The Riemann-Lebesgue Lemma

Transcription

1 Physics 215 Winter 218 The Riemnn-Lebesgue Lemm The Riemnn Lebesgue Lemm is one of the most importnt results of Fourier nlysis nd symptotic nlysis. It hs mny physics pplictions, especilly in studies of wve phenomen. In this short note, I will provide simple proof of the Riemnn- Lebesgue lemm which will be dequte for most cses tht rise in physicl pplictions. The simplest form of the Riemnn-Lebesgue lemm sttes tht for function f(x) for which the integrl where nd b re rel numbers, we hve Sometimes, the result of eq. (2) ppers in the form, f(x) dx <, (1) f(x)e ix dx =. (2) eix =. (3) Of course eq. (3) mes no sense when interpreted s stndrd it in mthemticl nlysis. However, if one interprets the it of eq. (3) in the sense of distributions, i.e. by treting e ix s generlized function, then eq. (3) cn be ssigned useful mening. 1 If we further ssume tht f(x) hs certin nice properties [e.g., sufficient (but not necessry) condition is tht f(x) is continuously differentible for x b], then it follows tht f(x)e ix dx = O, s. (4) Moreover, eqs. (2) nd (4) continue to hold if nd/or b, ssuming tht eq. (1) holds over the infinite intervl. We will present proof of eq. (2) under the ssumption tht f(x) is continuous over the closed intervl x b. The origin of eq. (2) in this cse is not too difficult to understnd. In the it of, the fctor e ix oscilltes fster nd fster such tht f(x)e ix verges out to zero over ny finite region of x inside the intervl. 1 For further detils, see J. Cmpos Ferreir, Introduction to the Theory of Distributions (Addison Wesley Longmn Limited, Essex, UK, 1997). 1

2 If we cn ssume tht f(x) is N-times differentible in the region x b, one cn derive eq. (4) simply by repeted integrtion by prts. Nmely, f(x)e ix dx = f(x) b i eix 1 f (x)e ix dx i = eib f(b) e i f() f b (x) i (i) 2 eix + 1 f (x)e ix dx (i) 2 N 1 ( ) = ( 1) n eib f (n) (b) e i f (n) () 1 +O, (5) (i) n+1 N+1 n= where f () (x) f(x) nd f (n) (x) d n f/dx n. For exmple f (b) is equl to the first derivtive of f(x) evluted t x = b, etc. Ting N = 1 in eq. (5), we see tht the leding term tht survives is of O(1/) s sserted by eq. (4). More generlly, we cn prove eq. (2) without the ssumption f(x) is differentible in the intervl. We do this by writing the integrl I() = in two different but equivlent wys: 2 nd I() = I() = +π/ π/ f(x)e ix dx+ f(x)e ix dx+ f(x)e ix dx, +π/ b π/ f(x)e ix dx (6) f(x)e ix dx. (7) By chnge of vribles, x = x π/, it is strightforwrd to verify tht π/ ( f(x)e ix dx = f x+ π ) e ix dx, (8) +π/ fter using e iπ = 1 nd dropping the primes fromthe x in the second integrl. Thus, writing I s one hlf the sum of eqs. (6) nd (7), nd employing eq. (8), it follows tht I() = 1 2 +π/ f(x)e ix dx π/ b π/ f(x)e ix dx [ ( f(x) f x+ π )] e ix dx. (9) 2 This proof is ten from Brin Dvies, Integrl Trnsforms nd Their Applictions, 3rd edition (Springer-Verlg, New Yor, 22) pp

3 We not te the it of. The men vlue theorem of for integrls sttes tht if f(x) is continuous nd bounded over closed intervl, x b, then f(x)dx = f(c)(b ), for some rel number c tht lies in the intervl c b. Applying this to the first two integrls in eq. (9), we immeditely conclude tht +π/ f(t)e ix dx = O, b π/ f()e ix dx = O, which vnish in the it of. Finlly, under the ssumption tht f(x) is continuous t ll points in the closed intervl x b, it follows tht π/t [ ( f(x) f x+ π )] e ix dx =. (1) This is true becuse function tht is continuous t ll points in closed, bounded intervl is uniformly continuous over the intervl. 3 Hence, one cn me the integrnd in eq. (1) rbitrrily smll by choosing sufficiently lrge. The it of eq. (1) is thus estblished, nd the Riemnn-Lebesgue lemm stted in eq. (2) is proven. Note tht the rgument bove does not necessrily imply tht π/ [ ( f(x) f x+ π )] e ix = O, (11) s. However, if the function f(x) is continuously differentible in the intervl, then we cn employ the men vlue theorem for differentible functions, which sttes tht f(b) f() = f (c)(b ), for some c between nd b. It follows tht f(x) f ( x+ π ) = π f (x+c), for c π. Hence, in this cse eq. (11) does hold, in which cse eq. (4) is stisfied. 4 3 For further detils, see e.g. Dvid Bressoud, A Rdicl Approch to Rel Anlysis, 2nd edition (The Mthemtics Assocition of Americ, Wshington, DC, 27) pp We lso obtin eq. (11) under slightly weer conditions in which the function f(x) stisfies the so-clled Lipschitz condition, f(x) f(y) M x y for ll x nd y in the intervl for some positive finite bound M. Indeed Lipschitz continuous function is uniformly continuous(lthough the converse is not necessrily true). A Lipschitz continuous function need not be differentible. On the other hnd differentible function whose derivtive is bounded on the intervl stisfies the Lipschitz condition. Thus, to estblish eq. (11) is is sufficient to require tht f() is Lipschitz continuous in the intervl. 3

4 The extension to cses where nd/or b is strightforwrd. For exmple, suppose tht Then, noting tht one cn write where f(x)e ix dx = ǫ < f(x) dx <. (12) b f(x)e ix dx+ǫ, f(x) dx, it follows [in light of eq. (12)] tht it is lwys possible to me ǫ rbitrrily smll by suitble (finite) choice of b. Hence, we cn use eq. (2) to conclude tht f(x)e ix dx =. (13) Finlly, it should be noted tht ting the rel nd imginry prts of eq. (2) under the ssumption tht f(x) is rel vlued function yields, f(x)sinxdx =, (14) f(x)cosxdx =, (15) with similr extension to cses where nd/or b. Tht is, cosx = sinx =, (16) where the it is interpreted in the sense of distributions [cf. discussion below eq. (3)]. If we further ssume tht f(x) hs certin nice properties [cf. comments bove eq. (4)], then the two integrls given in eqs. (14) nd (15) behve s O(1/) s. ByconsultingtbleofFouriertrnsforms, 5 onecnseemnyexmplesoffunctions tht stisfy eq. (1). In ll cses, you will find tht the corresponding Fourier trnsform stisfies eq. (2). It is interesting to loo for cses tht stisfy eq. (2) nd not eq. (4). For exmple, x ν 1 e x e ix dx = Γ(ν), for > nd Re ν >. ( i) ν Indeed, eqs. (1) nd (2) re stisfied for ll Re ν >, wheres eq. (4) is only stisfied when Re ν 1. 5 See e.g., F. Oberhettinger, Tble of Fourier Trnsforms nd Fourier Trnsforms of Distributions (Springer-Verlg, Berlin, 199). 4

5 Finlly, we provide the following interesting exmple tht contrdicts eq. (14). Choosing f(x) = sin(x 2 ) nd consulting I.S. Grdshteynn nd I.M. Ryzhi, Tble of Integrls, Series nd Products, 8th edition, edited by Dniel Zwillinger (Acdemic Press, Elsevier, Amsterdm, 215), we note the following result on p. 419 [formul number 5], ( ) sin(x 2 )sinxdx = 1 2 +π 2 πcos, 4 which does not vnish s. However, this is not surprising in light of the fct tht eq. (12) [with = ] is not stisfied. Indeed, even though sin(x 2 )dx = 1 π 2 2, is finite, it turns out tht sin(x 2 ) dx diverges. Thus, one of the ey ssumptions underlying the Riemnn-Lebesgue lemm does not hold in this cse. 5