Class Meeting # 16: The Fourier Transform on R n

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1 MATH COUSE NOTES - CLASS MEETING # Itroductio to PDEs, Fall 2011 Professor: Jared Speck Class Meetig # 16: The Fourier Trasform o 1. Itroductio to the Fourier Trasform Earlier i the course, we leared that periodic fuctios f L 2 ([ 1, 1]) (of period 2) ca be represeted usig a Fourier series: a 0 (1.0.1) f(x) = + a m cos(mπx) + b m si(mπx). 2 m=1 m=1 The = sig above is iterpreted i the sese of the covergece of the sequece of partial sums associated to the right-had side i the L 2 ([ 1, 1]) orm. The coefficiets a m ad b m represet the amout of the frequecy m that the fuctio f cotais. These coefficiets were related to f itself by (1.0.2a) a 0 = 1 f(x) dx, (1.0.2b) 1 1 a m = f(x) cos(mπx) dx, (m 1), 1 1 (1.0.2c) b m = f(x) si(mπx) dx, (m 1). 1 The Fourier trasform is a cotiuous versio of the formula (1.0.1) for fuctios ied o the whole space. Our goal is to write fuctios f ied o as a superpositio of differet frequecies. However, istead of discrete frequecies m, we will eed to use cotiuous frequecies ξ. Defiitio (Fourier Trasform). Let f L 1 ( ), i.e., f(x) d x <. The Fourier trasform of f is deoted by f, ad it is a ew fuctio of the frequecy variable ξ. It is ied for each frequecy ξ as follows: (1.0.3) f(ξ) = f(x)e 2πiξ x d x, where deotes the Euclidea dot product, i.e., if x = (x 1,, x ) ad ξ = (ξ 1,, ξ ), the j j ξ x = j=1 ξ x. I the above formula, recall that if is r is a real umber, the e ir = si r + i cos r. The formula (1.0.3) is aalogous to the formulas (1.0.2a) - (1.0.2c). It provides the amout of the frequecy compoet ξ that f cotais. Later i the course, we will derive a aalog of the represetatio formula (1.0.1). 1

2 2 MATH COUSE NOTES - CLASS MEETING # 16 emark The Fourier trasform ca be ied o a much larger class of fuctios tha those that belog to L 1. However, to make rigorous sese of this fact requires advaced techiques that go beyod this course. We will also use the followig otatio. Defiitio (Iverse Fourier trasform). Give a fuctio f(ξ) L 1 ( ), its iverse Fourier trasform, which is deoted by f, is a ew fuctio of x ied as follows: (1.0.4) f (x) = f( x) = f(x)e 2πiξ x d ξ. The ame is motivated as follows: later i the course, we will show that (f) = f. Thus, is i fact the iverse of the operator. The Fourier trasform is very useful i the study of certai PDEs. To use it i the cotext of PDEs, we will have to uderstad how the Fourier trasform operator iteracts with partial derivatives. I order to do this, it is coveiet to itroduce the followig otatio, which will simultaeously help us bookkeep whe takig repeated derivatives, ad whe classifyig the structure moomials. Defiitio If (1.0.5) α = (α 1,, α ) is a array of o-egative itegers, the we ie α to be the differetial operator 1 (1.0.6) α = 1 α α. 1 Note that α is a operator of order α = α + + α. If x = (x 1,, x ) is a elemet of C, the we also ie x α to be the moomial x α 1 (1.0.7) = (x 1 ) α (x ) α. The followig fuctio spaces will play a importat role i our study of the Fourier trasform. Throughout this discussio, the fuctios f are allowed to be complex-valued. Defiitio (Some importat fuctio spaces). C k (1.0.8) = {f : C α f is cotiuous for α k}, (1.0.9) C = {f : 0 C f is cotiuous ad lim f(x) = 0}. x We also recall the followig orm o the space of bouded, cotiuous fuctios f : C : (1.0.10) f C 0 = max f(x). x The L 2 orm plays a importat role i Fourier aalysis. Sice f is i geeral complex-valued we also eed to exted the otio of the L 2 ier product to complex-valued fuctios. This is accomplished i the ext iitio.

3 MATH COUSE NOTES - CLASS MEETING # 16 3 Defiitio (Ier product for complex-valued fuctios). Let f ad g be complexvalued fuctios ied o. We ie their complex ier product by (1.0.11) f, g = f(x)ḡ(x) d x, where ḡ deotes the complex cojugate of g. That is, if g(x) = u(x) + iv(x), where u ad v are real-valued, the ḡ(x) = u(x) iv(x). We also ie orm of f by ( ) 1/2 (1.0.12) f = f, f 1/2 = f(x) 2 d x. Note that this is just the stadard L 2 orm exteded to complex-valued fuctios. Note that, ad verify all of the stadard properties associated to a complex ier product ad its orm: f 0 ad f = 0 if ad oly if f = 0 almost everywhere g, f = f, g (Hermitia symmetry) If a ad b are complex umbers, the af + bg, h = a f, h + b g, h, ad f, ag = ā f, g (Hermitia liearity) f, g f g (Cauchy-Schwarz iequality) f + g f + g (Triagle Iequality) 2. Properties of the Fourier Trasform The ext lemma illustrates some basic properties of f that hold wheever f L 1. 1 Lemma (Properties of f for f L ). Suppose that f L 1 ( ). The f is a bouded, cotiuous fuctio ad (2.0.13) f C0 f L 1. Proof. Sice e ir = 1 for all real umbers r, it follows that for each fixed ξ, we have (2.0.14) f(ξ) f(x)e 2πiξ x d x f(x) d x = f L 1. Takig the max over all ξ, the estimate (2.0.13) thus follows. We ow prove that f is cotiuous. Give ɛ > 0, let B be a ball of radius cetered at the origi such that the itegral of f over its complemet B c is o larger tha ɛ : (2.0.15) B c f(x) d x ɛ. It is possible to choose such a ball sice f L 1. We the estimate

4 4 MATH COUSE NOTES - CLASS MEETING # 16 (2.0.16) f(ξ ) f(η) f(x) e 2πiξ x e 2πiη x d x + B f(x) e 2πiξ x e 2πiη x d x + 2ɛ. B Bc 2 {}}{ f(x) e 2πiξ x e 2πiη x d x Now sice e 2πir is a uiformly cotiuous fuctio of the real umber r o ay compact set, if ξ η is sufficietly small, the we ca esure that max x B e 2πiξ x e 2πiη x ɛ. We the coclude that the fial itegral over B o the right-had side of (2.0.16) will be o larger tha (2.0.17) max e 2πiξ x e 2πiη x x B B f(x) d x ɛ f(x) d x = ɛ f L 1. Thus, i total, we have show that if ξ η is sufficietly small, the f(ξ) f(η) ɛ f L 1 + 2ɛ. Sice such a estimate holds for all ɛ > 0, f is cotiuous by iitio. It is helpful to itroduce otatio to idicate that a fuctio has bee traslated. Defiitio (Traslatio of a fuctio). If C is a fuctio ad y is ay poit, the we ie the traslated fuctio τ y f by (2.0.18) τ y f(x) = f(x y). The ext theorem collects together some very importat properties of the Fourier trasform. I particular, it illustrates how the Fourier trasform iteracts with traslatios, derivatives, multiplicatio by polyomials, products, covolutios, ad complex cojugates. Theorem 2.1 (Importat properties of the Fourier trasform). Assume that f, g L 1 ( ), ad let t. The (2.0.19a) π (τyf) (ξ) = e 2 iξ y f(ξ), (2.0.19b) h(ξ) = τ 2πiη x ηf(ξ) if h(x) = e f(x), (2.0.19c) h(ξ) = t f(tξ) if h(x) = f(t 1 x), (2.0.19d) (f g) (ξ) = f(ξ)ĝ(ξ), (2.0.19e) If x α f L 1 for α k, the f C k ad α f(ξ) = [( 2πix) α f(x)] (ξ), (2.0.19f) If f C k, α f L 1 α for α k, ad α f C 0 for α k 1, the ( α f) (ξ) = (2πiξ) f(ξ), (2.0.19g) f(ξ) = (f) (ξ) ad (f )(ξ) = (f) (ξ).

5 MATH COUSE NOTES - CLASS MEETING # 16 5 Above, f deotes the complex cojugate of f; i.e., if f = u + iv, where u ad v are real-valued, the f = u iv. Proof. To prove (2.0.19a), we make the chage of variables z = x y, d z = d x ad calculate that (2.0.20) (τ y ) f (ξ) = f(x y)e 2πix ξ d x = f(z)e 2πi(z+y) ξ d z = e 2πiy ξ f(z)e 2πiz ξ d z = e 2πiy ξ f(ξ). To prove (2.0.19b), we calculate that (2.0.21) h(ξ) = e 2πiη x f(x)e 2πix ξ d x = f(x)e 2πix (ξ η) d x = f(ξ η) = τ η f(ξ). To prove (2.0.19c), we make the chage of variables y = t 1 x, d y = t d x to deduce that (2.0.22) h(ξ) = f(t 1 x)e 2πix ξ d x f(y)e 2πiy tξ t d y = t f(tξ). To prove (2.0.19d), we use the iitio of covolutio, (2.0.19a), ad Fubii s theorem to deduce that (2.0.23) (f g) (ξ) = ( ) ( ) e 2πx ξ f(x y)g(y)d y d x = g(y) e 2πx ξ f(x y)d x d } {{} e 2πiξ y f(ξ) = f(ξ) e 2πiξ y g(y) d y = f(ξ)ĝ( ξ). To prove (2.0.19e), we differetiate uder the itegral i the iitio of f(ξ) to deduce that (2.0.24) (ξ) (ξ) α f(ξ) = f(x) α e 2πix ξ d x = f(x)( 2πix) α e 2πix ξ d x = [( 2πix) α f(x)] (ξ). To prove (2.0.19f), we itegrate by parts α times ad use the hypotheses o f to discard the boudary terms at ifiity, thus cocludig that (2.0.25) ( ) 2πix ξ α (x) 2πix ξ α f (ξ) = α f(x)e d x = f(x)( 1) α e d x = f(x)(2πiξ) α e 2πix ξ d x = (2πiξ) α f(ξ). To deduce the first relatio i (2.0.19g), we compute that y

6 6 MATH COUSE NOTES - CLASS MEETING # 16 (2.0.26) f(ξ) = f(x)e 2πix ξ d x = f(x)e 2πix ξ d x = f(x)e 2πix ξ d x = f( ξ) = (f) (ξ). The secod relatio i (2.0.19g) ca be show usig similar reasoig. (2.0.19e) roughly shows that if f decays very rapidly at ifiity, the f is very differetiable. Similarly, (2.0.19f) roughly shows that if f is very differetiable with rapidly decayig derivatives, the f also rapidly decays. The Fourier trasform thus coects the decay properties of f to the differetiability properties of f, ad vice versa. I the ext propositio, we provide a specific example of these pheomea. More precisely, the ext propositio shows that the Fourier trasform of a smooth, compactly supported fuctio is itself smooth ad rapidly decayig at ifiity. Propositio Let f C c ( ), i.e., f is a smooth, compactly supported fuctio. The f is smooth ad rapidly decayig at ifiity i the followig sese: for each N 0, there exists a costat C N > 0 such that (2.0.27) f(ξ) C N (1 + ξ ) N. Furthermore, a estimate similar to (2.0.27) holds (with possibly differet costats) for all of the derivatives β f(ξ). I particular, f L 1 : (2.0.28) f(ξ ) L 1 = f(ξ) d ξ <, ad similarly for β f, where β is ay derivative multi-idex. Proof. Usig (2.0.19e) ad the fact that f is compactly supported (ad hece x α f L 1 ), we see that f is smooth. To prove (2.0.27), we use (2.0.19f), (2.0.13), ad the fact that α f L 1 < for ay differetial operator α to deduce that (2.0.29) (2πiξ) α f(ξ) = ( α f) (ξ) α f L 1 = C α, where C α is a costat depedig o α. I particular, if M 0 is a iteger, the by apply ( ) M ( ig (2.0.29) to the differetial operator M = ( 2 i=1 i ) M i.e., (2πi) 2M (ξ i ) 2 f(ξ) = ( M f) (ξ) C M ), it follows that (2.0.30) (2π ξ ) 2M f(ξ) C M for some costat C M (2.0.30). i=1 }{{} ξ 2M > 0. It is easy to see that a estimate of the form (2.0.27) follows from

7 MATH COUSE NOTES - CLASS MEETING # 16 7 (2.0.28) follows from (2.0.27) ad the fact that 1 (2.0.31) d ξ < (1 + ξ. ) +1 To see that (2.0.31) holds, perform the itegratio usig spherical coordiates o : 1 1 ρ (2.0.32) d ξ = ω dρ, (1 + ξ ) ρ=0 (1 + ρ) where ρ = ξ = j=1 (ξj ) 2 is the radial variable o, ad ω is the surface area of the uit ball i. By a simple compariso estimate, it is easy to see that the itegral o the right-had side of (2.0.32) coverges (the itegrad behaves like 0 ear ρ = 0, ad like 1 ear ). ρ 2 To show that similar results hold for for β f, we first use (2.0.19e) to coclude that β f(ξ) (2.0.33) = [( 2πix) β f(x)] (ξ). Furthermore, the fuctio ( 2πix) β f(x) also satisfies the hypotheses of the propositio. We ca therefore repeat the above argumets with β f i place of f ad ( 2πix) β f(x) i place of f. 3. Gaussias Oe of the most importat classes of fuctios i Fourier theory is the class of Gaussias. The ext propositio shows that this class iteracts very icely with the Fourier trasform. Propositio (The Fourier trasform of a Gaussia is aother Gaussia). Let f(x) = exp( π z x 2 ), where z = a + ib is a complex umber, a, b, a > 0, x = (x 1,, x ), ad 2 j 2 x = j=1 (x ). The (3.0.34) f(ξ) = z /2 exp( π ξ 2 /z). Proof. We cosider oly the case b = 0, so that z = a. The cases b = 0 would follow from a argumet similar to the oe we give below but requirig a few additioal techical details. We first address the case = 1. The by properties (2.0.19e)-(2.0.19f) of Theorem 2.1, we have that 2 i d 2 i 2π (3.0.35) f (ξ) = ( 2πixe aπx ) (ξ) = ( e aπx ) (ξ) = 2πiξf(ξ) = ξf(ξ). a dx a a We ca view (3.0.35) as d 2π (3.0.36) l f = ξ. dξ a Itegratig (3.0.36) with respect to ξ ad the expoetiatig both sides, we coclude that (3.0.37) f(ξ) = Cexp( πξ 2 /a.) Furthermore, the costat C clearly must be equal to f(0).

8 8 MATH COUSE NOTES - CLASS MEETING # 16 We ow compute f(0) : (3.0.38) f(0) = 1 {}}{ 2 e πax e 2πiξ0 dx = a 1/2. Note that you have previously calculated this itegral i your homework. Combiig (3.0.36) ad (3.0.38), we arrive at the desired expressio (3.0.34) i the case = 1. To treat the case of geeral, we ote that the properties of the expoetial fuctio ad the Fubii theorem together allow us to reduce it to the case of = 1 : (3.0.39) f(ξ) = We have thus show (3.0.34). exp( πa x 2 )exp( 2πiξ x) d x ( ) ( ) = exp πa (x k ) 2 exp 2πi ξ j x j d x k=1 j=1 { = exp ( πa(x j ) 2) } exp( 2πiξ j x j ) d x j=1 = { exp ( πa(x j ) 2) exp( 2πiξ j x j ) dx j j=1 = ( ) a 1/2 exp π(ξ j ) 2 /a j=1 = a /2 exp ( πa 1 = a /2 exp( π ξ 2 /a). (ξ j ) 2 j=1 ) } 4. Fourier Iversio ad the Placherel Theorem The ext lemma is very importat. It shows that the Fourier trasform iteracts icely with the L 2 ier product. Lemma (Iteractio of the Fourier trasform with the L 2 ier product). Assume that f, g L 1. The (4.0.40) f(x)g(x) d x = f(x)ĝ(x) d x. Alteratively, i terms of the complex L 2 ier product, we have that (4.0.41) f, g = f, g.

9 MATH COUSE NOTES - CLASS MEETING # 16 9 Proof. Usig the iitio of the Fourier trasform ad Fubii s theorem, the left-had side of (4.0.40) is equal to (4.0.42) f(ξ)g(x)e 2πiξ x d ξ d x. By the same reasoig, this is also equal to the right-had side of (4.0.40). To obtai (4.0.41), simply replace g with ḡ i the idetity (4.0.40) ad use property (2.0.19g). The ext theorem is cetral to Fourier aalysis. It shows that the operators ad are iverses of each other wheever f ad f are ice fuctios. Theorem 4.1 (Fourier iversio theorem). Suppose that f : C is a cotiuous fuctio, that f L 1, ad that f L 1. The (4.0.43) (f) = (f ) = f. That is, the operators ad are iverses of each other. Proof. We first ote that { } (f) (4.0.44) (x) = f(y)e 2πiy ξ d y e 2πix ξ d ξ. Note that the itegral i (4.0.44) is ot absolutely coverget whe viewed as a fuctio of (y, ξ). Thus, our proof of (4.0.43) will ivolve a slightly delicate limitig procedure that makes use of the auxiliary fuctio (4.0.45) φ(t, ξ) = exp( πt 2 ξ 2 + 2πiξ x). Note that (2.0.19b) ad Propositio together imply that (4.0.46) φ(y) = t exp( π x y 2 /t 2 ) = Γ(t, x y), where (4.0.47) Γ(t; y) = t exp( π y 2 /t 2 ). Also ote that Γ(t, y) is just the fudametal solutio of the heat equatio with diffusio costat D = 1. I particular, we previously showed i our study of the heat equatio that 4π (4.0.48) for all t > 0. We ow compute that Γ(t, y) d y = 1

10 10 MATH COUSE NOTES - CLASS MEETING # 16 (4.0.49) (Γ(t, ) f)(x) = Γ(t, x y)f(y) d y = φ(t, y)f(y) d y = φ(t, ξ)f(ξ) d ξ = exp( πt 2 ξ 2 )f(ξ)exp(2πiξ x) d ξ Durig our study of the heat equatio, we showed that the left-had side of (4.0.49) coverges to f(x) as t 0. To complete the proof of the theorem, it remais to show that the right-had side coverges to d (4.0.50) f(ξ)exp(2πiξ x) ξ = (f) (x) = (f) ( x) as t 0. To this ed, give ay umber ɛ > 0, choose a ball B of radius cetered at the origi such that (4.0.51) f(ξ) d ξ ɛ. B c Above, B c deotes the complemet of the ball. It is possible to choose such a ball sice f L1. We the estimate (4.0.52) exp( πt 2 ξ 2 )f(ξ)exp(2πiξ x) d ξ f (x) = exp( πt 2 ξ 2 )f(ξ)exp(2πiξ x) d ξ f(ξ)exp(2πiξ x) d ξ exp( πt 2 ξ 2 ) 1 f(ξ) d ξ 1 max Bc {}}{ exp( πt 2 ξ 2 ) 1 f (ξ) d ξ + exp( πt 2 ξ 2 ) 1 f(ξ) d ξ ξ B B max exp( πt 2 ξ 2 ) 1 f L 1 + f(ξ) d ξ ξ B B c max exp( πt 2 ξ 2 ) 1 f L 1 + ɛ. ξ B As t 0, the first term o the right-had side of (4.0.52) coverges to 0. I particular, if t is sufficietly small, the the right-had side of (4.0.52) will be o larger tha 2ɛ. Sice this holds for ay ɛ > 0, we have thus show that the right-had side of (4.0.49) coverges to the expressio (4.0.50) as t 0, i.e., that it coverges to (f) (x). Sice, as we have previously oted, the left-had side of (4.0.49) coverges to f(x) as t 0, we have thus show that (f) (x) = f(x). It ca similarly be show that (f ) (x) = f(x). This completes the proof of (4.0.43).

11 MATH COUSE NOTES - CLASS MEETING # The ext theorem plays a cetral role i may areas of PDE ad aalysis. It shows that the Fourier trasform preserves the L 2 orm of fuctios. Theorem 4.2 (The Placherel theorem). Suppose that f, g : C are cotiuous fuctios, 1 that f, g L L 2, ad that f, ĝ L 1. The f, ĝ L 2, ad (4.0.53) f, g = f, ĝ, i.e., the Fourier trasform preserves the L 2 ier product. I particular, by settig f = g, it follows from (4.0.53) that (4.0.54) f L 2 = f L 2. Proof. By applyig (4.0.41) with g replaced by ĝ, we have that (4.0.55) f, ĝ = f, (ĝ). By the Fourier iversio theorem (i.e. Theorem 4.1), we have that (ĝ) = g, ad so the right-had side of (4.0.55) is equal to (4.0.56) f, g. We have thus show (4.0.53).

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