Uniform continuity of sinc x

Transcription

1 Uniform continuit of sinc Peter Haggstrom August 3, 03 Introduction The function sinc = sin as follows: is well known to those who stud Fourier theor. It is defined sinc = sin = { = 0 0 otherwise () The integral of sinc() over the whole real line or the half real line causes some calculus students big problems because it looks like it should be simple but it isn t. Thus sinc d = π so that 0 sinc d = π due to the fact that sinc is even. Don t even think about using integration b parts or a substitution. It takes comple integration theor or some relativel heav dut classical analsis to do it properl. The oscillator nature of sin is responsible for the difficulties. Serious students of analsis or calculus will want to know the complete proof of wh the integral converges but others ma be happ with a more practical (ie this works ) understanding. If ou are doing Fourier Theor ou reall do need to know how to derive the integral because it is such a fundamental part of the theor, although I won t present a proof here as I am more interested in the uniform continuit aspect of the function. The sinc function appears in lurid green flurorescent form on the front of Fr s electronics store in Sunnvale, California. Students of Fourier Theor at Stanford can go and pa homage before it in their spare time! Here s the link: The sinc function appears in a wide varier of contets including signal processing and probabilit theor. For instance, the uniform probabilit distribution on ( a, a) has densit f() = a and hence has characteristic function sinc(aξ). Recall that the characteristic function of f is defined as:

2 ψ(ξ) = e iξ f() d () Hence in the case of the uniform distribution where a we have: ψ(ξ) = a a e iξ a ] eiξ a cos(ξa) + i sin(ξa) [cos( ξa) + i sin( ξa)] f() d = = aiξ a aiξ = sin(ξa) = sinc(ξa) (3) ξa The graph of the sinc function is set out below: The uniform probabilit distribution is a species of the bo function that is a fundamental part of Fourier Theor. The bo function is also used in filtering signals. Perhaps one of the most important applications of the sinc function and one which is of immense interest to electrical engineers is the role it plas in the Sampling Theorem in Fourier Theor. The Sampling Theorem can be epressed this wa. If the Fourier Transform F (ω) of a signal function f(t) is zero for all frequencies above ω B (ie the signal is band limited ) then f(t) can be uniquel determined from its sampled values f(t) = (t nt ) n= f(nt ) sinc T where T = B. This is quite a remarkable theorem and more information can be found here: Nquist-Shannon_sampling_theorem

3 In quantum theor the sinc function arises in the analsis of a localized wave function of rectangular shape for a free particle []. The wave function is given b ψ() = Bk e ik dk where the wave number k is taken to be a continuous variable. The coefficients are B k = ψ() e ik dk b virtue of Fourier theor. If the rectangular function has width b then: B k = b b e ik d = kb sin( ) = b sinc( kb k ) (4) The phsical significance of this is that the wave function includes contributions of equal magnitude from positive and negative values of k. The consequence is that the average momentum of the particle is zero and the average position of the wave packet will not move right or left. So how do ou prove uniform continuit? The onl people who reall care about this are people who specialise in analsis and Fourier theor in particular, but if ou happen to inhabit those worlds even briefl ou reall do need to know how to prove uniform continuit unless ou want to look like a complete dunce! The wa the function is defined in () makes it clear that sinc is continuous on the whole real line and hence it is continuous on an closed, bounded interval and so must be uniforml continuous on all such closed intervals no matter how large. So the function is indeed uniforml continuous on the entire real line. Note here that we cannot argue that because the product of two continuous functions is continuous (and hence replicate the same logic about uniform continuit on that basis) because blows up at = 0. However, we know from calculus or analsis courses that sin lim 0 = and so sinc is continuous at = 0 and it is obviousl continuous everwhere else. Now an ɛ δ proof is trickier as indeed the normall are for uniform continuit because it is a global rather than local propert of a function. Recall that to prove that a function f() is uniforml continuous, if ou are given an ɛ > 0 ou must come up with a δ > 0 such that for an, satisfing 0 < < δ ou will have f() f() < ɛ. First consider <. We know b the Mean Value Theorem that there is an ξ between and such that: We see that: sin sin ( sin ξ ) = = ξ ξ cos ξ sin ξ ξ (5) 3

4 since ξ > Thus for ɛ > 0 and 0 < < ɛ ie our δ = ɛ sin ξ cos ξ sin ξ ξ cos ξ + sin ξ ξ ξ ξ + ξ < (6) sin we have: = ξ cos ξ sin ξ ɛ < < ξ = ɛ (7) In the interval (, ) we can use Talor s theorem to approimate the derivative in the mean value epression b epanding about zero. The Talor series epansions are as follows: f(ξ) = sin ξ = ξ ξ3 + R(ξ) (8) 3! g(ξ) = cos ξ = ξ! + R (ξ) (9) Recall that the general form of a Talor series epansion about 0 looks like this: f() = f(0) + f (0) +! f (0) + + n n! f (n) (0) + n+ (n + )! f (n+) ( ) (0) where (0, ). The last term in (0) is the Lagrange form of the remainder and in the case of both sin and cos the absolute value of the derivative term in the remainder is bounded b (since the n th derivative is a multiple of sin or cos ). Thus: R(ξ) ξ 4 4! and R (ξ) ξ 3 3! () Thus the derivative in (5) is bounded as follows: ξ cos ξ sin ξ ξ( ξ =! + R (ξ)) (ξ ξ3 3! + R(ξ)) ξ ξ = ξ3 3 + ξ R (ξ) R(ξ) ξ ξ ξ R (ξ) + R(ξ) ξ = ξ ξ 4 ξ 3 3 ξ 3 + ξ 6 + ξ 4 4 ξ < since ξ < () 4

5 Going back to (5) we see that if we choose 0 < < δ = ɛ we will have: sin sin = ξ cos ξ sin ξ < < ɛ (3) ξ So if we choose as our global δ = ɛ sin, whenever 0 < < δ it will follow that sin < ɛ. In this proof we needed to get different estimates for the two tailed intervals and the middle hump interval. This is not uncommon as the following demonstration shows. Note that the interval ou choose in relation to a proof of uniform continuit makes a difference to the tpe of estimate ou need to make. For instance, if we were onl interested in the interval (0, ), we could use the fact that on this interval sin ξ ξ (in ξ cos ξ sin ξ ξ ξ cos ξ ξ ξ cos ξ ξ fact this holds for 0 ξ π/) so that the derivative cos ξ. Thus the absolute value of the derivative is in the interval (0, ] so that sin sin = ξ cos ξ sin ξ ξ < ɛ. In this case we needn t use Talor s theorem. 3 Proving that if f is Riemann integrable on ever interval [a, b] and f(t) dt converges, then the Fourier transform ˆf is uniforml continuous We take ɛ > 0 as given. Now because f(t) dt converges, we can make the tails of the integral arbitraril small b going far enough out, just like with a convergent infinite series. This means that there is a R(ɛ) > 0 (obviousl how far we need to go out will depend on how small we make ɛ) such that t R(ɛ) f(t) dt < ɛ 4. That deals with the tails. To make an estimate of the middle rump we observe that because f is Riemann integrable on [ R(ɛ), R(ɛ)] it must be bounded on that interval (it will be continuous due to the Fundamental Theorem of Calculus) so we can sa that sup t [ R(ɛ),R(ɛ)] f(t) = B(ɛ). Now to show that the Fourier transform ˆf converges uniforml we have to show that is some δ such that whenever 0 < ξ ν < δ we have ˆf(ξ) ˆf(ν) < ɛ and to do this we break the relevant integral into the tails and the rump as follows, recalling that the Fourier transform of f is ˆf(ξ) = f(t) e πiξt dt: 5

6 ˆf(ξ) ˆf(ν) = t R(ɛ) t R(ɛ) f(t) (e πiξt e πiνt ) dt f(t) (e πiξt e πiνt ) dt + f(t) (e πiξt e πiνt ) t R(ɛ) Now if we choose ξ ν f(t) dt + R(ɛ) dt + R(ɛ) R(ɛ) R(ɛ) R(ɛ) f(t) (e πiξt e πiνt ) dt f(t) (e πiξt e πiνt ) sup f(t)(e πiξt e πiνt ) t [ R(ɛ),R(ɛ)] ɛ + R(ɛ) B(ɛ) 4 sup (e πiξt e πiνt ) t [ R(ɛ),R(ɛ)] dt ɛ + R(ɛ) B(ɛ) sup ξt νt ɛ t [ R(ɛ),R(ɛ)] + 4π R(ɛ)3 B(ɛ) ξ ν (4) ɛ 8π R(ɛ) 3 B(ɛ)+ then the last inequalit will become: ɛ +4π R(ɛ)3 B(ɛ) ξ ν = ɛ + 8π R(ɛ)3 B(ɛ), ξ ν This establishes the uniform continuit of the Fourier transform. < ɛ + 8π R(ɛ)3 B(ɛ) ɛ 8π R(ɛ) 3 B(ɛ)+ Note that the Mean Value theorem is used in (4) for if we let g(t) = e πit so that there eists a λ between ξ and ν ( all of which are in [ R(ɛ), R(ɛ)] ) such that: < ɛ = ɛ (5) g(ξt) g(νt) = e πξit e πνit = g (λt) ξt νt = πiλ e πiλt t ξ ν π λ t ξ ν (6) Hence sup t [ R(ɛ),R(ɛ)] π λ t ξ ν π R(ɛ) ξ ν because λ is also in [ R(ɛ), R(ɛ)]. If we define f() = for [, ] and f() = 0 otherwise then f clearl satisfies the condition of absolute integrabilit since f(t) dt = dt =. However, ˆf(ξ) dξ does not converge. So what is ˆf(ξ) dξ? It is none other than the absolute integral of the sinc function. Thus: ˆf(ξ) = f() e πiξ d = ] e πiξ d = e πiξ = sin(πξ) πiξ πξ (7) To demonstrate that ˆf(ξ) dξ does not converge we make the following estimates: 6

7 n+ (n+) ˆf(ξ) dξ n k+ 3 8 k+ 8 = n ˆf(ξ) dξ n (k ) > k+ 3 8 k+ 8 n ξ dξ > n 4 (k ) as n (8) (k + ) In making these estimates a couple of points are worth making:. First, we ignore the negative tail in the integral which certainl ensures that the integral of ˆf(ξ) will be at least as big as the series of positive tail terms.. Second, we choose the intervals of integration so that ou get sin(π(k + 8 ) ) = sin(πk + π 4 ) and sin(π(k [ ) = sin(πk + 3π 4 ) ) ] etc. The reason for doing this is that for ξ in the interval sin(πk+ π 4 ) ), sin(πk+ 3π 4 ) ) the value of sin(πξ) sin( π 4 ) =. 3. Third, the term ξ in the integrand in each sub-interval dominates the rectangular bo formed b using the right-hand ξ-value, eg, the first bo has area 4 > (k+ 3 8 ) 4 (k+). All subsequent boes have the same width of 4. 4 Concluding remarks Although we have established b laborious ɛ δ techniques that sinc is uniforml continuous, if we had known some Fourier theor we could have ridden on the coat tails of a general theorem about the uniform continut of Fourier transforms to get to the same result. The oscillator nature of the sin function is sufficient in combination with to ensure that the integral sinc d converges but the deca of is not sufficientl fast to ensure that the integral of the absolute value of the function converges. 5 References [] A P French and Edwin F Talor, An Introduction ot Quantum Phsics, Van Nostrand, 987, pages [] T W Körner, Fourier Analsis, Cambridge Universit Press,

8 6 Appendi Note that for [ π/, π/], sin. The following diagram which deals with sin on the interval [0, π ] demonstrates the inequalit. 8