### Create a StudySoup account

#### Be part of our community, it's free to join!

Already have a StudySoup account? Login here

# Analysis MAT 201C

UCD

GPA 3.88

### View Full Document

## 70

## 0

## Popular in Course

## Popular in Mathematics (M)

This 34 page Class Notes was uploaded by Otilia Murray I on Tuesday September 8, 2015. The Class Notes belongs to MAT 201C at University of California - Davis taught by Staff in Fall. Since its upload, it has received 70 views. For similar materials see /class/187429/mat-201c-university-of-california-davis in Mathematics (M) at University of California - Davis.

## Reviews for Analysis

### What is Karma?

#### Karma is the currency of StudySoup.

#### You can buy or earn more Karma at anytime and redeem it for class notes, study guides, flashcards, and more!

Date Created: 09/08/15

Chapter 5 Fourier Transform 51 Fourier Transform in L1 De nition Let f E LICK The Fourier transform of f7 denoted by 70 f or is a function de ned on R given by foe A arrmam Proposition Some properties of the map f gt gt 1 Linear fltaf y Rn 6 2 ik39 afw gmd a 727rikmfd 727rikmgd Rquot Rquot af f9 2 Translation Let Thf 7 h7 7Thf 727rikmf 7 hd 727riyk6727rihkfydy Rquot Rquot 773k armW 3i Scaling Let ame Rn e2rkmflt gtdw HamilAh ywy let 3 Andy dm Rn 590 An Ak 4 6 HOUR and We 3 M1 gRnleerwaneAwwzwz Her l k Rn e Wk39m wW lLocation of 27r is not standard across texts or disciplines like A 7 1 filea MFWwe 00W The coef cient is needed to have symmetry of f and filv Note that Whenever f 2 0 in this case oo Mm fltogt f Hm De nition We introduce the notation COOK E CR f vanishes at That is for all e gt 0 I Z Z 6 is compact Clearly COOK C CO Rn COOK Proposition 1096 f Note that fg 6 L1 we have f 6 g E Lli Ref Lieb and Loss pl 130 Proof By Fubini s theorem f ggtltkgt R R frI 7 ygtgltygt dydw e2 k39ltygtfltw 7 y dw amk39ygwy Rquot Rquot kmk Usefulness of Fourier transform is it converts differentiation to integration and vice versa Proposition If 170 6 L1 lal g 1 then 1 Cl and 604 72mmw Proof We first justify n l lal 1 Using LDCTi Assuming If E Lli Show f E C1 and 8k 727rizfAi De ne gzk fze 27 39 39k then 99 727mm 99 1 i 7 i lt i 6k 27rzze and g S E L De ne k k S g I n 7 g at Ma kn 7 k be the difference quotient in kl We have S ELli lhnrl S Slip 6k Thus by LDCT Pk lim lim hnzdz gdz 727rize 27rm39kfzdzi We can increase a by induction and use partial derivative for higher dimensioni By direct computing aa k 6 AM foeWm do Ani27rim fme 27rik39m do am0405 Proposition If f 6 Cl and 8 E L1 for lal S l and 80 E COOK for lal S171 then 1907 g ikV kl Proof We check for n 1 lal 1i 0 Do 530k fze 2mlk39xdz fze 2mlk39x 7 fz727rike 27rik39xdzi Recall we assume f vanishes at iooi So the boundary terms vanishi Theorem RiemannLebesgue Lemma fL1R C COOK That is7 for all f E L1R 7 we have A 0 as A 00 and is continuous Proof For the continuous it is easy to use LDCT to show is continuous for L1R 7 hm 727rikmf 727rikomfz kako pointwise7 e hik39mfl S E Lll lirnknkO fAkol Assume rst f E C1 N Cal So f has 1 derivative and compact support use this to construct an approxi mating sequencel Recall if f E C1 N CC by previous fact7 we have lkl is bounded since 8f 27rikfk7 and thus must have compact support Have result for f E C1 N CC dense in Lll Use fact if fn A f in L17 then fn A f uniformlyl Furthermore C0 is complete in the uniform norml So fn A f uniformly to a limit point also in Co Example Let f me then 1 e2m bk 7 727riak 27rik Note f E LICK A 0 as W A 007 and fAE L2Rl However7 L1Rl Example X711 X711 17 M e L1Ri A 7 sin2 27rk 7 2 Proposition Fourier transform of Gaussian is again a Gaussian That is7 let gm exp77r 42 then k xii2 exp77r w A Note that as A A 0 we have a jg approximation Proof By scaling w gt gt t it suf ces to consider the simple case A 1 Since 91cc expw W H expm 7 now it remains to show the result for n 1 Since 91 E L1 R7 glow 7Wxik2ei7rk2dz eiwlklz 77racik2dz 91khk R R where hk ei x lkvdzl R Now we claim the above hk E l by showing dh 2 d 2 06 72m Rz 2k exp77rz 2k dz z Eexp77rz 2k dz 0 and h0 ff 6 ng 1 D Proposition Let g 6 L1R i We have fltygtgltygtdy fltygt ltygtdyl Rquot Rquot Proof Simple application of Fubinii write it out D 52 Fourier Transform with L2 Theorem Plancherel s Theorem 1 If f e L1Rnn FOR then k 6 FOR and my 2 The map f gt gt f has a unique extension to a linear map from L2 Rn which is an isometry7 iiei7 Plancherells theorem holds even if f L1R i 3 Parsevalls formula g where 7 d ltf79gt Rn f917 17 Proof 1 If f E L1R L2lR 7 is bounded and hence A is de ned Note that gy7 k exp767r E L1R3 i n A 2 2 we expeewlkl we fk 2exp767rlkl2dk m eimlk dk Rn aimmung emik39y wdyE Mlklgdk n n Rquot 727rikEiyfmmeiewlkl2dkddy Rn mm expemkl2e2iik39ltm ygtdk dwdy Rquot Rquot Rquot 2 fwWe 2 XP MWMV Rquot Rquot 6 Note that 2 6art2 exp gydyjeg g Ru 6 as e A 0 by C00 approximation theoremi Thus A A fmW5n2exp77rl yl2dmdy HRn muf n AM 1 2expe7rlkl2dk a Rn by monotone convergent theoremi To conclude7 we have ML Hit 2 Assume f E L2R but f L1R i Since L1 N L2 is dense in L27 so there exists E L1 N L2 such that 7 A 0 Use isometry to show is Cauchy On the other side we have k 2 dk H19quot ka2 m 7 M2 Mfr fH2 m 4H2 0 So is Cauchy in L27 which is complete So fhave a convergence subsequence7 denote still fj A f in L Note the de nition of fdoes not depend on the subsequence hm hm llfjll2 Jam jam Furthermore one can use this subsequence to show that f gt gt ffor f E L2 is continuous and linear 3 By polarization identity 1 ltf79gt1Hfgll2illfiyllr17iHfH271iHgH2 1 A A 1 llf ll 239 fi9 27lt17igtl l27lt1igtw ltf7 gt 1 Note that f gt gt fis an isometry on L2R i Moreover it is a unitary transformation ie it is an invertible isometryi De nition Inversion Formula Let f E L2R we have fvltwgt 7 few Amounts to switching i gt gt 7i in original formula fwv 7 R em39m wk Now we need to prove f Proof Let g x are Gaussiani Note if fj E Ll LQ with the previous proposition and 731 x 317 we have a me 7 AM my 7 wgtfjltygtdy R Akfjkmdk by lettingky7m g kfjkeik39mdh Now we consider f E L2R and construct such that fj A Using de nition and Fubini giltkgtfltkgtewmdk7 explt7mkl2gtemm 6 2 ik39yfjydydw Rquot Rquot Rquot 0471 quot 7 explt7mkl2gtamltwgtfjltygtdkdw Rquot Rquot Here we want to use f 6 L2 not fj E L1 N L2 by Holder since y 7 6 L2 bounded 57W 17fjydy 7 1 17fydy S NAME 7 Mb Hfj fll2 v 0 Rquot Rquot as j A 00 Now have my 7 wfydy 7 giltkgtfltkgtewmdk Rquot Rquot for f E L Now AM my 7 wfydy 7 fw by C00 approximationi Note g A as A A 0 by LDCTi Thus giltkgtfltkgtemmdw7 fltkgte2mdw7ltfgtv 1Rquot Rn Now we know Fourier transform is de ned as L1 functions with range contained in L L2 functions with range L2 constant function in L but not f of f 6 L1 What about L17 Do we have an estimate H lq 07 w Observe that the duality of 104 is necessary otherwise My M a co by scaling Fact we need I lt p S 2 Note that we have already done for p I p 2 If p gt 2 using scaling for Gaussian yields lgHfllp we 53 HausdorffYoung Inequality In the rest we introduce the HausdorffYoung Inequality where the proof relies on RieszThorin Interpolation Theorem convexity inequality which in turn relies on a fact from complex theory ref Folland Lemma Phragmen Lindelof Maximum Principle or Threeline Lemma Let F be a continuous function de ned on SzziyEC OSISI and F is analytic inside the strip If for all g S M0 lFl S M1 then lF2l MOHM In other words let log 1 kg sup z y E R then is convex on 01 Proof Assume M0M1 gt 0 else take sequence toward 0 By considering F G 7 lt2 MOHMIZ it s suf cient to prove the result with M0 M1 1 If S I and lGl S I we have S I and thus S lMg szl S Mg fo Now let M0 M1 1 If lim y H00 Fz 0 uniformly in z Maxmod principle implies the result Fnltzgt Fltzgtelt1 gtn anzl mm w eiyzneWLDn 3 mm m effn a 0 uniformly in I as n A 00 and S1 anl S I we have S1 Now taking n A 00 and thus get the result D Theorem RieszThorin Interpolation Theorem Let p0 p1 qo y ql T L7O Rn H Um Rn with norm M0 T L171 Rn A U Rn with norm M1 then T is bounded from LPR A LqR with norm Mp MOHME where 1 P0 P1 4 40 41 Remark For Fourier transform 7 lt m1 f L1 a U M0 1 W l lfH2 szlaL x M11 Now interpolate f L17 A Lq S MolieMle One can use RieszThorin interpolation to get Young s Inequality for convolutions Tf R f1 7 wow linear and bounded ifg E Lpl 1 directly apply RieszThorin interpolationi Theorem HousdorffYoung lnequality For 1 lt p lt 2 f E LNR L1R we have H Equality is achieved if and only if p S 0 Hfllp where 05 p1pp 1p f0 anP lt177A17gt 717 is Gaussian where a E C c E C and A symmetric real positivede nite Proof The proof relies on RieszThorin lnterpolation Theorem convexity inequality which in turn relies on a fact from complex theoryi Need fact Converse of Holder inequality Hhuqsuplthggtl Ham1 Using this fact we can write Mm supltTfggtl Hpr M 1 Because Mpq supf 0 i Note that p q lt 00 because p0 p1 qo 41 f and y can be approximated q by simple function fr ZanA 17 99 Zkade 1 k recall f 6 L17 1 g E Lq Hgllq 1 For 0 S Rez S1 de ne 1 ilizz 1 ilizz 102 P0 P1 W2 46 4 1 in the end 2 9 is what we want Extend f and g to S 2 E C 0 S Rez S1 awn Z Z aj P9Pl eiargW XA I7 j W m2 ZlbquW M eiargaywxggy k Note that Since f 6 Lquot for j 01 we have M 6 LP W 6 L44 we 6 Lg and same for derivatives simple functions f 6 L17 9 E Lqi So Tab 6 L47 by Holder7s inequality and Tab ng HTfH Hgl Compute Z wat po aj PWWoMl eiarglt07gtXA7 1M1 f 179 j H f PwVPO p0 179po 1 p0 llfllpo A Chapter 2 L79 Spaces 21 De nition of LP Spaces De nition Let Q be a measure space With positive measure it 1 S p lt 00 Wald Q A C l f is Mmeasurable and lflp is integrable or Msummable L Qdt Q A C l f is Mmeasurable and there exists k lt 00 such that S k Mae z E Q If f 6 L17 dc we de ne the Lpnorm of f by 117 Hpr vi a For the case p 007 if f E L Qdt7 we de ne With essential supremum unlike the usual supremum7 esssup ignores sets of measure zero Hflloo esssup WM 169 the Lmnorm of Remark L17 is a vector space Why Note that glp S 217 1 lflp lglp by the convexity of t gt gt tp7 and af By E L17 if g 6 L177 Note that is only a seminorm for functions We Will use equivalence classes of functions to de ne L17 that is7 f N y if 91 except on a set of measure zero Picking if f E L17 makes no sense Thus is a norm for the equivalence classes of functions for 1 S p S 00 Need Minkowski s inequality triangle inequality Hf pr m M which fails if 0 lt 20 lt1 The triangle inequality is equivalent to convexity of the norm 0 S A S 17 HM 1 Mgllp S A Hfllp 1 A HgllpA Note that the statement HfIHoo 0 f10 fails if esssup is replaced by sup Example Note that 0 lt p lt 17 L17 is not a norm because the triangle inequality fails Let a gt 07 b gt 07 0 lt p lt 1 for t gt 0 t1 1 gt a tp 1 Integrate from 0 to b b b tp 1dtgt atP 1dt 0 0 a 1217 gt H b e a 7 gt a b and thus Example Let E and F be disjoint sets of positive measure 1 iszE 13117 b FlP x 1 L114 14 OifI E Exercise 1le XFHp 2 HXEHp HXFHpA De nition Dual index Let 1 S p S 00 p is called the dual index if or p 17 p 007 and vise versa 22 Elementary Inequalities De nition K E R is called convex if for all 0 S A S 17 Az17y K forallzyEK De nition A real function f is called a convex function on a convex set K if fz17y S 1 7 for all Ly E K and 0 S A S 1 1f equality never holds for I f y then they are called strictly convex In this section7 we assume a b 2 07 p 2 1 if there is no additional assumption stated Lemma Convexity of t gt gt tp a bp S 2p 1ap b Proof Observe the convexity oft 17gt tp7 we have n1 p 1 1 lt71 71 lt gt72a 212 2 By rearranging these term we proved the lemma Remark The above lemma is a special case of Jensen s inequality Lemma Let 0 lt t lt 17 then 11 g to 1 711 Proof Since a blquot b at is concave up7 we have m tf117tf0 ie7 a blquot S ta17tb Lemma For 1 S p S 2 0 lt b lt a we have a 12 a 7 In 2 Zap pp 71ap 2b2 Proof Let t g and thus 0 lt t lt 1 Now what we need to prove is 1 1 H171 2 2 pp71t2 Observe that lt1 i t 7 1 w we have 1tplitp2 lt2pkgtt2k 160 m p L 7 2 2k 72pp 1t 2 21 162 2 2 pp 7 12 and thus we complete the proof D Lemma 1 1 ab S 7a jbpi p 17 Proof Observe that t gt gt e is convex7 and 1 1 ll ab eln ab eln alnb eiln ap rln bP S 161ml lelnbp lap lbp p p p P Corollary Let s gt 01 We have ab lt65 ltewbgt Corollary Let y be continuous7 07 and strictly increasing for z 2 01 Then ab A mm 1 ar1ltygtdy Equality holds if and only if b a1 Lemma Let 0 ltMg1 ml 6 R We have Mp la W S 001 Mp S la W lalp Proof First we consider t 6 01 Applying the convexity of the map t gt gt la tb p for xed ab 6 R we ave 1atb1p 1 7 ta tabp g 17mm tla b1 Equivalently7 by rearranging terms we have 1 3 WWW W7 S lablp lull Moreover7 p a ta tb ti 1 t 7 7 777 tb 7b a 1t1t1t 1 11lta 1ta l t lt7 tbp 7b 1t a i1tia Equivalently7 by rearranging terms again we have 1 W7 i la W S WWW W7 For the case t E 717 07 we just apply the transform 2 A 7b in above and we obtain the same result 1 Wu laiblp S gltlatblpila S lablpilalpl 23 Basic Inequalities in L1 Theorem Holder7s inequality If p p are dual indices7 Hngl S Hpr Hngw Equality holds if A lfzlp71 ae for l S p lt 00 Note that it is called CauchySchwarz inequality if p p 2 Remark If f1 fm are functions on Q With fj E L177 and i 17 then 177 ycde H Hfij j1 j1 Why Let f I fl7 g I fj and apply Holder7s inequality and by induction Proof of Hb39lder s inequality It is trivial that p 00 or p 00 think of it If 0 or Hng 07 then 0 ae or 91 0 ae So assume 3e 07 HgHP 3e 0 Also trivial so 10 4 Let p 17 fltzgt gltzgt llpr 7 Hpr To prove Holder7s inequality use lemma With a and b as on previous lemma Av 1471 31 p p p p 117 17 lp p 17 WIN WIN 7 1 9 I lt 1 WIN 1 MIN 7 7 Hpr Hpr Hpr Hpr 0 llpr 0 Hpr Integrate over 9 d f9 was mm M S 3 3 1 Hpr Hpr 0 0 The Holder7s equality holds directly from akbli Aa 17 Ab if a 12 Theorem Minkowski s inequality For g 6 L176 du7 Hf9Hp S Hpr Hpr f0r1 p S 00 Proof We use Holder7s inequality to prove Minkowski s inequality If p l7 lf9ldMSlfldMlgldu If f g 0 ae7 than this is trivial ln general7 my m lgllfglp 1 m fgp 1 m Mm fmdtsmwgv dwgHMgW W Hpr M ngv p 117 HfH M lt wng dM 7 1 7 1 Note that p71p 7 p and 1 7 17 7 5 Leave Hng on the r1ght hand s1de we have 17 1116ng ltlfyldegt WWW D Theorem 1f is a Cauchy sequence in LWQ then there exists a unique f E L17 such that A f converges strongly Moreover there is a subsequence with the properties 1 is dominated by some positive function Fz 6 LP ie S Fz ae z for all k 2 fik A pointwise ae Proof It is suf cient to show convergence for a subsequence and this will give convergence for full sequence and uniqueness for the limit Let fik be a subsequence that converges strongly to f in LPG as k A 00 Since by the triangle inequality flk A rm LP Cauch Se Hence yAq A 6 6 llfz39 fllp S llfz39 fikllp llfik fllp lt i g 5 Now we construct such subsequence So choose i1 gt n 1 1 llfilif llgi7 7 Hfik fik71llpggik lkgtzk71gtgt21gtn De ne l Fl8 lfirl Z lfMI 12171901 161 is positive and monotonically increasing l 1 HFal S Hfil 11 2 S Hfil 1117 1 161 By monotone convergence theorem Fl converges to F and F E L17 pointwise ae Note that fik1 7 1 fika 7 fik 7 this sequence converges absolutely ae z to Furthermore lfik w S so by the Dominated Convergent Theorem fik A pointwise ae and 6 L17 Also Hfz39k pr S HFIHp Hfllp and thus fik A f strongly in L17 1 Remark Note that A f in L17 does not imply A f pointwise ae See HW4 in MAT201b Theorem L17 is a Banach space Exercise In general L17 gZ L4 for all p f L For example 9 0 00 db dz 1 a gt 0 1 faz01 E L17 if and only if p lt 071 In this case lflp blows up too fast at 0 lflp gets worse as p increases 2 faz1oo E L17 if and only if p gt 1 1 In this case lflp decays too slowly at 00 lflp gets better as p increases Remark Higher dimension Let 7 then dz lgnill Tnild l n 2 where lgnill 27 is the surface area of the unit ball in R ie mB T 1B0 T lgnill T F 24 Some Useful Inequalities of LP De nition A support plane to a graph of a function f K A R at a point z E K is a plane in Rn1 that touches the graph at z and that nowhere lies above the graph Remark If f K A R is convex then there exists a support plane such that fy 2 fr V yer Where the vector V E R exists such that V y 7 z locates on the plane and projects on y R suppo t plaQ f y fltzgtVltyezgt y Rn I Figure 21 A support plane of J at z The support plane at z is called tangent plane if it is unique lf J is convex the existence of a tangent plane at z is equivalent to differentiability at I Also a convex function J is continuous and thus Jof is measurable for any measurable function De nition Let f E L1Q De ne the average of f on Q by 1 f I 7 f an lt gt an 9 Theorem Jensen s inequality Let J R A R be convex and f 6 L1 Then lt10 fgt 2 MW Moreover assume J is strictly convex at then the equality holds if and only if f is constant Proof Since J is convex its graph has at least one support line support plane in R1 at each point Thus there exists a constant V E R such that JO 2 JltfgtV ti ltfgtgt for all t E R That says J o f the negative part of J o f is in L1 and thus f9 J o f an is wellde ned could be NOW substitute be t and integrate over 9 we have 0 Q JfIdMZ Q JltfgtdMV39 9 fzdM7ltfgtQ a Q Jltfgtd and thus by dividing 6 we hage J o 2 Assume J is strictly convex at then is a strict inequality either for all t gt or t lt Suppose f is not constant then 7 takes on both positive and negative values on sets of positive measure Which contradicts With the previous sentence Holder s inequality is a direct result of convexity and Minkowski s inequality is a result of Holder Theorem Hanner7s Inequality Let g 6 L1 Q7 l S p S 2 Then 17 17 lt1 HfyHZHf7yH 2 impugn leHp7 Hyllpl 17 17 lt2 Hf ng Hf 7 9H lepr 7 Hf 7 all 217HfHZHyHZ A When 2 S p lt 007 the inequalities are reversed Remark We see that 2 improves the triangle inequality by convexity at t 7gt ltlp Why Let A Hfng B 79H for simplicity We have 2 HfyHZ 7 2 W lABlp lA7Blquot 217HfHZHyH 2 consequently we have HfyH 2P m if m 7 Hal Note that for p 27 l and 2 are equality lg or p l7 1 is just triangular inequality 2 can be derived from 1 by substituting f 7gt f 97 g 7gt f 7 g and do some calculus 25 Differentiability of Norms Theorem Gateaux Directional Derivative Assume g 6 L Z Q7 l lt p lt 00 The function de ned on R by M2 7 Hf tall 7 was we dz 9 is differentiable7 and its derivative at t 0 is given by d p S d S P p72 7 L 167me 7 EN 0 7 x was fltzgtgltzgtfltzgtgltzgtgt dz t0 t Remark Note that lflp72 f is well de ned for l lt p even When f 0 in Which case it equals 0 Also7 W f M f 7 LP 9 Remark By direct computing informally With chain rule assume g are realvalued function7 dHft W7 dm gt72 lt M 7 dww lt M dt 9177aQ I gI I7 th I gI I 7Qplf t9 l lfIt9Il 7 109 was 7 we fa tgltzgtgltzgt dz gI dI and by setting t 0 we have d p 7 p72 dtwmup 070Qlfl was t 26 Convexity De nition K is called a conveI set in LPQ if g 6 K implies tf l 7 tg E K for 0 S t S 1 De nition K is called a norm closed set in LPQ if 9 is Cauchy in K7 implies that its limit 9 is also in K Theorem Projection onto Convex Sets Let K be a convex set and a norm closed set in L17 Q7 l lt p lt 00 Let f K and de ne the distance D dlstfK 112 Hf 7 yllp Then there exists h E K such that hllp D7 and every 9 E K satis es Rey7h7 lf7hlp 2 Mo 9 Proof Proof for 1 lt p S 2 only WLOG we assume f 0 translate otherwise and 0 f K et hj E K C L17 be a minimizing sequence ie thHp 7gt D We claim hj is Cauchy First we have th hkllp S thllp Hhkllp A 2D By convexity of K hjhk E K implies hj hk E K and thus 1 39 7 39 lt 7 v D glgl llf yllp glgl llgllp 7 H201 l hkl 17 and thus th thp 2 2D So now we have th hkllp 7gt 2D as jk 7gt 00 Suppose for a contradiction th 7 th 2 b gt 0 By uniform convexity 2 I 17 17 ltth hklp MM 7 My llama 7 HM 7thpl 217thHZHthZ 7 2WD that is l2D bl l2D 7 bl g 2WD Again since t gt gt lQD tlp is strictly convex l2D zlp l2D 7 zlp gt l2D z 7 2D 7 zlp 7 2P1DP unless I 0 which implies b 0 So th 7 thp 7gt 0 which says hj is Cauchy Since K is norm closed hj has a limit h E K To verify Re g7h77 lf7hlp 2 ago 9 we x 9 E K and set 9 l 7 th tg E K by convexity for 0 S t S 1 De ne with f 0 as before NO Hf 7ng2 2 DP while N0 DP Since Nt is differentiable by Gateaux Derivative we have that NO 2 0 which is exactly the above formu a D De nition A space is uniformly convex if for all 2 2 e gt 0 there exists 6 gt 0 such that 1 Hz 7 2 6 implies S 1 7 6 De nition A space is uniformly smooth if the limit 7 7 1mm all M 70 t exists 27 Continuous Linear Functionals and Weak Convergence A closed bounded subset of R is compact but not so for LPQ even if Q R even if Q C R and compact We need a new notion of convergence to have bonded sequence to be compact De nition De ne L LPQ 7gt C is a linear functional if Laf1 W2 aLf1 bLf2 for every f1f2 E LPQ ab E C L is a continuous linear functional if I102quot LU as fi A f in L for every sequence 6 Lquot L is a bounded linear functional if S k for k lt 00 a constant Note that the boundness is equivalent to continuous for linear functionals why De nition The dual of LNG denoted by LNQ 7 is L L is a continuous linear functional on LP is a normed vector space with operator norm check it lLH sup lLfzl mfl De nition Let E LNG We say converges weakly to f7 denoted by 4 f7 if A for all L E LNG Proposition Strongly convergence implies weak convergence Proof If 7 pr A 07 we have lLfiLfl 111027 fl S MN 111 7 fllp A 0 What does LNQ look like Hg 6 LpQ7 we de ne Lg E LPQ by Lgf dz for f E LNG We claim that Lg is a continuous linear function Holder7s inequality says Lgf is bounded So Lgf is a continuous linear functional in LNQ By Riesz Representation Theorem7 we have Lp is given by Lg where g 6 LP for 1 S p lt 00 Thus7 any 9 6 LP is identi ed by Lgf 9 9f d with the element Lg E To conclude7 LNQ Lp for 1 S p lt 00 There are three basic mechanisms showing that weak converge cannot imply strong converge 1 Functions may run away77 to in nity fig 91 I Let us compute Lfk Assume fk 6 L176 for 1 S p lt 00 Take hz 6 LP Q7 more speci cally7 we take h 6 Cam ok since CEO is dense in Thus dz A 0 as k A 00 because suppfk supph 0 eventually E0 Even if Q is compact7 we may still lose Functions may oscillate Einkz 0 S z S 1 otherwise By computing we have 1 1 d 1 sinkzhz dz i lt77coskzgt hz dz 0 0 dz k 1 cosUcz z dz zero boundary terms 0 and thus 1 S hz dz A 0 as n A 00 0 3 Functions may concentration or go upon spout klngiDI g E L Choose p 2 91 1 Theorem Linear Functionals Separate If f E LNG and 0 for every L E LNG then f E 0 are Consequently if A h A g in LNG then f g are Proof 1 lt p lt 00 De ne 91 7 lfzlp 2z if M d 0 0 if fz 0 Note if f E LNG then 9 6 LP G since 7 7 17 7 Hdug Q mzw 2 fltzgtl dz 3 9 W 1 dz m lt co For 91 de ned above let Lglthgt Q dltzgthltzgt dz Recall we know Lg de ned as above satis es Lg E LNG The fact that L is an integral it is linear By Holder7s inequality it is bounded and thus continuousi Now compute Lam QQIVI dz Q lfrlp 2rfz dz Q My dz HfH since 0 for every L E LNG and Lg E LNG we have Lgf 0 that is f 0 are p 1 De ne 7 91 7 zlfzl fltzgt do otherwise i i Need to know 9 E L G obviousi De ne Lgf dz 7 7 WI 7 7 o 7 Lgf 79 WWW dz sz dz M1 and thus f 0 are p 00 Assume the measure is a nitei Set A1 1 lf1lgt0z a niteness guarantee that there exists a measurable set E C A such that 0 S MB lt 00 Set 91 7 SW WIN 21erise A Now 9 E L1Gi De ne Lglthgt Adltzgthltzt and thus 7 7 fltzgtfltzgt I7 I I LacyQTI d Amnd Sof0aiei onGi El Theorem For 1 S p S 00 the L17 norm is wetlka lower semicontinuous That is if fj A f in LNG then 1113ngth 2 Hprz If p 00 we need the extra assumption that the measure M is a nitei Moreover if fj A f in LNG then 11ng Hfij Hprz 10 Proof For 1 S p lt 00 let gW m m 0 LAWA awmd 7 otherw1se By the weak convergence of fj we have llfllg 11m Lgfj 11m ij dz S lijxgglfllyllp Hfjllpv jam Jam where MAWWW RN that is HgHP Hfllgili Therefore p d 17 d 17 I AMEN I Hfllp HfHZ S limin HfHZ 1 Hfjllp jam and the result holds for l S p lt 00 For p 00 assume a gt 0 Consider A5 169 z gtaiei By a niteness there exists a sequence Bk with MBk lt 00 such that Ae BkTAei otherwise Am lr6amp 3k As before 9 E L1Q and thus MAE Bk liminf 2 Vlim ghefj dz by Holder7s inequality jam Jam 9 dz by assumption of weak convergence As Bk 2weaMampan Putting it all together we have MeanmmemmzweoanBm Jam Recall a we have llfll00 7 6 S 1311ng llfjlloo Thus llfllm S liJHiiogfllfjlloo We skip the proof of the equality when strong convergence It needs uniform convexityi 1 Theorem Uniform Boundedness Principle Let 6 L17 1 S p S 00 If is bounded for every L E LWQ then S C for some constant 0 lt C lt 00 Remark The theorem shows that weakly convergent sequences are at least normboundedi Proof We prove by contradictioni Suppose is unboundedi WLOG we suppose llfjllp 4j by the following reason taking sequence Hfjk Hp 2 lek and modifying ij fjkilj c Hfc Hp then clearly HF Hp lek and Lij Lfjk4j lt llfjkllpi Thus if Lock is bounded so is Liji lflep Q 711 HfjHi l De ne an E C recursively with lanl 1 as follows 01 1 an be chosen by requiring 0 f9 Tnfn db to have the same argument as S jaj f9 ijn dbl Thus Tm n71 V 1772 Zs ajandt 23419 Tm dM 3419 MW d W 4 n wwp g Lh Zg 39aj Tjh dt 11 9 which is obviously continuous by Holder s inequality and the fact that llTij 1 Thus k v 00 v Zg ajijk dt 7 Z 3 4k j1 jk1 k 13 1 4 23 k4k73 k4k L 7 7 H00 as kaooi 1 2 3 F1 De ne lLfkl 2 1 3 This contradictions the hypothesis that is bounded 1 28 The Dual of 1109 We come to the identi cation of LpQ the dual of LpQ for 1 S p lt 00 This is F Riesz s Representation Theoremi Note that L Q 2 L1 Theorem For 1 S p lt 00 LPQ is Lp in the sense that every L E LPQ can be written as W Q vltzgtfltzgtda f e mm for some unique vz 6 LP For p 1 we need to assume that 94 is a nitei Note that for all 1 S p S 00 we have HLH llvllp Proof For 1 lt p lt 00 Fix L E LPQ and de ne K E LpQ Kf Lf0 Note that K is convex linearity iiei let fg E K we have if 1 tg tLf1 tL9 0 Note that K is norm closed continuity iiei let gn be a Cauchy sequence with limit 9 E Ki Since L is continuous we have 0 Lgn A Lg 0 if 9 A 9 because 9 is Cauchy and L17 is complete 0 K is a convex norm closed set in L171 Use projection onto convex sets Take f K iiei f 0 we know there is a h E K which minimizes the distance between K and f and Rekudu S 0 foreverykEK with WE WI MIMI WI 31 12 Apply the linearity of Ki If k E K then 716 E K which implies ReukSOand Reu7k 0 Reuk0i Also if k E K then ik E K which implies Reuik 0andthus lmukgo uk0i Let g 6 L176 be arbitrary Write g 91 92 with 91 fihy 92 yigr Note that Lf 7 h is nonzero and 92 E K since Lltg2gt My 7 M91 My 7 Lltggt 0 M91 My and thus uyu91uy2u91LyA where 7M7 1 7 1772 7 7 7 1 7 p A7 WW L imlf M f hf hwiwim f hl 0 Now we have vz is wellde ned since A f 0 and v 6 LP since because f h E L Thus the existence of vz is constructed To show uniqueness x 9 6 L176 suppose we have 1 f w and My v97 My wg v wy 0 gltzgt lviwlPJwiw 0v7wglviwlquot gt0 HLH Observe that for all g E LWQ Let and thus a contradiction Show Hva where Lfl sup fELP9gtllfllP51 Observe that lLfl Slvfl S llvllp Hfllp S Hvllpw Note that Holder7s inequality states that if A lvlpil are then equality is achieved ow we prove for p l wit additional condition that 9 M is a nitei Case 1 M9 lt 00 we have LTQ C for 7 2 p by using Holder7s inequality W 1 g WWp pTsthIWfHTwo A A A gt 13 lLfl S Cllflh Q lfl S WWI llfllp for all p 2 1 If f 6 L161 then f E for p 2 1 Moreover L has a restriction to LPG which is continuous in L1Qi By the previous proof for all p gt 1 there exists a unique vpz 6 LP such that W vprfrdz for allf E LWQ But f E L17 implies that f 6 LT for 7 2 p ie on a nite measure set we have LT Q L17 for 7 2 10 So we can choose up independent of p which yields 1 E L pi 6263 So there exists a unique Lp for 1 lt p lt 00 and we have f lt 00 De ne MI gm was k otherwise Note that S 6 L1 fk A f in L1 vfk A of in L11 By LDCT m 1131wa m vf for v E L f 6 L11 So the Riesz representation theorem holds for p 1 with the assumption 6 lt 00 Case 2 Now consider M9 00 Since we assume that Q is a nite we can write 9 r Qj disjoint Mnj lt 00 C8 j 1 Take f 6 L161 we have 1 212m mac gm I 6 91 otherwise By M j lt 00 and previous argument there exists a unique 1111 1 E 9 such that Lfj Q vjf LELWQ The vjls all come from the same L C is xed so S C for all j De ne 111 1111 1 E Q o countably additive iiei MU 91 ZM917 j1 j1 LU vf 2491va 9 F1 The uniqueness is omitted here 1 29 Convolution De nition Let fg R A C De ne the convolution of f g to be f ggtltzgt R m 7 wow Remark Note f 96 g g 96 f by change of variable The product is de ned if the two integral on RHS makes sense For example 1 if f E LPGR g E LpRn then apply Holder7s inequality 14 2 if fg E LICK then f 95 91 6 L1R i HW consequence of Young s inequality for convolution More is true as Youngls inequality showsi Theorem Young s Inequality Let p q r 2 1 1 i 2 f e LNR g 6 L4 Rn h 6 LTR 1 Then mow hgtltzgtdz R A A fltzgtgltz 7 ygthltygtdydz 1 1pr M Hm Note that 1 is not sharp there exists C173 lt 1 Proof The proof depends only on Holder7s inequalityi De ne I AM Analtz7ygt ltz7ygtvltz7ygtdzdy May fIpTyryyqT my gltz 7 ygt4PhltygtTP my fltzgtPqhltygtT 14 Note that by the setting of p q 7 we have 1 1 1 pg1 where the primes denote the dual indicesi Applying Holder7s inequality for three functions we have I S Hle H llp Hvllq where 17 17 HaltgtHT laltzygtvdzdygt lfrlplyriylqdzdygt Rquot n Rquot n 1r lfrlplytlqdzdtgt bylettingtqiy Rquot Rquot mg mg Similarly 1117 Tp 7 pq Tq H llpillgllq HQHT 7 Hillqillfllp NWT 7 and thus Hallw ll llp Hillq Hfllp Hyllq thlr which yields the Young s inequality 1 Corollary Young s lnequality for Convolutions Let g 6 L4 h E L Then 9 96 h E L5 and Hg hlls S Hyllq llhll r where1i Proof We know from Holder that there is equality given a relationship It gives us an idea by letting fx 67191 19 hxlpp 91 provide that g 6 h em lg hl in Youngls inequality which yields lfzyhzldr lghlldz ghp dz Hamil 11pr 11qu My 15 where 117 117 I Ll Hmp1 m Mhvd H9Mb MM Rearranging these terms we have Ll Hy hHZ S Ma hHZ 11qu thlw Thus Hy hlls S Hgllq thlr where Remark Note that for a special case that 7 1 and s 417 we have My th S 11qu My 210 Approximation by 000 Functions Lemma Let j E L1 Rn with Rn jzdx 1 For 6 gt 07 de ne mma Then law39de 1 and lljllLl lljellLl jg de ned above is often referred to as a molli eri Theorem Approximation by C00 Functions Take f E NOR 1 S p lt 00 De ne fa je fl Then fa 6 EUR llfellp S llj lll llfllp lljlll llfllpv and fa H fin Lp Furthermore if jg 6 CC00 Rn7 then f5 6 C00 Rn and Dafe Dal f jg DD f in distribution sense Proof We rst simplify life by claiming it is suf cient to assume 1 j7 f have compact support7 suppj C B31 2 f E L 7 and assume p 1 above For claim 17 we de ne 131 CX ltRIjI for some 0 lt R lt 00 and C gt 0 satis ng ij 1 and inHl lt 6 De ne j 6 7LjRze7 which has supp jER Q I z lt Re Note that the number 1mi m is independent of 6 By Youngls inequality7 Hkfij hHUei 3 h H evme b ltllj lel1llfllp lt 6 i dIHiiRh 16 Thus Hmf7pr lljefj5Rpr llj f7fllplt6626 provided that if we can prove Hie f7pr lt 6 for small enough 6 Henceforth we shall omit the R and just assume that j has support in a ball of radius R Similarly by adding another error 06 we can assume f has compact support For claim 2 using Youngls inequality and LDCT we can also replace f by fk X f lthkzfz for hk suf ciently large at the cost of an additional error 6 that is Hie 16 fllp S Hie fije kap llje fk kap ka fllp S lljlh llfifkllp Now f E L S h implies Hje e S h and that We f 7 ip lt2hgtW H f 7 le V f has compact support from end of claim Now we prove the theorem under these claims above We apply the fact from Chapter 1 is that simple functions are dense in L1R Chl Theorem 118 Fact that we need is that there exists a simple function Holderls inequality applied flglp dz flglpil lgldz S Hng ng le So we can assume f E L and 2 i 00 F 2 chHJ Hj E collection of half open rectangle in R j1 such that HF 7 fllp lt 6 for some 6 small By Youngls inequality we have We 96 F 7 jg lt 6 llfe fll1lljef fll1lljequotf j Fll1llj F Fll1llFifll1lt6 Since F is sum of characteristic functions of rectangles it suf ces to show We 9 XH 7 XHH1 7gt 0 as e 7gt 0 Recall that jg has supp Q B35 let 7 Re make 7 small that 6 mA U A lt 7 llth where A IEH distzHC lt7 AIEHC distzHltT Consider jg XHltIgt 7 If z A U A jg XHltIgt XHI jg Xmas R 721 7 yXHydyA If I E A U A by changing of variables lieIXHI XHWl Wm mac 7y 7 XHltzgtgtdy M1 then Hie XHW XHIH1 S W47 U 14 lljlll lt 5 and thus we prove that f5 7 f in L17 Now we show that if jg E GEOGR then f5 6 C R and a a 811 7 811 f Since the support of jg is compact the difference quotient is uniformly bounded in h and of compact support and it is obviously bounded by some xed Lp function That is WHIQQ Ame ajeziyfydy e 811 as h A 0 by LDCTi V Note that the above formula also implies f5 6 CHE and by induction since 3 E C00 R that f5 6 C R i El Bounded functions in L17 are weakly compact l lt p lt 00 The proof relies the fact that LpLp 1ltpltooi Note A space X where X X is called re exive Note that L1 and L are not re exive Example Let qo on 01 We know follfjldz l and thus fj E Lli Suppose a weak limit existed that is fj A 0 since we want 7 A 0 for all hz E L But 1 E L f fji 1 7A 0 Theorem L17 is separable for l S p lt 00 but L is not Proof Read the textbook 1 By the separability of L17 We know therefore there exists countable dense subset which approxi mates f E L17 Q l S p lt 00 Therefore we show the BanachAlaoglu theorem which says bounded sequence in L17 Q l lt p lt 00 a separable space are weakly compact has weak limiti Theorem BanachAlaoaglu Theorem for LP Let Q C R measurable sequence of functions bounded in LWQ l lt p lt 00 Then there exists a subsequence and f 6 L176 such that fnj A f in LPG 076711007 4 LU as j A 00 Proof Note it is sufficient to show the existence of a subsequence such that fnIyIdI lt 12 z jzdzgt converges where 415139 E is a countable dense subset in Lp which approximate g E Lpi We prove by diagonalization argumenti De ne oi fjz 1zdr By Holder is bounded and a complex number Since bounded sequence of C have convergent subsequence called subsequence still 1 and 1 A all Pass to a further subsequence and de ne c fjr 2zdr Note fj are the subsequence from above Since is a bounded sequence it has a convergence subsequence called still 3 A 02 Continue inductively cl A all Note that cl is a subsequence of 31 l 2 kl De ne our nal subsequence Fk to be the kth element of the kth subsequence and thus qu5lgtclaskgtooi Because Fk is a diagonal subsequence use density of to show f Fkg converges as k A 00 Call the number F169 converges to Lgi Since F169 is linear and bounded its limit is as well By Riesz representation theorem Lg ffgdz for some f 6 L176 9 6 LP We use the fact that Lp L17 re exive spacer Note that L1 excluded because LIV L but L D Lli D 18 211 Approximation by 030 Functions Lemma Approximation by CC00 Functions 1f 9 Q R K E Q is compact then there exists JK 6 CC00 such that 0 S JKz S1 for all z E Q and JKz 1 for z E K Proof By construction x K C Q compact There exists 6 gt 0 such that 169 lziyl lt26forsomeyEKCQ De ne KIEQ lziylSeforsomeyEK and note that K C Q compact Let E C R suppj C I z S 1 0 S S 1 for all z and fjzdz 1 De ne JKz jg XK z and JK has the correct properties D Corollary As a consequence of the lemma 1 there exists 9192 6 Cam 0 S 91 S 1 such that llmjaoogj1 1 for all z E Q 2 Let where E Cm 1f1 S p lt 00 A f in LpQ then hi A f in LPQ1f1lt p lt 00 A f in Lquot then hi 4 f in Lquot Proof For 1 use the lemma to construct jKi where K s are a sequence of compact sets approximating Q is there exists K1 C K2 C K3 C C Q compact and for any I E Q I E KMW for some integer Now we obtain 91 JKI For 2 the strong convergence of hi to f is a consequence of LDCT The weak convergence is also a consequence of LDCT provided we recall that Lp with 1 lt p lt 00 and that the functions of compact support are dense in Lph 1 The next theorem says that convolutions of function in dual L17 are continuous Theorem If f E LNR 9 E LpRn where pp gt 1 are dual index Then f 96 9 is continuous and f 95 A 0 as A 00 is for every 6 gt 0 there exists R5 such that sup W gzl lt 6 MgtR which interpret last part as rapid decay at in nity Proof Note that Holder s inequality gives us that 96 is de ned and nite for all I f mm f1 7 mm s M Hng de ned as y for every I By approximation by CC00 functions there exists f5 95 such that 7 g S 6 and HQ 95Hp S 5 Note that Hf 97 f5 y llm Hf 7 f5 g f5 giyts lm S Hf 7 MP Hgllp Hf llp H9 175 S 5H9Hp HfllpL is uniform bounded Since f 96 9 approximated by f5 96 95 6 CC00 and f 96 9 is uniformly approximated by smooth functions we have f 96 9 is continuous By constructing R5 based on the compact support of f5 96 95 directly we have that f 96 9 has to decay rapidly which means sup w 9Il lt e MgtR 212 More Integral Inequalities De nition Gaussian function We call 9 R 7gt C bounded function exponential of quadratic and linear orm 91 exp dry AIgt iltryBIgt ltk71gt 0 Where AB symmetric real matrices A positive semide nite ltzAzgt 2 0 for all z E R k E C If g E LPGR then A is positive de nite Theorem Young s lnequality Sharp Let pq 7 2 l i 2 f 6 L177 9 6 L117 h 6 LT We have Rn fr9 MIMI S 0mm lpr Hqu HhHT Where 0 opoqom with of plPplP 0 1017 With equality holding iff f g h are Gaussian functions f1ta eXP nlt1 a7 AW agtiltk71gt 91 tb exp itlt1 571495 i W WWW hz t0 exp 7Tltz 7 cAz 7 5 iltk Where ta tbtc E C abc k E R a b c A real symmetric positive de nite matrix Remark There are many symmetric properties in this inequalities Let7s denote the integral in the Youngls inequality by Ifzgz Then by change of variables and Fubinils theorem 1fryyr7hr 1yr7 fr7hr 1fr7hr7yr 10400744 fr Theorem HardyLittlewoodSobolev Inequality Let 107quot gt 1 0 lt A lt n 117 g 2 f E NOR h E LWR Then there exists a sharp constant C Cnp such that fltzgtwz7y hltygtdzdylcltnApgtHprHhH Rquot Rquot W75memm 7 7 271 In casepi 7 7 27k Where clt A gt 0 A we 7 a we n7 10 n 7r 7 7 Fn 7 lquotn and equality holds for h constf Where 161 m2 7 z 7 a 2 n 6 c L1 6 R 0 Remark Note that HLS inequality is like Youngls inequality ifgz zlik Lq Rn for any 4 but 91 is in Lg called LqWeak That is EUR f 1 ltfgtqw lt 00 Where 1 ltfgtqw sup a z z gt a q Dtgt0 But the expression ltfgtqw is not a norm One can show if q gt 1 Hf q 7suprr1q wwz A A 20 is a norm7 where A is any measurable set in R and 01ltfgtw Hf c2ltfgtw Apply ChebysheV ineq for the rst ineq7 Thm 114 for the second Note that for 91 lzli q n we have A n n W 1 7 n7 A n i X7 Remark The Weak Young s inequality states that for g 6 L2 Rn and l lt p q lt 007 1 i 2 g 6 EUR Hy Rn Rn fltzgtgltz 7 ygthltygtdzdyl mt Hg w Hm where the sharp constant 1 1 sn quot Kmm E T 0 We can also View the HLS inequality as the statement that convolution is a bounded map L20 X L17 A L5 replacing s by 7 Pick h 6 190 lg flp p we have 1 n 1 1 Hm st i 0W l9 Hpr 4 Proof of HLS inequality without sharpness Assume WLOG fh 2 0 and l and HhHT 1 Use layercake representatlon fltzgt magma hltzgt xmwmdb z X x igttzdt A X E ltCzc quot1dci 0 0 For lzli gt t7 we have ltt 1Ai Let c t lp 7 we have c t7 iAc A ldc dti Now we compute I fzlriyl lhydzdy Rquot Rquot AA 0 0 Cik71XfgtaIXhgtbyXlxiylltcI Mdrdydadbdc gA c A 11abcdadbdc 0 0 0 vltagtwltbgtultcgt 1W maxvltagtwltbgtultcgt wbAM Xhgtb1dz vaAMXfgtazdz uc 5n and the norm of f7 h can be written as with HfHZp ap1vltagtda1 HhHZT br1wltbgtdb1 0 0 Do 5 integral rst assume va 2 101 and the other case is similar 2 11 12 cikil a b Cdc S C A 1wbu5dc Cikilwa mawc 0 ultcgtsvltagt ultcgtgtvltagt 21 Chapter 8 Partial Differential Equations 81 Introduction De nition PDE the generic notation fDD u 0 Where f function space A R or C lal is called the order of differential equation equation Example Au 0 in R u R A C L A 21 3 is a linear operator Most interesting examples of PDE tend to be nonlinear systems of PDE PDE courses in general start With scalar linear PDEls l Elliptic Laplace7s equation Au 0 Poissonls equation Au 2 Parabolic Heat equation Au 7 Au 0 3 Hyperbolic Wave equation Eu 7 Au 0 Differential operators are unbounded operators That is taking derivatives does not preserve continuity differentiability etc Properties of the function spaces changes Methods for dealing With unbounded operators 1 Weak solutions distributions 2 Green s functions 82 Green s Functions Associated to a differential operator is a natural domain Let A be a differential operator DA is the subset of the function space that we want to consider Example Aku u ODE k 123 4 DA may be many different things 1 DA1 u e 0201 u0 ul 0 2 DA2 0201 3 DA3 H201 4 DA4 u e H20 1 u0 ul 0 Greenls function act as an inverse to certain differential operators Work now Where DA C H H for Hilbert space like L2Q Want DA to be density de ned in H If DA is not dense in H modify A 0 on DAi De nition A DA C H 7gt H is a densely de ned unbounded linear operator on H if its adjoint A DA C H 7gt H is the operator With DA y E H there exists 2 E H with Amy lt172 for all z E DA If y E DA de ne A y 2 Where 2 is the unique element such that Amy lt172 for all z E DA Just like dual space Note that boundary conditions for A come from DA and DA gives boundary conditions for A We call A is selfadjoint if A A7 ie7 A z A1 for all z E DA and also DA DA symmetn39c if A Az A1 for all z E DA7 A is an extension of A if DA C DA Example Computing adjoints formally Let Au an 121 cu Where a E C2O1l7 a E 0107 17 a E CQ Let lt be the usual inner product A v 51 7 Evy Ev For all uv E C2O1l7 ltvAu 7 ltAvu 7 MW 7 M b 7 aVH 1 7 0 just proof by integration by parts 1 lt1 Au an bu cudz 0 Possible domain 17207 1f the above a 127 c are smooth enough7 then the above formula Greenls formula holds for all uv E H2017 not formal anymore The proof is If uv E H2017 there exists un vn E C 07 1 such that un 7gt u and vn 7gt v in H207 Formula holds for umvn 1 lt1 Ann 7 AZ un 1127 vgun b 7 aV 0 pointWise converges since7 by Sobolev embedding7 un7 vn are C1 Assume Au f7 u0 u1 07 and given f 01 7 C is continuous We want to nd 1 ultzgt 7 gltzygtfltygtdz 0 Where g is called the Greens function of the above system Let WA u E 0210711 1 MO W1 0 then G CO1 7gt DA de ned by 1 GM 7 gltzygtfltygtdy 0 is the inverse of A idea Au f7 Cf u Write the equation for the greens function in terms of 517 ie7 gzy must solve Agryy5r7y7 907y 91yy 0 Formally 1 1 Aultzgt 7 Agltz7ygtfltygtdy 7 we 7 yfydy 7 fa and boundary condition u0 A1901 yfy 07 u1 A191yyfy 0 To conclude if Agzy 6ziy and g0y gly 0 then Auz where e already saw this for Laplace equations Au f in 9 u 0 on 9 We e ne Greens function 7A9 6z 7 y in Q gzy 0 on 89 Cglog n2 I7 7 g ycnlziyl2 n n7 2 For 9 suf ciently nice and Green s operator for Lu is a compact selfadjoint operator on L2Q then L has a complete orthonormal set of eigenfunctions Example For Laplace equation this holds 7A has a complete set of orthonormal eigenfunctions Start with iAu Aui Now multiply by u and integrate we have 0Q7A7A1u dIQlt7VuVuqul27A2ugtdzAltqul27A2ugtdz since f9 7V uVu 0 by divergence theorem with zero boundary termsi Thus we must have qul2 7A2u 0 Since u f 0 we have A 2 0 But if A 0 we have Vu 0 in Q and thus u is a constant in 9 Note that u 0 on 89 we have u E 0 which is a contradiction So we must have A gt 0 83 Wave Equation Applications of Fourier transform BfuiAu0 zeR t R 140570 f1 E 67 o gltzgt called the Cauchy problem for the wave equation FACTl Find the solution in terms of the Fourier transform um t Ulttgtltfyggt COSltDtgtf with D a symbol A Founex multiplxex A DME ZWlEl h5 Now take Fourier transform to 631 7 Au 0 in 1 variables A 7 n a a 7 2A 7 2 2 A Au 7 7727r251 7 74w u7 1 we get 63 47 is mat 0A For each 5 xed this is an ODE in t The general solution is 35 t acos27r t bsin27r ti Recall that by initial data u50 AWE 0 So plug in these condition we have 77 7 375 a7 5 ems and we get the general solution sin27r t ms 7 Rs coslt2w m t as 2 5 So uzt tV is What we mean by the notation FA T2 Consider 8211 icQAu 0 With speed c 1 for us If fg are supported in z E R S M then ut is supported in z E R S M t look rst at f part of the solution f has support in lrl S M ultz7tgt 008Dtf co v cosew m tgtfltsgtgtv 27ril lt 727mm A V A A gum mm ltfltz H fltz 7 m So if suppf C S M then supp it C S M t ow see 9 part for n l and obtain in the general solution in n 1 Not in Fourier form By Fourier inversion formula so A 2Mltwtgtl l 27ririrl l 9ds ltfltztgtfltzitgty 9 part 00 A 27riztl l S e2m m4gi d 00 A k zt 6mm dk LEWT 5 L gtH 2 1 zt 00 I 1 zt 7 ke27r kdkds gsds 2 x7 700 2 m4 D7Alembert7s solution to wave equation n 1 1 was fz t m 7 t g 7t gltsgtds FACTS De ne the energy for wave equation 1 Et i lam qulet Claim Et E0 Where 1 l E0 7 launch 413 de 7 m2 wme 2 R 2 R We Will prove that for all t Et E0 furthermore lim law ztdz taco R E 0 2 E 0 2 aul vul The equipartition of energy Will follow from our Fourier representation of the solution and a modi cation of the RiemannLebesgue lemma Brodsky Energy method very standard technique for understanding linear nonlinear PDE ldea in general take equation multiply by some number of derivatives of u or just u and integrate Assume u E GEOGR Equipartition of energy 0 8211 7 Au hu dz 8211tu dz 7 Au hu dz Rquot Rquot Rquot Where 2 1 2 1 d 2 a 10811 dz 7 848111 dz if Btu dz Ru 2 W 2 dt Rn 7 Auatu dz aggluatailu dz 13 law R R 2dt R 1d 2 dzi qul dz 2 7 li 2 2 ii 071 Btu Auu d172dt Rn hu HVul dz7thti Thus Et E07 iiei7 energy for wave equation conservedi Also7 Et 7 E0 7 lgl2 7 Vfwm Now we want to show E 0 lim law ztdz taco Ru 2 Recall RiemannLebesgue Lemma f 6 L17 then fis continuous and A 0 as A 00 Recall Plancherel L2 Note that a is in FACTL 8 3 727T lkl sin27r lkl tfk cos27r lkl ak am 7 4772 W sin227r w am 7 cos227r w t k2 7 477 w mm W t cos27r w ammo Apply triangular doubleangle iclentity7 and apply Plancherel Btu2dz 7 6L 2dk w W lVM2 12 z A 7 1 2 2 A 2 1 A 2 7 g 4 m me 5 we die cosOlTr w t k2 7 4772 W PM 7 4 w sin27r w t cos27r w tfk kdk Since g 6 L27 we have E L2 and P7 27 E Lli Hence 6L1 z quot x cos47r w t k2 7 4772 W P dk 7 0 as t A 00 84 Weak Solution Weak form of the wave equation Assume u E C2 Rn so all the derivatives in wave equation make sense aguiAuFzt 16R tgt0 149570 1 0 7 lt gt at 17 i g I Now take 45 E C8 Rn1 and 45 q5zt has compact support in z iiei7 0z is not necessarily zeroi Multiply 45 by F and integrate over space and time an M W Rn ma 7 Au M The at part 41581211 dzdt 7 845811 dzdy 41507 aw dzdt tgt0 Rquot tgt0 Rquot Rquot 8124511 dzdt 7 qb0z tu dzdti zgto 1Rquot Rquot3 Oxu dam Rquot The A part the boundary terms go to zero and thus quu dzdt 7 A415 u dzdti zgt0 1R tgt0 1R So tgt0 Rn 7 Au dzdt tgt0 Rn 7 A gtu dzdt 7 RW 0zgzdz AM 81 0zfzdz for 45 E GEOGR BUR makes sense for any g 6 DRn uF E 39DRwrl So a weak of distributional solution to the wave equation is given by that form above

### BOOM! Enjoy Your Free Notes!

We've added these Notes to your profile, click here to view them now.

### You're already Subscribed!

Looks like you've already subscribed to StudySoup, you won't need to purchase another subscription to get this material. To access this material simply click 'View Full Document'

## Why people love StudySoup

#### "There's no way I would have passed my Organic Chemistry class this semester without the notes and study guides I got from StudySoup."

#### "When you're taking detailed notes and trying to help everyone else out in the class, it really helps you learn and understand the material...plus I made $280 on my first study guide!"

#### "I was shooting for a perfect 4.0 GPA this semester. Having StudySoup as a study aid was critical to helping me achieve my goal...and I nailed it!"

#### "It's a great way for students to improve their educational experience and it seemed like a product that everybody wants, so all the people participating are winning."

### Refund Policy

#### STUDYSOUP CANCELLATION POLICY

All subscriptions to StudySoup are paid in full at the time of subscribing. To change your credit card information or to cancel your subscription, go to "Edit Settings". All credit card information will be available there. If you should decide to cancel your subscription, it will continue to be valid until the next payment period, as all payments for the current period were made in advance. For special circumstances, please email support@studysoup.com

#### STUDYSOUP REFUND POLICY

StudySoup has more than 1 million course-specific study resources to help students study smarter. If you’re having trouble finding what you’re looking for, our customer support team can help you find what you need! Feel free to contact them here: support@studysoup.com

Recurring Subscriptions: If you have canceled your recurring subscription on the day of renewal and have not downloaded any documents, you may request a refund by submitting an email to support@studysoup.com

Satisfaction Guarantee: If you’re not satisfied with your subscription, you can contact us for further help. Contact must be made within 3 business days of your subscription purchase and your refund request will be subject for review.

Please Note: Refunds can never be provided more than 30 days after the initial purchase date regardless of your activity on the site.