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Fourier Analysis

by: Otilia Murray I

Fourier Analysis MAT 129

Otilia Murray I
GPA 3.88


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This 12 page Class Notes was uploaded by Otilia Murray I on Tuesday September 8, 2015. The Class Notes belongs to MAT 129 at University of California - Davis taught by Staff in Fall. Since its upload, it has received 22 views. For similar materials see /class/187438/mat-129-university-of-california-davis in Mathematics (M) at University of California - Davis.

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Date Created: 09/08/15
Notes on different modes of Convergence De nition Suppose fn is a sequence of functions in L2a b 0 We say that fn converges to f pointwise if for each x 6 11 ms 7 M as n 7 oo ie 7 7 0 as n 7 00 This means for any 6 gt 0 and any x 6 11 there is an integer N gt 0 which depends on e and x such that lf m 7 lt e for all n gt New 0 We say that fn converges to f uniformly if sup WW 7 fMl H 0 as n H 00 2992 this means for any 6 gt 0 there is an integer N gt 0 such that lf m 7 lt e for all n 2 N and for all z 6 11 0 We say that fn converges to f in norm if an7fll H0 asn oq where is the norm on the vector space L2a b By the de nition of the norm this means 17 mm 7 fm2dz a 0 as n a 00 De nition A set E of real numbers E C R is said to have measure zero if V6 gt 0 there exist open interals 112 with lengths 112 such that ECU and szlt e 739 739 This de nition can be generalized to sets in Rk For example suppose E C Rk then I of lengths 7 in the de nition are replaced by rectangles Rj with areas r72 Example Any countable set has measure zero To see this suppose E 1 2 For any xed 6 consider the open intervals I m7 7 e271zj 62j1 each of length 627 Clearly E is contained in the union of all Ij and the total length of the intervals is 00 e e 00 1 e Z a 5 Z a 539 71 70 This means the rational numbers form a set of measure zero De nition We say that a statement S is true almost everywhere or for almost all x if S is true except for a set of measure zero For example fz 0 almost everywhere ae means fz 0 for all z outside of a set of measure zero fn 7 f ae means fnz 7gt fz for all z outside a set of measure zero OBSERVE that if two continuous functions are equal ae then they are identically equal Some remarks about the different modes of convergence gt lt Pointwise convergence does not imply convergence in norm and viceversa However if fn a f uniformaly then clearly fn a f pointwise and also fn a f in norm as stated in Theorem 32 in our text gt lt gt lt Suppose fn a f in norm then i fn a 1 almost everywhere 11 Hle H Hill iii lt fnggt gt lt fggt and ltgfngt gt ltgf gt foranyg L2ab De nition A sequence 1 in a norm vector space is called a Cauchy sequence if Hun 7 va a 0 as m n a 00 In other words for any 6 gt 0 there is an integer N gt 0 such that anivmll lt e for all 77171 2 N A norm vector space V is said to be complete if every Cauchy sequence converges in norm to a limit point in V Known result from Math 25 o In Ck and Rk for any k12 a sequence converges if and only if it is Cauchy This implies Ck and Rk are complete spaces De nition A vector space equipped with inner product lt gt and associated norm lt 121 gt is called a Hilbert space if it is complete with respect to convergence in norm Examples Ck Rk L2a b are all Hilbert spaces MAT 129 Fourier Analysis Supplementary Notes II by Naoki Saito The Fourier Inversion Theorem The Fourier transform 3 was de ned initially on LlOK a space of integrable functions and Sr LlOK a BOOK OOK LOOOK However 1 the Fourier transform of f 6 L1 may not be in L1 An example x Xk x sinc 7 5117 L1 CO The Inverse Fourier Transform For f 6 L1 x e2Wigm d5 00 The Fourier Inversion Theorem Ifboth f and f are in L1 then f f 1 almost everywhere There are many functions in L1 whose Fourier transforms are also in L1 one needs only a little smoothness off for necessary decay off as 5 a 00 An example Iff C OZOK f and f are both in L1 then rJrf 727r 2 6 BOOK This boundedness implies that Ol 52 This in turn implies that f 6 L1 The Fourier Transforms on L2 The previous remark leads to the L2 theory of the Fourier transforms In general simply assuming 1 C L2 is not enough xe 27ri m dx may not converge sin 7rx 7rx An example x sincx 6 L2 but not in L1 We will overcome this problem as follows De ne a subspace of L1 3C Q 1 C L1 f 6 L1 We rst note that for such functions we can have the Parseval equality f g ltf gt as well as the Plancherel equality Also for any 1 6 3C 1 f 6 BOOK as the remark after the Fourier inversion theorem This implies that both 1 and f are also in L2 ie 3C C L2 because 1 C L1 O BO implies f C L2 thanks to the theorem LP LT C Lq for 0 lt p lt q lt x g 00 which in turn can be proved by Holder s inequality Now the point is that 3C is also dense in L2 We can proceed as follows for any 1 6 L2 we can nd a sequence fn C 3C such that ani ig a 0 as n a 00 fn C 3C means that C 3C Now using the Plancherel equality to this sequence we can see 7 Hfn 7 meZ a 0 as m7 n a 00 In other words is a Cauchy sequence in L Since L2 is complete there exists the limit of in L2 and we de ne this limit as f the Fourier transform of f 6 L2 The Plancherel Theorem For any g 6 L2 g ltf gt and Hsz Finally we can use all these facts for computing the Fourier transform of L2 functions as follows Suppose we set x x where f 6 L2 Then g i An example x sincx 6 L2 ThCII I So 1 PjE rigj Suppose LD H I fltgL Satisfy Hag A003 amok 2 Q6279 frf jg 4x fiat 4x 7 A V 143 nggrpclion Ly may Mag omoL amtNae We 5 b w L V Mugyou 1ng cox b F35 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bmlcnm mlodw mq m cm 94 myrcg mw as Z my chm m We ance we 4 1 H74 Wok Lwa eraba M413 W oh am I I Z 9C dDhCX G h1gt an gt PL1 fill66 th a i 3 UH Us iqnl an 60 th bf be Q aejztgj W cm 5 so V Kitty r WA E e39Zt Marf0 bfctww wi Fact115 1 123quot ms W s LynLtBI O 7 2gtrn22 5lt V H W tnoL bit 10 n ALB 7 5 94 16 abut fj bICf k z t zw fr e L e Z 6 MerQI 971111 EQ2JC3 t t 2 3 LAWw c3 3 CD S e bltt Ea tf gkwds D 7 xrE f39ef f f 21 1 72 J 47140 or khcwowmq OCl x 30W TC Show dgtm7lt7h is Of konorm a 4amp52 5 Lw m9gt with WCXgt gt0 Lam ilnuws 0V7 al ja L L mgw Eggma Wmea Lacx dx p f h ltC n00 g fim39 n 00 39 quot quotCK r 0L AO 503 n 49 V Heme j ZI m A Of runmay Q34 To show Ra 0W7 4 1553 SW0 5 M 53 Cm jIeszLa ogt SHEA tim ltch n 7920 2 O Cov 011 7 Hm ltK 20 DCOY GUMmy m gwjxamgmampwmam footo m Emu gt fogtb cue Wm 9 g I t O in BOOWMQ W 2410 ULCJC CW mayo o MLf7 af x XQEOgi 1090 L Eggx5 7 K Gig 9 Mt E780 O 5mm Liowzue mblem 0 WSde S W WOW mom m 2 J56 eigxnws s m f n2 1 lt mm Esin gmgt Juo39f m 176 01 0 M It 2 I 213 ig n gx 076 XVXg9inVZ1Z1X 017C u 0 J LWMJ I II 1 2m 39 I WEJ J Exsmi xdx O HT i M 3 14 d5 3ampx n1 I 2 if at 3 my Qwa gt m 2 3mg rijj 2 4 g gt083 obj M2 1L o o gm cosh2 9ih gn Q nnj l L L Q 2 VHT 2m 2quot K xxTL IC 3mgg z xd7 j z xw w 99 72 1 NF 2 Q 39 2 L a 39 gt aggrst EJMWM drx M L y L L h WSW 9m F W W T 39 wg i quot W WWW 2 Mg wg ffzm j cOSVE 6395 2 M 7quot M rm I rm 1 0m 2 gawk 9W W 3 So ltub Sngt I I 1 4m E h CA 2 2 0 A 397 v nr 00 1080 1090 Earl Ex Sing 296 11 00 WT x 0 n7 3m0ltgt 2 In M 811 7 HZ 7 1 7T 00 x M Now LLCK C 2 Eric gm 176 I13 90 MUMO Z bh0gt ESt rly li9 quotq H H g m f m g M n3 139 h1 Hth gl w x m quot gymizgimz 3 W 1 knoll utgxo0 7 U co W


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