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by: Dr. Kurtis Nolan
Dr. Kurtis Nolan
GPA 3.55


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This 66 page Class Notes was uploaded by Dr. Kurtis Nolan on Sunday September 6, 2015. The Class Notes belongs to N 1 at University of Texas at Austin taught by Staff in Fall. Since its upload, it has received 16 views. For similar materials see /class/181627/n-1-university-of-texas-at-austin in Nursing and Health Sciences at University of Texas at Austin.

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Date Created: 09/06/15
The geometry associated With viewing a surface angle of incidence i angle of emittance e phase angle g With respect to surface normal N T he normal refers to the direction that is perpendicular to the surface H750 nm 44 UAU V J 375 nm The wave nature of light Light like sound can differ in its wavelength phase and amplitude 10 Frequency 1 in cycles per second 5 10 2 10 l l 10 l 1024 r Cusmic rays Gamma rays X rays vwn uv V I l Inlvared l Radar FM 4 4 Ligm Television wart wave olet Blue Green Yellow R d I l x Broadcast 400 0 600 700 Power Wavelength mu 1 A 1 micron 1 meter 1 km l y l l l 1 y l I 10 quot2 10 393 10 1 10 10B Wavelength x in centlmeters The range of Visible wavelengths of light out of the entire electromagnetic spectrum RELATIVE RADIANT POWER WAVELENGTH X um Examples of daylight radiant power spectra normalized at 550 nm The curves are labeled by the color temperature the temperature of a black body radiator that would produce approximately the same power spectrum 0 049 am P F NPM mum a m Re ectance o 5 o 00 0 o o 5 3000quot Juniper full lea o quotnun 400 450 500 550 600 650 Wavelength nm o o o 39M 39w lxgt NU39IKAUIbUUI 0 15 11 Re ectance rout 0 05 U l 400 45 0 500 Wavelength nm 550 600 650 Grass young and green O 400 450 500 550 60 650 Wavelength nm Wheat straw 0 400 450 500 550 500 650 Wavelength nm Re ectance spectra of some natural materials From Krinov 1947 0 60 Time ms Typical sequence of photons emitted from a 10W imensity lighl source Photons and the Randomness of Light Energy of a photon J1 ch Photon interarrival time PW st17e Number of photons in a fixed time interval 059 2 102 20l2m aw Formulas describing the randomness of light absorption and emission The number of photons absorbed in a xed time interval is described by the Poisson probability density function PROBABILITY O5 39 39 00 005 I o I J I H H n n n 1 O 5 I0 I5 20 NUMBER OF ABSORBED PHOTONS Examples of the Poisson density that describes the randomness of the light absorption and emission Posterior chambev Limba zone Conjunctiva Canal 0 Schlemm Clllal39y musde Aw Ora serrate x y LVisual axis 1 SC EISA Chumid Lamina cnbrosa Major anatomical structures in the human eye Formation of images by an aperture pupil Without a lens Formation of images with a lens The distance from a lens to the point Where an image is in shaIp focus di depends upon the distance of the source to the lens d0 5 n 10 1333 60dzupters f are n n w 53 Reduced eye model of the human eye PUPIL DIAMETER mm 4 DEGROOT 1552 o o SPRING a STILES 948 o I u I l I I I I I I I I n ao 4 2 o 2 4 e LOG LUMINANCE cam2 Geisler W S 1989 Sequential idealobserver analysis of Visual discriminations Psychological Review 96 267314 08 I I I I I 02 39 39 FRACTION TRANSMITTED PHOTONS O 4 00 I I I I I 400 450 500 550 600 650 700 WAVELENGTH NM Wyszecki G and W S Stiles 1982 Color Science Concepts and Methods Quantitative Data and Formulae New York Wiley Examples of sine wave gratings of different contrasts 7 Output Input Ah u A u 4 u Ah uv AD uv l uv Phu Pau Ru Phuv Pauv Buy Measuring the transfer function of an optical system The amplitude transfer function is the ratio of the output contrast to the input contrast as a function of sine wave grating spatial frequency M427K Handout Linear Systems Preliminaries Salman Butt April 4 2006 LINEAR SYSTEMS Our next topic is linear systems These systems describe a set of simultaneous equations in the language of linear algebra Linear algebra is well understood eld whose tools are computationally effective in solving dif cult problems We will be exploiting this fact to solve differential equations We will be concerned with rst order linear systems ie the differential equations in our system are all rst order If each of the differential equations in our system is linear7 we say the system is linear otherwise it is nonlinear Given a linear system7 we naturally encode it into a matrix For example7 the set of equations m aubv y cudv can be identi ed with the matrix equation Now this all requires being familiar with linear algebra7 so let us review some of its basic facts Recall the transpose7 conjugate7 and adjoint of a matrix A When are two matrices equal When all their entries are equal 0 will often denote the zero matrix and the zero vector in addition to the number zero 7 context will make it clear which zero we are talking about Two matrices of the same dimensions are added component wise To multiply a matrix by a number scalar multiplication simply multiply each entry by the number Scalar multiplication satis es the distributive laws Multiplication of matrices is the tricky bit Firstly given two matrices A B the product AB is only de ned if the number of columns of A is equal to the number of rows of B Thus if A is an m x 71 matrix B must be an n x r matrix for some r The product AB is then an m x r matrix How do you multiply matrices Like so Exercise Compute the following products when possible The associative and distributive laws hold for matrix multiplication ABC ABC and AB C AB AC But the commutative law does NOT hold AB 7 BA The identity matrix will be denoted by I and it satis es AI IA A We say a square matrix is invertible or nonsingular if and only if there exists a matrix B such that AB BA I If such a B exists we will write it as A 1 observe that if such a B exists it is unique If no such matrix exists we say A is noninvertible or singular A matrix is nonsingular if and only if its determinant in non zero Exercise Are the following matrices invertible or not If they are invertible compute their inverse We differentiate and integrate a matrix function ie a matrix whose entries are functions component wise Exercise Compute the following derivative and integral The standard rules of differentiation hold for matrix functions as well We will often think of vectors as simply n x 1 matrices7 ie as column vectors Thus the above discussion can be specialized to vectors Exercise Multiply the following matrix and vector The Banach space LP by E Odell The 111 International course of Mathematical Analysis in Andalucia Huelva Spain September 2007 Our goal is to explore the structure of the small subspaces of LI mainly for 2 lt p lt oo discussing older classical results and ultimately presenting some new re sults of HOS We will review rst some Banach space basics By Lp we shall mean Lp01 under Lebesgue measure m Unless we say otherwise X Y7 shall denote a sep arable in nite dimensional Banach space X g Y means that X is a closed subspace of Y X g Y means that X is Cz somorphz c to Y ie there exits an invertible bounded linear T X H Y with HTH HT lll g C IfX i Y we shall say X is isometch to Y X gtY means X is C isomorphic to a subspace of Y De nition A basis for X is a sequence Q X so that for all 1 E X there exists a unique sequence ai Q R with 1 aim ie limn 2271 aidg x Example The unit vector basis cl5 is a basis for 19 1 g p lt 00 Of course 6139 61421 where 6m 1 if 2 j and 0 otherwise De nition g X is basic if is a basis for E the closed linear span of Proposition 1 Let Q X Then 1 is basic i a i y 0 for alli and for some K lt 00 all n lt m in N and all a g R 71 m HZWH SKHZWZ39H 1 1 In this case is called K basic and the smallest K satisfying 1 is called the basis constant of 2 is a basis for X i 1 holds and X is called monotone if its basis constant is 1 The proof of this and other background facts we present can be found in any of the standard texts such as LTl AK D FHHSPZ The paper A0 contains further background on LI spaces De nition A bounded linear operator P X gt X is a projection if P2 P In this case if Y PX then X Y 69 Ker P Writing X Y Z means that Y and Z are closed sub spaces of X and every 1 E X can be uniquely written xyzforsomey E Y z E Z lnthis case Pxy de nes a projection of X onto Y Y is said to be comple mented in X if it is the range of a projection on X Y is Ccomplementcd in X if g C If F g X is a nite dimensional subspace then F is complemented in X If X is isomorphic to 2 then all Y Q X are complemented but this fails to be the case if X 00 2 by a result of Lindenstrauss and Tzafriri LTZ Now from Propostion 1 if is a basis for X then setting PM aim 2271 may yields a projection of X onto E linear span of 0314 Moreover the Pn s are uniformly bounded and supn PnH is the basis constant of Not every Banach space X has a basis but the stan dard ones do The Haar basis for LP 1 g p lt 00 The Haar basis hiO is a monotone basis for LP ill E l h2 1012 11271 ha 1014 11412 7 M 11234 1341 To see this is a monotone basis for LP is not hard Via Proposition 1 We need only check a couple of things First W f 012119 3 Win 2 R Z 1 Z if Where Z 2n 9 From real analysis these functions over all n are dense ian1 pltoo Secondly to see 1 holds with K 1 it suf ces to show for all n 01 g R n1 71 H Zaz39hzllp S H Zaz39hz39llp 1 1 5 d i 2i 72J with left half D and right half D supporting the Haar function h 1D 1197 then for all a b E R HalDHP Help th or llalpllp S H01 0119 a b1Dllp This is an easy exercise This reduces to proving if D l is a dyadic interval De nition Basic sequences and are Cequz valent if there exist A B with A 1B g C and for all ai Q R 1 EHZWH S HEWall S BHZWH This just says that the linear rnap T gt with Tm yz for all z is an onto isomorphism with HTH HT lH S C Proposition 2 Perturbations Let be a normal ized Kbasic sequence in X and let Q X satisfy 0 1 gm n M Then is CAeqnivalent to where CA i 1 as A i 0 If in addition is complemented in X by a projection P and A lt m then is complemented in X by a projection Q where gt as A i 0 Notation If and are C equivalent basic sequences we write g De nition Let be basic is a block basis of ifyiOforalliandforsome0n0 ltn1 ltn2 lt and In E R7 31239 Eyin 11551 Note is then automatically basic with basis constant not exceeding that of If is a normalized K basis for X we de ne the co ordinate or bi0rthogorial functionals Via ajxj a1 From Proposition 1 we obtain 3 2K and so for all ai 1 llmlloo S H Emirill 3 EM walla In other words Z aia iH is trapped between the Co and 1 norms of all From Proposition 2 we obtain Proposition 3 Let X have a basis and let C SX E a E X 1 be weakly hall ie gt 0 for all xquot E X Then given 5139 i 0 there exists a sub seqaeriee 0f and a block basis Q SX 0f with lt ez for alli In particular given 5 gt 0 we can choose to be 1eeqaivalerit to a normalized block basis of De nition A basis for X is Karieoriditiorial if for all Zalag E X and all 51 i1 ll Zazwzll 3 KM Ewell 8 It is not hard to show is unconditional iff for all 1 Z aim E X and all permutations 7r of N x Z a a m iff for some C lt 00 all Zola E X and all M Q N H ZieMaixzH 3 CH This just says that the projections PM M Q N given as above are well de ned and uniformly bounded Easily the unit vector basis 6 is a 1 unconditional basis for 19 1 g p lt oo or co Fact The Haar basis is an unconditional basis for Lp if 1 lt p lt 00 This is a more dif cult result see Bu1 if p y 2 For p 2 is an orthogonal basis H in ll 2W More generally if is a normalized block basis of then H Zamlb 2 WW It is easy to check that is not unconditional in 12 L1 For example if 21 h1h2h3h5 h9h17 9 is the sequence of left most his then l Z illl1 Whileforsome cgt0 1 leh TL HD lr 3 H12 on 1 leh De nition A nite dimensional decomposition FDD for X is a sequence of non zero nite dimensional sub spaces of X so that for all 1 E X there exists a unique sequence With 5L Z39 E E for all i and 1 2 xi As With bases the projections Pna PM Z 5L Z39 are uniformly bounded and supn is the basis constant of the FDD Also for n g m if Hmme xi then the 13mm s are uniformly bounded and supngm 13mm is the projection constant of the FDD is monotone if its basis constant is 1 and bimonotonc if its projection constant is 1 A blocking of an FDD for X is given by GZ Fjm711gt for some 0 no lt n1 lt Cl is then also an FDD for X 10 A basis also may be regarded as an FDD with From Proposition 3 we see that if 1 lt p lt 00 and 311 Q SLP is weakly null equivalently fE 31139 gt 0 for all measurable E Q 0 1 then some subsequence is a pertur bation of a block basis of and hence is unconditional just like for bases block bases of unconditional bases are unconditional This fails in L1 by a deep new result of Johnson Maurey and Schechtrnan Theorem 4 JMS There exists a weakly null sequence Q SL1 satisfying for alle gt 0 and all subsequences 31 Q is 1 eequivalent to a block basis of Eli Now lets x 2 lt p lt 00 and let KP be the un conditional constant of in LP We shall list what we consider to be the small subspaces of LP These are also subspaces of LP for 1 lt p lt 2 but as we shall note shortly the situation there as to what constitutes small is more complicated LP contains the following srnall subspaces 0 1 isornetrically If g SLP are disjointly sup ported then HEW3H lzamt pdt1p Zlailplxit pdt1p Z ai pya Also is 1 cornplernented in X Via Pm x Where are the functions naturally biorthogonal to as Signkcmwilp l 0 2 isornorphically Via the Radernacher functions Tn are 1 valued independent random variables of mean 0 Khintchin s inequality For 2 lt p lt 00 12 2 ran H Zanan H Zanan 3 BPltZlanl212 Forlltplt2 ApZlanl212 g H Znnrnup S Zan ran an 21 2 The constants AP Bp depend solely on p 0 2 isometrically via a sequence of symmetric Gauss ian independent random variables in S LP 0 2 191 isometrically For this we use that Lp r11 LAO analpg 1p and LAO 1 1961 r11 Lp01 More generally if we partition 01 into disjoint intervals of positive measure 0311 then Lpun i Lp and Lp i Z Lp1np Hence Lp contains also 0 Z gp 2 2 p E 5L Z39 6 2 for allz and HailH129 lt oo isometrically 13 Our topic will be to characterize when X g LI 2 lt p lt 007 embeds isornorphically into or contains iso rnorphically one of the four spaces 19 2 p 2 or 2p Now some remarks are in order here First it is known that Lqup ifp lt q 3 2 XgY rneans X is C isornorphic to a subspace of Y Thus Lp contain q if p lt q lt 2 so is this small Secondly we have Proposition 5 Let X g 19 1 g p lt 00 Then for all e gt 0 there exists Y Q X with Y Ii 19 and Y is 1 ecomplemented in 19 This is due to Pelczynski Every normalized block basis of ei in 19 is 1 equivalent to ei and 1 cornplernented in 19 as is easily checked Then one uses perturbation as in Proposition 2 Some other classical facts are i The 19 spaces are totally incornparable ie for all XQKPY qpqX76Y ii For13pqltooLqlt gtLpiflq2or1 p qlt2 Also qQLpiff1 p qlt2orq2 14 8 0 0 1 A319 Zagxnwfst 0 2 ABPgtPZ met2 15 by the triangle inequality in Lpg ABppZ Edgy2 This gives the upper 2 estimate Similarly APH Zanxnr 2 1Zawiltsgtp2d8 0 2 f1 lanlplwn8lpd8 2 map 0 using M g H52 The argument is similar for 1 lt p lt 2 D The technique of proof integrating against the Rademacher functions yields Proposition 7 For 1 lt p lt 00 there exists Cp so that if Q SLP ts Auncondz ttonal then for all ai lt1 H atly99 12 lalelxnltsl2p2ds The expression on the right is the so called square function By A g B we mean A 3 CB and B 3 CA 119 Corollary 8 S2 Let Q SLP 1 lt p lt 00 be unconditional basic Then is equivalent to a block basis gm of Sketch By 1 it follows that if is a block basis of hi with on 01 then N By a perturbation argument we may assume each ib z39 E lthjgt Then it is easy to construct the ys Indeed given a simple dyadic function a and any n one can nd 3 E lth gt so that lyl D We are now ready to begin our investigation announced previously if X g LP 2 lt p lt 00 when does X contain or embed into one of the 4 small subspaces of LP7 namely 19 2 19 2 or 2p We begin with a result from 1960 Theorem 9 Kadets and Pelczynski Let X Q LP 2 ltplt 00 ThenXNEQ H2 N Hp onX ie for some C lt HJJHP lt CHCUHQ for all a E X Moreover there is a projection P LP gt X 17 Sketch First note that if 1 E SLP and mt 2 e 2 5 then 3 Hpr 1 g 5 32Hng Indeed 12 Mast mom 12 2 xt2dt 29512 H9428 The direction requiring proof is if X N 2 then H2 N H MP on X If not we can nd Q SX 5L Z39 i 0 so that for all e gt 0 lirnnm xn 2 e 0 From this we can construct a subsequence and disjointly supported fi g SLp with iirnz 0 Hence by a pertur bation argument a subsequence of is equivalent to the unit vector basis of 19 which contradicts X N 2 The projection ontoX with Hpr g CHCL HQ for 1 E X is given by the orthogonal projection P L2 gt X acting on LP For 3 6 LI HPpr S CHMH2 S CHEH2 S CHEHp Remarks The proof yields that if X g LP7 2 lt p lt 00 and X 74 2 then for all e gt 0 pligt5X Moreover if Q SLP is weakly null and e limz39 then a sub sequence is equivalent to the 19 basis if e 0 and the 2 basis if e gt 0 In the latter case we have essentially assuming say is a normalized block basis of with e for all Za12HZW 12 S pngBpZa Pelozynski and Rosenthal PR proved that if X 5 2 then X is GHQ complemented in LP Via a Change of den sity argument Our next result shows that if X does not contain an isomorph of 2 then it embeds into 19 The argument uses Pelozynski s decomposition method Proposition 10 P Let X be a complemented sub space of p 1 g p lt 00 Then X N 19 Proof 19 N X V for some V Q 19 Also X N 19 W for some W Q X by Proposition 5 Finally 19 N 1 19 and moreover 19 N 19 19 1 The latter is proved by splitting ei into in nitely many in nite subsets Thus twdaeaom XVMMXoVMay quoXow oWovom NXeMeXewyeWevemn NXNWQNWQNX D A consequence of this is that if is any blocking of into an FDD then Hmp N 19 lndeed each H n is uniformly complemented in 1731 for some mm7 hence Z Hmp is complemented in 3171 19 Theorem 11 J01 Let 2 lt p lt 00 X Q LP Then Xlt gt p ltgt 2 lt7Lgt X If g lt7Lgt X then for all ggtaxifay The scheme of the argument is to show if 2 lt7Lgt X then there is a blocking of the Haar basis into an FDD so that X Hnp in a natural way 1 Z 51 71 6 En gt E Hnp Since Hnp N 19 we are done We won t discuss the specifics here of this argument but rather will sketch shortly the proof of a stronger result First we note the analogous theorem for 1 lt p lt 2 which has a different form Note the Theorem would also hold for 2 lt p lt 00 and unlike 1 lt p lt 2 the constant K need not be speci ed Theorem 12 Jo Let X Q LP 1 lt p lt 2 Then Xlt gt p if and only if there exists K lt 00 so that for all weakly hull Q SX some subsequehce is K equz ooleht to the unit vector basis of 19 These results were uni ed using the in nite asyrnp totic garne weakly null trees machinery which we will dis cuss after stating Theorem 13 Let X Q LP 1 lt p lt 00 Then Xlt gt p l every weakly null tree in S X admits a branch equiv alent t0 the unit vector basis of 19 A tree in SX is maaeTOO C SX where Twn1nkhEN n1 lt ltnk areinN A node in T00 is all xa7nngtnk where 05 n1 nk or 05 0 The tree is weakly null means each node is a weakly null sequence A branch is given by ib z39 n177m for some subsequence of N It is worth noting that7 just as in Proposition 3 if X Q Z a space with a basis and maaeToo Q SX is a weakly null tree then the tree admits a full subtree yaaeTOO so that each branch is a perturbation of a block basis of By fall subtree we mean that yaaeTOO maaETl where T Q T00 is order isomorphic to T and if ya ama then l704l lozl length of oz ln1nkl h Remarks The conditions for a general re exive X A Every weakly null sequence Q X has a sub sequence K equivalent to the unit vector basis of 19 and B Every weakly null tree in SX admits a branch equivalent to the unit vector basis of 19 are generally dif ferent It is not hard to show that 13 actually irnplies B For some C every weakly null tree in S X admits a branch C equivalent to the unit vector basis of 19 Also B gt A by considering the tree ragga where xmhwk Jink Indeed the branches of ma coincide with the subsequences of But in LP one can show that A and B are in fact equivalent Thus Theorem 13 encompasses both Theorerns 11 and 12 Theorem 13 follows from Theorem 14 OS Let 1 lt p lt 00 let X be reflexive and assume that every weakly null tree in S X admits a branch Cequiualent t0 the unit vector basis of 19 As sume X Q Z a reflexive space with an FDDE Then there exists a blocking 0f so that X naturally embeds into Imp The conclusion means that for some K and all 1 E X 1 2x Jim 6 F we have lle 5 Z HamelP We shall outline the steps involved in the proof First we give a de nition De nition Let be an FDD for Z Let 5 61 6139 i 0 A sequence Q 32 is a 5skz39pped block sequence Wrt if there exist integers 1 3 161 lt 1 lt 2 lt 2 lt so that Hzn Pn7gnznll lt on for all 71 Here for 1 2x1 5L Z39 E El Piggy Emma Thus above the skipping is the PE terrns is almost a block basis of With the Ekn alrnost skipped Now let X g Z be as in the statement of Theorem 14 Step 1 There exists a blocking of and 5 so that every 5 skipped block sequence wrt in SX is 2C equivalent to the unit vector basis of 19 To obtain this one rst shows that the weakly null tree hypothesis on X is equivalent to S having a winning strategy in the following game for all e gt 0 The in nite asymptotic game Two players S for subspace and V for vector alternate plays forever as follows S chooses n1 E N V chooses x1 E SX EZZZn1 Thus the plays are n1 x1 n2 x2 S wins if E AC e E all normalized bases C e equivalent to the unit vector basis of 19 S has a winning strategy means that 3 711 V 1 E SX Ez39z392n1l 3 n2 V x2 E SX E AC e V wins if E AC e 25 V has a winning strategy means that V 711 El 31 E SX V n2 3 x2 E SX g AC 8 Now these two winning strategies are the formal negations of each other but they are in nite sentences so must one be true Yes if the game is determined which it is in this case since Borel games are determined Ma Now if V had a winning strategy one could easily produce a weakly null tree in S X all of whose branches did not lie in AC 8 So S has a winning strategy Then by a compactness argument one can deduce Step 1 20 could be any C 5 here The next step is a lemma of WB Johnson Jo which allows us to decompose any 1 E S X into almost a linear combination of d skipped blocks in X Step 2 Let K be the projection constant of There exists a blocking of Cl E CZquFLNi N0 0 lt N1 lt satisfying the following 26 For all 1 E S X there exists Q X and for all i there exists ti 6 N 1N to 0 t1 gt 1 satisfying a 1 Z CUj b ml lt 6i or llP1t xz will lt z39llwz39ll 0 llp i will lt 5r d lt 1 e HngH lt 6i Moreover the above holds for any further blocking of which rede nes the Ns Remark Thus iii 6 SX we can write 1 2 xi Q X where if B Z 6139 then mpg is a d skipped block sequence wrt Also the skipped blocks Ga are in predictable intervals ti 6 NZ1N And 2 Hxil lt 26 To prove Step 2 we have a LemmaVegt0VN N3ngtNsothattfxEBX 1 231139 31139 6 Ci then there exists t E N n with 25 1 HytH lt e and dist EWX lt 5 1 27 Proof If not we obtain 30 E B X for n gt N failing the conclusion for t E N Choose 310 i y E BX and let t gt N satisfy HPgoo H lt 82K Choose 30 from 310 so that n gt t and HP MyW lt EZK Then HaifaW 21H S Wishn 20H llP oo H lt 8 8 lt 5 2K 2K 39 Also 71 n 5 5 New W HPEc gt m HealH lt i i g This contradicts our choice of 30 D To use the lemma we select N0 0 lt N1 lt N2 lt so that for all 58 E B X there exists ti 6 NFl7 Ni and 2139 E X with lt 55 and HP mx lt 51 Set 5L Z39 3113 zi z 1forz gt 1 Then 2n gt 1 and the other properties bd hold as is easily checked if K 1 2 1 lt 63 D Now let be the blocking obtained in Step 2 It is not hard to show that if 58 Z 5L Z39 is as in Step 2 then Pg Hib Z39Hp1p provided 5 is small enough But this is not the decomposition given by 1 232 31139 E E However we do have 5 N 13827172 yr 1 31 and le PCZV 1N SEEM which yields Kg Hyillp1p by making the appro priate estirnates D Returning to X g Lp 2 lt p lt 00 we have seen that one of these holds 0 X N 2 0 X lt gt p o p GB 2 c gt X The latter comes from Theorems 9 and 11 If X 00 2 and X lt7Lgt p then X contains a subspace isomorphic to 2 so X N 2 Y Now Y also contains p or else X N 2 and in fact cornplernentably as a perturbation of a dis jointly supported Q SLP so p 2 lt gt X 29 Our next goal will be to characterize when X lt gt 19 2 and if not to then show that 2p lt gt X First we recall one more old result Theorem 15 J02 Let X g LP 2 lt p lt oo Assume there exists Y Q 19 2 and a quotient onto map Q Y gtX ThenXlt gt p B 2 This is an answer of a sort to when X lt gt 19 2 but it is not an intrinsic characterization The proof however provides a clue as to how to nd one The isomorphism X lt gt 19 2 is given by a blocking of so that X naturally embeds into 2170169 Emmi H22 pea Before proceeding we recall some more inequalities Theorem 16 R Let2 lt p lt 00 There exists KP lt 00 so that if is a normalized mean zero sequence of in dependent random variables in Lp then for all all Q R K lp 12 HEW p 3 Dar v Dawns 3O Note that in this case lt gt p 2 Via the embed ding Eel5L3 W ml2 Mam e a e9 62 The next result generalizes this to martingale difference sequences eg block bases of Theorem 17 Bu2 BDG Let 2 lt p lt 00 There e117 z39sts Op lt oo 80 that if is a martingale dz erehce se quence m LP with respect to the sequence of oalgebms 7371 then H 22239 2 WW H ZEElt21gt12lt Recall something we said earlier Suppose that Q SL1 is weakly null Passing to a subsequence we obtain which by perturbing we may assume is a block basis of hi Passing to a further subsequence we may assume 5 E lirnz exists If e 0 a subsequence of is equivalent to the unit vector basis of 19 by the KP 31 arguments Otherwise we have essentially 82 WW H 2 23 H 2 0ltpgtZiaii212 using the fundamental inequality Proposition 6 Thus embeds into 19 2 with as a block basis of the natural basis for 19 2 P Johnson Maurey Schechtman and Tzafriri obtained a stronger version of this dichotomy using Theorem 17 Theorem 18 JMST Let 2 lt p lt 00 There exists DP lt 00 with the following property Every normalized weakly null sequence in LP admits a subsequenee satisfying for some w E 0 1 and all all Q R HM Wzimrgttvwltzwgt 12 P We are now ready for an intrinsic Characterization of when X Q LP embeds into 19 2 Theorem 19 HOS Let X Q LP 2 lt p lt 00 The following are equivalent b Every weakly null tree in S X admits a branch satisfying for some K and all ai MM 5 me 1 12 R Dalr pv ZlalFHnHS C Every weakly null tree in S X admits a branch 2 satisfying for some K and Q 0 1 and all ai H Zam 5 Z ai pYpv Z a wgym d There exists K so that every weakly null seguenee in S X admits a subsequenee satisfying the con dition in 1 MM 5 Dani HM Zwaip1pv Z vhf2 where e 11m Z 2 33 Condition c just says that every weakly null tree in SX admits a branch equivalent to a block basis of the natural basis for 19 2 discussed more below Conditions b and c do not require K to be universal but the all weakly null trees hypothesis yields this The latter m near equalities in b and d come from the fact that every weakly null tree in SLP can be rst pruned to a full subtree so that each branch is essentially a normalized block basis of Condition d is an anomaly in that usually every 7 sequence has a subsequence is a vastly different condi tion than every tree admits a branch Here the special nature of LI is playing a role The embedding of X into 19 2 will follow the clue from Theorem 15 by producing a blocking of and embedding X naturally into 21 e 2 H H22 Thus if a Earn 5 E Hm then HCCH N Z llwnllp1p V lexnll 12 The proof of b gt a is much like that of Theorem 14 We produce a blocking of hm so that X naturally embeds into Hmp H22 N 19 2 In fact we obtain a more general result A basis is 1sabsyrnrnetric if it is 1 unconditional and SamH Zawmll for all ai and all n1 lt n2 lt Theorem 20 Let X and Y be Banach spaces with X reflexive Let V be a space with a 1sabsyrnrnetric nor malized basis and let T X gt Y be a bounded linear operator Assume that for some C every nor malized weakly null tree in X admits a branch satisfying HZanrn X 9 HEW Vv HTZanrn Then if X Q Z a reflexive space with an FDDE there exists a blocking 0f so that X naturally embeds int0 GOV Y if a 2a ib z39 E GZ then a I gt Ta 6 Gav Y ly39 35 This is applied to V 19 Z LP and Y L2 where T X gt L2 is the identity rnap So we obtain b gt a and clearly a gt c Indeed suppose that X Q 19 2oo Then given a weakly null tree in X sorne branch is a perturbation of a normal ized block basis of the unit vector basis for 19 2 Thus if Hyz39llrp Cz39 and lela 71 then H Earlill Z lailPlCZ39WP V la l2w 12 Frorn Proposition 6 ll Earlillwpem Z Z laz39lp1pa hence Ejmrf v jmrwa 2 Eat21 E 2 Zam 2lltziargtwv Daml To see c gt b we begin with a weakly null tree in S X and p 2p choose a branch satisfying the c condition H WWYWV Zmrmrf 36 Now we could rst have pruned our tree so that each branch may be assumed to be a block basis of hi by perturbations We want to say that for some K Hm Dani HM Kl We have 2 by the fundamental inequality If this fails we can nd a block basis gm of gin kn kn 7 2 2 7 gm 7 E discl With E 1111 a1 7 1 2 ikn11 ikn1l k 119 and Z loll vanH2lt2 71 ikn1l But then from the c condition yn is equivalent to the unit vector basis of 2 and from the above condition and the KP argument a subsequence is equivalent to the unit vector basis of 19 a contradiction Note that b gt d since if is a normalized weakly null sequence and we de ne maaeToo by xmlmnk xnk then the branches of maaeToo coincide with the subse quences of Note that the condition d just says we may take the weight w in JMST to be him HailH2 It remains to show d gt b in Theorem 19 and this will complete the proof of Theorem 21 The idea is to use Burkholder s inequality using d on nodes of a weakly null tree following the scheme of JMST to accomplish this That argument will obtain a branch scan ozn m1 mn with H EmWV waa where wl Org limn Hxan7n H2 using d E 12 Our next goal is to show that if X Q Lp and X dos not embed into 19 2 then X contains an isomorphic copy of 2p The idea will be to use the failure of d to show 2p lt gt X In the KP argument we obtained a sequence Q SX with the xs becoming more and more skinny lirnm zrjzl Z 5 0 for all e gt 0 and then extracted an 19 subsequence of almost disjointly supported functions Here we want to replace 5L Z39 by a sequence of skinny K isornorphic copies of 2 Theorem 21 Let X Q LP 2 lt p lt oo IfX does not embed into 19 2 then 2p lt gt X We want to produce XZ Q X XZ 5 2 where two things happen First for all e gt 0 there exists 2 so that if 1 E SXZ then Z 5 lt 55 Secondly we need that XZ is not too skinny narnely each B X2 is puniforrnly integrable De nition A Q Lp is puntformly integrable ifV e gt 0 36gt0VmE lt6VzEAwehayefElzlplte 39 Lemma Assume for some K and all ii there exists link Q 5X with rm an i 0 and is K Z equivalent to the unit vector basis 0M2 Then 2p lt gt X Sketch of proof Note that if y air has norm 1 then assuming as we may that is a block basis of Z hi and en then 12 M2 Zailln H3 5 Ken So we have a sequence of skinny K 2 s inside of X We would like to have if y E then they are essentially disjointly supported so N HynHP1p as in the KP argument Unlike in KP we cannot select one yn from each Z and pass to a subsequence We need to x a given for large n so it is skinny enough based on the earlier selections of subspaces and also so that its unit ball is p uniforrnly integrable so that future selections of will be essentially disjoint from it To achieve this we need a sublernrna Sublemma Let Y Q LP 2 lt p lt 00 with Y N 2 There exists Z Q Y with 32 puniforrnly integrable 4O This is proved in two steps First showing a nor malized martingale difference sequence with puniformly integrable has A aim Z a g 1 also puniformly integrable by a stopping time argument The general case is to use the subsequence splitting lemma to write a subsequence of an 2 basis as xi yz39 z where the are a puniformly integrable perturbation of a martingale difference sequence and the 25s are dis jointly supported and then use an averaging argument to get a block basis where the 25s disappear D The subsequence splitting lemma is a nice exercise in real analysis Given a bounded Q L1 there exists a subsequence Q with 5L Z39 y zi yZAzl 0 is uniformly integrable and the 25s are disjointly supported Now we return to condition d in Theorem 19 and recall by JMST eyery weakly null sequence in S X has a subsequence with for some w E 01 am 25 dip pr ai212 Hz 21 l 21 l and d asserts that for some absolute C w 3 limZ Now clearly we can assume that w 2 limZ and if d 41 fails we can use this to construct our 2 s satisfying the lemma and thus obtain 2p lt gt X Indeed d fails yields that we can take a normalized block basis of a given failing the condition for a large C to obtain r3 2 basis yet remains small D So we have the dichotomy for X g LI 2 lt p lt oo Either 0XltH p 2 or 39 Z 2pcgtX39 In the latter case using Lp is stable we can get for all e gt 0 15 2 2 Cgt X 39 The theory of stable spaces was developed by Kriv ine and Maurey X is stable if for all bounded x71 g X7 limlim ymH limlim ymH provided both limits exist They proved that if X is stable then for some p and all e gt 0 19 5X They also proved Lp is stable 1 g p lt oo We have obtained in our proof that if X lt7Lgt p 2 then for some K and all e gt 0 there exist Xn Q X Xn ii a and if an e X H212 1i leanP1p Us ing Lp is stable we can choose Yn Q Xn7 Yn 13 2 for all n In fact we can get 2p complemented in X Via the next result We note rst that if Q SL1 is K equiyalent to the unit vector basis of 2 then as mentioned earlier by PR it is CK cornplernented in Lp by some projection P Also P must have the form true for any projection of any space onto 2 Pm Where is biorthogonal to and is weakly null in Lp 1 1 Proposition 22 For all n let 31 be a normalized ba sic sequence in LP 2 lt p lt 00 which is K eqaivalent t0 the unit vector basis of 2 and so that for gm 6 yf ll 2a 5 Dwarfp Then there exists sabseqaences Q ygm for each n so that ni 6 N is complemented in LP 43 Proof By PR each is CK cornplernented in Lp Via projections P 2m y lxyq Passing to a sub sequence and using a diagonal argument and perturbing we may assume there exists a blocking H7731 of hi in some order over all n m so that for all n m suppy 1 suppy b g H7211 This uses 321 3 0 and 3131 g 0 in Lpl as m gt 00 for each 71 Set Py 2mm y lyy We show P is bounded hence a projection onto a copy of Z 2p Let y 2mm yn m yn m 6 HQ HPyHHZZyZ yn7m1gl N lyglyn7m 2P21P Now 12 2 iyzltyltnmgtgti2 Hagmu OltKHyltnH Where yn 2m yn So HPyH 0ltKgtZ Hynllp 3 C UOHyH Remarks The proof of Proposition 22 above is due to Schechtman He also proved by a different much more complicated argument that the proposition extends to 1 lt p lt 2 ln HOS the proofs of all the results are also consid ered using Aldous Ald theory of random measures We are able to show if 2plt gtX g LI 2 lt p lt 00 then given 5 gt 0 there exists Y p 5X Kln 2 lt 1 5 and moreover there exist disjoint sets An Q 0 1 With for all n y E Y HylAnH Z 1 52 HyH and Yn n E N is 1 8 CP 1 complemented in Lp Where Op is the norm of a symmetric normalized Gaussian random variable in LP This is best possible by GLR We can also deduce the J02 result X g LI 2 lt p lt 00 and X is a quotient of a subspace of 19 2 gt X lt gt 19 2 by showing that such an X cannot contain 2 2p39 We shall prove something more general namely that 2amp19 is not a quotient of a subspace of KP q When pq gt 1 and p y q By duality it Will be enough to 45


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