , x * ) = || x * || . If xe and || x + M || = a \u0026gt; 0 , then a 1x + M || = 1. But also || xo + M = 1. ( Why ? ) Since dim / M = 1 , there is a ẞ in F , | B | = 1 , such that\u0026nbsp;..."},{"page_id":"PA134","page_number":"134","snippet_text":"... \u003cb\u003eAssume\u003c/b\u003e that is separable and let { x } be a countable dense subset of ball . For each n let D ,, = { xeF : | a | \u0026lt; 1 } . Put X = II , D ,; X is a compact metric space . So if ( ball * , wk * ) is homeomorphic to a subset of X , ball X\u0026nbsp;..."},{"page_id":"PA137","page_number":"137","snippet_text":"... \u003cb\u003eAssume\u003c/b\u003e X is completely regular . If x1 x2 , then there is an ƒ in C ( X ) such that f ( x1 ) = 1 and f ( x2 ) = 0 ; thus dx , ( ƒ ) # dx2 ( ƒ ) . Hence A is injective . To show that A : X → ( A ( X ) , wk * ) is an open map , let U be\u0026nbsp;..."},{"page_id":"PA138","page_number":"138","snippet_text":"... \u003cb\u003eassume\u003c/b\u003e that is a compact space and л : X → is a continuous map such that : ( а \u0026#39; ) л : X → л ( X ) is a homeomorphism ; ( b ) ( X ) is dense in ; ( c \u0026#39; ) if feC ( X ) , there is an ƒ in C ( ) such that ƒ ° л = ƒ . Define g : A ( X )\u0026nbsp;..."},{"page_id":"PA140","page_number":"140","snippet_text":"... \u003cb\u003eassume\u003c/b\u003e that C , ( X ) is separable . Thus ( ball C , ( X ) * , wk * ) is metrizable ( 5.1 ) . Since X is ... \u003cb\u003eassumed\u003c/b\u003e that x , xm for nm . Let A = { x ,: n is even } and B = { x ,: n is odd } . Then A and B are disjoint closed\u0026nbsp;..."},{"page_id":"PA142","page_number":"142","snippet_text":"... \u003cb\u003eassume\u003c/b\u003e that K is not a singleton ) , U. Let Uo be a chain in u and put U1 = { U : Ueo } . Clearly U , is open , and since o is a chain , U。 is convex . If U。= K , then the compactness of K implies that there is a U in 。 with UK , a\u0026nbsp;..."},{"page_id":"PA143","page_number":"143","snippet_text":"... \u003cb\u003eAssume\u003c/b\u003e ( 7.6 ) is false . That is , \u003cb\u003eassume\u003c/b\u003e there is an open convex subset V of such that ext KV but VOKK . Then VOKE and is contained in a maximal element U of U. Since K \\ U = { a } for some a in ext K , this is a contradiction . Thus\u0026nbsp;..."},{"page_id":"PA149","page_number":"149","snippet_text":"... \u003cb\u003eassumed\u003c/b\u003e to be linear or affine . This is a small part of nonlinear functional analysis . To begin with , recall ... \u003cb\u003eassume\u003c/b\u003e that Rd , 1 ≤ d \u0026lt; ∞o . If K = { xeR : || x || \u0026lt; r } , then the result is immediate from Brouwer\u0026#39;s Theorem\u0026nbsp;..."},{"page_id":"PA153","page_number":"153","snippet_text":"... \u003cb\u003eassumed\u003c/b\u003e that there is an integer m such that T ( xo ) #xo for 1 \u0026lt; k \u0026lt; m and T ( xo ) = xo for m \u0026lt; k \u0026lt; n . Let To = ( T1 + ... + Tm ) / m . Then xo = To ( xo ) 1 = == n ... · [ T1 ( xo ) + ··· + Tm ( xo ) ] + n- -m n Hence 1 To ( xo )\u0026nbsp;..."},{"page_id":"PA169","page_number":"169","snippet_text":"... \u003cb\u003eassume\u003c/b\u003e that A * is invertible and show that A is invertible . Since A * is an open mapping , there is a constant c \u0026gt; 0 such that A * ( ball * ) = { x * X * : || x * || ≤c } . So if xe , then || Ax || = sup { | \u0026lt; Ax , y * \u0026gt; ] : y\u0026nbsp;..."},{"page_id":"PA173","page_number":"173","snippet_text":"... \u003cb\u003eassumed\u003c/b\u003e that M≤1 . Hence { Ax , } cl A ( ball ) . Since A is compact , there is a subse- quence { x } and a y ... \u003cb\u003eassume\u003c/b\u003e that is separable ; so ( ball X , wk ) is a compact metric space . So if { x } is a sequence in ball there is\u0026nbsp;..."},{"page_id":"PA174","page_number":"174","snippet_text":"... \u003cb\u003eAssume\u003c/b\u003e A is a compact operator and let { y } be a sequence in ball y * . It must be shown that { 4 * y * } has a norm convergent subsequence or , equivalently , a cluster point in the norm topology . By Alaoglu\u0026#39;s Theorem , there is a y\u0026nbsp;..."},{"page_id":"PA175","page_number":"175","snippet_text":"... \u003cb\u003eassume\u003c/b\u003e that F is equicontinuous and F≤ball C ( X ) . Let \u0026amp; \u0026gt; 0. For each x in X , let U , be an open neighborhood of x such that | f ( x ) − f ( y ) | \u0026lt; ɛ / 3 for f in F and y in U. Now { U : xeX } is an open covering of X. Since X is\u0026nbsp;..."},{"page_id":"PA184","page_number":"184","snippet_text":"... \u003cb\u003eassume\u003c/b\u003e R is reflexive ; thus ball R is o ( R , R * ) - compact . Therefore C = A ( ball ) is weakly compact in . By ( a ) , cl W is weakly compact . The next theorem , as well as the preceding lemma , are from Davis , Figiel , Johnson\u0026nbsp;..."},{"page_id":"PA185","page_number":"185","snippet_text":"... \u003cb\u003eassume\u003c/b\u003e that T is weakly compact and put W = T ( ball ) . Define R as in Lemma 5.3 . By ( 5.3d ) , R is reflexive . Let A : R → Y be the inclusion map . Note that if xeball , then TxeW . By ( 5.3a ) , || Tx || \u0026lt; 1 whenever x 1. So B : X\u0026nbsp;..."},{"page_id":"PA187","page_number":"187","snippet_text":"... \u003cb\u003eassume\u003c/b\u003e that the underlying field of scalars IF is the field of complex numbers C. It will be \u003cb\u003eassumed\u003c/b\u003e from Section 3 until the end of this book that all vector spaces are over C. This will also enable us to apply the theory of analytic\u0026nbsp;..."},{"page_id":"PA192","page_number":"192","snippet_text":"... \u003cb\u003eassume\u003c/b\u003e that { a } is a sequence in G such that a , → 1 . Let 0 \u0026lt; 8 \u0026lt; 1 and suppose || a1 || \u0026lt; 8. From the preceding lemma , a1 ( 1 − ( 1 − a ) ) ̄ 1 = Σo ( 1 - a ) = 1 + ( 1 - a , ) . Hence = 1 - - 1 || = k = 1 ( 1 - - k = 1 \u0026lt; 8\u0026nbsp;..."},{"page_id":"PA193","page_number":"193","snippet_text":"... \u003cb\u003eassume\u003c/b\u003e that is a Banach algebra . Let be a Banach algebra and let M be a proper closed ideal . Note that AM becomes an algebra . Indeed , ( x + M ) ( y + M ) = xy + M is a well - defined multiplication on AM . ( Why ? ) 2.6 . Theorem\u0026nbsp;..."},{"page_id":"PA194","page_number":"194","snippet_text":"... \u003cb\u003eassume\u003c/b\u003e that ✓ has an identity . If I is a left ideal of , say that I is a modular left ideal if there is a u in such that A ( 1 − u ) = { a — au : ae } I ; call such an element u of ✓ a right modular unit for I. Similarly , define\u0026nbsp;..."},{"page_id":"PA196","page_number":"196","snippet_text":"... \u003cb\u003eassume\u003c/b\u003e that ( A - a ) , → g . Then 8fn - fm || || ( A - α ) ( fn -ƒm ) | = || ( A — α ) ƒ „ — ( A - a ) . Thus { f } is a Cauchy sequence . Let f → f . Then g = lim ( A — α ) ƒ „ = ( A - x ) f ; hence geran ( A - a ) . Let ( A - a )\u0026nbsp;..."},{"page_id":"PA207","page_number":"207","snippet_text":"... \u003cb\u003eassume\u003c/b\u003e that λedo ( a ) . Since int o ( a ) ≤ into ( a ) , it suffices to show that eo ( a ) . Suppose 140 ( a ) ; there is thus an x in ✓ such that x ( a - 2 ) = ( a− 2 ) x = 1. Since Aedo ( a ) , there is a sequence { 2 } in Co ( a )\u0026nbsp;..."},{"page_id":"PA218","page_number":"218","snippet_text":"... \u003cb\u003eassumed\u003c/b\u003e that every Banach algebra is over C. Also \u003cb\u003eassume\u003c/b\u003e that all Banach algebras contain an identity . A division algebra is an algebra such that every nonzero element has a multiplicative inverse . It may seem incongruous that the\u0026nbsp;..."},{"page_id":"PA219","page_number":"219","snippet_text":"... \u003cb\u003eassume\u003c/b\u003e that λeo ( a ) ; so a 2 is not invertible and , hence , ( a ) is a proper ideal . Let M be a maximal ideal in such that ( a - 2 ) AM . If heΣ such that M = ker h , then 0 = h ( a - 2 ) = h ( a ) - 2 ; hence σ ( α ) = Σ ( α )\u0026nbsp;..."},{"page_id":"PA222","page_number":"222","snippet_text":"... \u003cb\u003eassume\u003c/b\u003e that has no identity and let 1 be the algebra obtained by adjoining an identity . For a in A , let o ( a ) be the spectrum of a as an element of , and show that o ( a ) = { h ( a ) : he } { 0 } . Also , show that the maximal\u0026nbsp;..."},{"page_id":"PA224","page_number":"224","snippet_text":"... \u003cb\u003eassume\u003c/b\u003e that xeK for every U in U. Now { xu : UEU } is a net in K. Since K is compact , there is a point x in K such that xx . But x1yu → e . Since multiplication is continuous , y = x ( xuyu ) x . Therefore if W is any neighborhood of\u0026nbsp;..."},{"page_id":"PA227","page_number":"227","snippet_text":"... \u003cb\u003eassume\u003c/b\u003e that h : L1 ( G ) → C is a nonzero homomorphism . Since h is a bounded linear functional , there is a \u0026amp; in L ° ( G ) such that h ( ƒ ) = { ƒf ( x ) 4 ( x ) dx and || 4 || ∞ = || h || = 1. If f , geL1 ( G ) , then_h ( f * g )\u0026nbsp;..."},{"page_id":"PA230","page_number":"230","snippet_text":"... \u003cb\u003eassume\u003c/b\u003e that G is compact . Hence I L1 ( G ) since m ( G ) = 1. Suppose yer and the identity for г , then there is a point x , in G such that 7 ( xo ) # 1 . Thus fxxx ) dx = √ xxx + xo ) dx [ = 7 ( xo ) | 7 ( xx1 ) dx = 7 ( x0 ) [ 7 ( x )\u0026nbsp;..."},{"page_id":"PA235","page_number":"235","snippet_text":"... \u003cb\u003eassume\u003c/b\u003e that ✓ has an identity . By Exercise VII.8.1 , || h || = 1. If a = a * and tεR , | h ( a + it ) 2 \u0026lt; || a + ... \u003cb\u003eassume\u003c/b\u003e that a is hermitian and let §1 . Elementary Properties and Examples 235."},{"page_id":"PA253","page_number":"253","snippet_text":"... \u003cb\u003eassume\u003c/b\u003e that π ( 1 ) = 1 ] . Let P1 = ( 1 ) . Show that P , is a projection and H1 = P1H reduces л ( ) . If л1 ( a ) = л ( a ) | Н1 , show that 1 : A → B ( H1 ) is a representation . 2. Show that the representation in Example 5.4 is a\u0026nbsp;..."},{"page_id":"PA265","page_number":"265","snippet_text":"... \u003cb\u003eassume\u003c/b\u003e that λess - ran ( 4 ) . So there is a set △ in 2 with μ ( X \\ △ ) = 0 and 2 not in cl ( $ ( A ) ) ; thus there is a 8 \u0026gt; 0 with | ( x ) - 2 | ≥ 8 for all x in A. If y = ( p − 2 ) ̄ 1 , ЄL ( μ ) and M1 = ( M ¿ — ¿ ) ̄ 1\u0026nbsp;..."},{"page_id":"PA270","page_number":"270","snippet_text":"... \u003cb\u003eassume\u003c/b\u003e that V : Lu1 ) → L2 ( 2 ) is an isomorphism such that VN Nμ2 . Put = V ( 1 ) ; so VEL ( 2 ) . For convenience , put N ; = Nu ,, j = 1,2 . It is easy to see that VNV - 1 = N and VN ** V - 1 = N ** . Hence Vp ( N1 , N ) V - 1 = p\u0026nbsp;..."},{"page_id":"PA271","page_number":"271","snippet_text":"... \u003cb\u003eassume\u003c/b\u003e that N is compact and let \u0026amp; \u0026gt; 0 . Put ( z ) = z 1 % , ( z ) ; so EB ( C ) . Since N is compact , so is No ( N ) . But No ( N ) = { zz ̄1x1 , ( z ) dE ( z ) = E ̧ . Since E , is a compact projection , it must have finite rank\u0026nbsp;..."},{"page_id":"PA280","page_number":"280","snippet_text":"... \u003cb\u003eassume\u003c/b\u003e that contains the identity operator . Let Mv { ran A : Ae } and let P = the projection of onto M. Show that SOT - clA = A \u0026quot; P = PA \u0026quot; . = 11. Formulate and prove a polar decomposition for operators between different Hilbert spaces\u0026nbsp;..."},{"page_id":"PA290","page_number":"290","snippet_text":"... \u003cb\u003eassumes\u003c/b\u003e a knowledge of Exercise 2.21 . ) Let N = √zdE ( z ) be a normal operator with scalar - valued spectral ... \u003cb\u003eassume\u003c/b\u003e that all Hilbert spaces are separable . Every normal operator on a Hilbert space of dimension at least 2 has\u0026nbsp;..."},{"page_id":"PA291","page_number":"291","snippet_text":"... \u003cb\u003eAssume\u003c/b\u003e that N is * -cyclic ; thus N = N. Suppose feL \u0026#39; ( u ) and ffydu = 0 for every in P ( μ ) . If it can be shown that fødμ = 0 , then the Hahn - Banach Theorem implies that EP ( u ) . This is the strategy we follow . Let f = gh for\u0026nbsp;..."},{"page_id":"PA305","page_number":"305","snippet_text":"... \u003cb\u003eAssume\u003c/b\u003e A is closable and let k。e ( dom A * ) + . We want to show that ko = 0 . Thus ko → 0e [ gra A * ] + = [ J gra A ] 11 = cl [ J gra A ] = J [ cl ( gra A ) ] . So 0 → - k 。= J * ( k 。→ 0 ) ĒJ * J [ ( gra A ) ] = cl ( gra A )\u0026nbsp;..."},{"page_id":"PA306","page_number":"306","snippet_text":"... \u003cb\u003eassume\u003c/b\u003e that f ( x ) = -ig ( t ) dt for all x . Therefore ƒ is absolutely continuous and f ( x ) → f ( x ) uniformly on [ 0,1 ] ; thus f ( 0 ) = f ( 1 ) = 0 and f \u0026#39; = - ige L2 ( 0 , 1 ) . So ƒe and fg = fg = ƒ if\u0026#39;egra A ; that is , Ae\u0026nbsp;..."},{"page_id":"PA313","page_number":"313","snippet_text":"... \u003cb\u003eassume\u003c/b\u003e hA * h 1 gra A + . Since h A * h 1 gra A , for every f in dom A , 0 = ( h⇒ A * h , ƒ → Af ) = \u0026lt; h , f \u0026gt; + ( A * h , Af ) . So ( A * h , Af ) = - \u0026lt; h , f ) for every ƒ in dom A. This implies that A * hedom A * and A * A * hh\u0026nbsp;..."},{"page_id":"PA317","page_number":"317","snippet_text":"... \u003cb\u003eassume\u003c/b\u003e that U is a partial isometry as in ( b ) . It follows that ker ( 1 — U ) = ( 0 ) . In fact , if ƒeker ( 1 − U ) , then Uf = f ; so || ƒ || = || Uƒ || and hence feinitial U. Since U * U is the projection onto initial U , ƒ = U\u0026nbsp;..."},{"page_id":"PA348","page_number":"348","snippet_text":"... \u003cb\u003eAssume\u003c/b\u003e that a is left invertible , so there is a b in such that ba = 1 . b | -1x . Thus for any x in A , || x || = || bax || \u0026lt; || b || || ax || , and hence || ax | In particular , this is true whenever xeC * ( a ) . Because a is normal\u0026nbsp;..."},{"page_id":"PA355","page_number":"355","snippet_text":"... \u003cb\u003eAssume\u003c/b\u003e B is invertible . Without loss of generality we may \u003cb\u003eassume\u003c/b\u003e that ind A = - ∞ since the alternative situation is covered in Case 1. If M = B ( ran A ) = ran BA and N = B ( [ ran A ) + ) , then the fact that B is invertible\u0026nbsp;..."},{"page_id":"PA356","page_number":"356","snippet_text":"... \u003cb\u003eAssume\u003c/b\u003e that A is a left semi - Fredholm operator . By definition there is an operator X : H \u0026#39; → H and a compact operator Ko : H → such that XA = 1 + Ko . Thus X ( A + K ) = 1 + ( K。 + XK ) and so A + K is left semi - Fredholm . To\u0026nbsp;..."},{"page_id":"PA365","page_number":"365","snippet_text":"... \u003cb\u003eassume\u003c/b\u003e that λ = 0. We may also \u003cb\u003eassume\u003c/b\u003e that ker A is finite dimensional . Indeed , if this is not the case , then ker A * must be finite dimensional and the proof that follows applies to A * . But observe that if the conclusion of the\u0026nbsp;..."},{"page_id":"PA372","page_number":"372","snippet_text":"... \u003cb\u003eassume\u003c/b\u003e that ƒ is continuous at x 。 and let { x ; } be a net in X such that x ; → xo in X. If V is open in Y and f ( x ) EV , then there is an open set U in X such that x。EU and f ( U ) V. Let io be such that x , EU for i \u0026gt; io . Hence\u0026nbsp;..."},{"page_id":"PA374","page_number":"374","snippet_text":"... \u003cb\u003eassume\u003c/b\u003e that T2 is larger than T1 . ( a ) If F is T - closed , F is T2 - closed . ( b ) If f : Y → ( X , T2 ) is continuous , then f : Y → ( X , T1 ) is continuous . ( c ) If f ( X , T1 ) → Y is continuous , then f : ( X , T2 ) → Y\u0026nbsp;..."},{"page_id":"PA375","page_number":"375","snippet_text":"... \u003cb\u003eAssume\u003c/b\u003e FELP ( u ) * . Case 1 : μ ( X ) \u0026lt; ∞ . Here XL ( u ) for every △ in N. Define v ( A ) = F ( x ^ ) . It is easy to see that v is finitely additive . If { A , } with A , 24 , 2 ... and 14 , = , then n γραμ 1 / p Caullo = [ [ ixa\u0026nbsp;..."},{"page_id":"PA379","page_number":"379","snippet_text":"... \u003cb\u003eassume\u003c/b\u003e that μ ( A ) \u0026lt; ∞o . It follows that μ ( A , ) \u0026lt; ∞ for every n . ( Why ? ) Let \u0026amp; \u0026gt; 0 and for each n1 let { E \u0026quot; ) , En be a partition of A , such that μ ( E ) \u0026gt; | μ ( A ) - ε / 2 \u0026quot; . Then N Ž mn N Σ IMKAJ \u0026lt; 2 [ 2+ \\ MESI ] Σ 2\u0026nbsp;..."}],"search_query_escaped":"assume"},{});</script></div></div></div><script>(function() {var href = window.location.href;if (href.indexOf('?') !== -1) {var parameters = href.split('?')[1].split('&');for (var i = 0; i < parameters.length; i++) {var param = parameters[i].split('=');if (param[0] == 'focus') {var elem = document.getElementById(param[1]);if (elem) {elem.focus();}}}}})();</script>