MIT OpenCourseWarehttp://ocw.mit.edu 18.102 Introduction to Functional Analysis Spring 2009 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.� � 87 LECTURE NOTES FOR 18.102, SPRING 2009 Lecture 14. Tuesday, March 31: Fourier series and L2(0, 2π). Fourier series. Let us now try applying our knowledge of Hilbert space to a concrete Hilbert space such as L2(a, b) for a finite interval (a, b) ⊂ R. You showed that this is indeed a Hilbert space. One of the reasons for developing Hilbert space techniques originally was precisely the following result. Theorem 12. If u ∈ L2(0, 2π) then the Fourier series of u, (14.1) 1 � cke ikx , ck = � u(x)e−ikxdx2π k∈Z (0,2π) converges in L2(0, 2π) to u. Notice that this does not say the series converges pointwise, or pointwise almost everywhere since this need not be true – depending on u. We are just claiming that 1 � (14.2) lim |u(x) − 2πcke ikx|2 = 0 n→∞ |k|≤n for any u ∈ L2(0, 2π). First let’s see that our abstract Hilbert space theory has put us quite close to proving this. First observe that if e�k(x) = exp(ikx) then these elements of L2(0, 2π) satisfy � � 2π � 0 if k = j(14.3) e�ke�j = exp(i(k − j)x) = �0 2π if k = j. Thus the functions e�k 1 ikx (14.4) ek = = e �e�k�√2π form an orthonormal set in L2(0, 2π). It follows that (14.1) is just the Fourier-Bessel series for u with respect to this orthonormal set:-(14.5) ck = √2π�u, ek� =1 cke ikx = �u, ek�ek.⇒ 2π So, we alreay know that this series converges in L2(0, 2π) thanks to Bessel’s identity. So ‘all’ we need to show is Proposition 21. The ek, k ∈ Z, form an orthonormal basis of L2(0, 2π), i.e. are complete: (14.6) ue ikx = 0 ∀ k = ⇒ u = 0 in L2(0, 2π). This however, is not so trivial to prove. An equivalent statement is that the finite linear span of the ek is dense in L2(0, 2π). I will prove this using Fej´er’s method. In this approach, we check that any continuous function on [0, 2π] satisfying the additional condition that u(0) = u(2π) is the uniform limit on [0, 2π] of a sequence in the finite span of the ek. Since uniform convergence of continuous functions certainly implies convergence in L2(0, 2π) and we already know that the continuous functions which vanish near 0 and 2π are dense in L2(0, 2π) (I will recall why later) this is enough to prove Proposition 21. However the proof is a serious piece of analysis, at least it is to me!� �� � � 88 LECTURE NOTES FOR 18.102, SPRING 2009 So, the problem is to find the sequence in the span of the ek. Of course the trick is to use the Fourier expansion that we want to check. The idea of Ces`aro is to make this Fourier expansion ‘converge faster’, or maybe better. For the moment we can work with a general function u ∈ L2(0, 2π) – or think of it as continuous if you prefer. So the truncated Fourier series is 1 �� u(t)e−iktdt)e ikx (14.7) Un(x) = (2π (0,2π)|k|≤n where I have just inserted the definition of the ck’s into the sum. This is just a finite sum so we can treat x as a parameter and use the linearity of the integral to write this as (14.8) Un(x) = Dn(x − t)u(t), Dn(s) = 21 π � e iks . (0,2π) |k|≤n Now this sum can be written as an explicit quotient, since, by telescoping, 1212)s(14.9) (2π)Dn(s)(eis/2 is/2) = e i(n+− e)s − e−i(n+. So in fact, at least where s = 0, 1212i(n+ )s − e−i(n+2π(eis/2 − e−is/2) )se(14.10) Dn(s) = and of course the limit as s 0 exists just fine. →As I said, Ces`aro’s idea is to speed up the convergence by replacing Un by its average n1 � (14.11) Vn(x) = Ul. n + 1 l=0 Again plugging in the definitions of the Ul’s and using the linearity of the integral we see that � n1 � (14.12) Vn(x) = Sn(x − t)u(t), Sn(s) = n + 1 Dl(s). (0,2π) l=0 So again we want to compute a more useful form for Sn(s) – which is the Fej´er kernel. Since the denominators in (14.10) are all the same, n n1 21 2)sis/2 − e−is/2)Sni(n+(s) = e )s e−i(n+(14.13) 2π(n + 1)(e − . l=0 l=0 Using the same trick again, n1 2)s i(n+1)s= e − 1is/2 − e−is/2) i(n+e(14.14) (el=0 so (14.15) 2π(n + 1)(eis/2 − e−is/2)2Sn(s) = e i(n+1)s + e−i(n+1)s − 2 =⇒ sin2( (n+1)1 2 s)Sn(s) = . n + 1 2π sin2( 2 s )� � � � 89 LECTURE NOTES FOR 18.102, SPRING 2009 Now, what can we say about this function? One thing we know immediately is that if we plug u = 1 into the disucssion above, we get Un = 1 for n ≥ 0 and hence Vn = 1 as well. Thus in fact (14.16) Sn(x − ·) = 1. (0,2π) Now looking directly at (14.15) the first thing to notice is that Sn(s) ≥ 0. Also, we can see that the denominator only vanishes when s = 0 or s = 2π in [0, 2π]. Thus if we stay away from there, say s ∈ (δ, 2π − δ) for some δ > 0 then – since sin is a bounded function (14.17) |Sn(s)| ≤ (n + 1)−1Cδ on (δ, 2π − δ). Now, we are interested in how close Vn(x) is to the given u(x) in supremum norm, where now we will take u to be continuous. Because of (14.16) we can write (14.18) u(x) = Sn(x − t)u(x) (0,2π) where t denotes the variable of integration (and x is fixed in [0, 2π]). This ‘trick’ means that the difference is (14.19) Vn(x) − u(x) = Sx(x − t)(u(t) − u(x)). (0,2π) For each x we split this integral into two parts, the set Γ(x) where x − t ∈ [0, δ] or x − t ∈ [2π − δ, 2π] and the remainder. So (14.20) � � |Vn(x) − u(x)| ≤ Γ(x) Sx(x − t)|u(t) − u(x)| + (0,2π)\Γ(x) Sx(x − t)|u(t) − u(x)|. Now on Γ(x) either |t − x| ≤ δ – the points are close together – or t is close to 0 and x to 2π so 2π − x + t ≤ δ or conversely, x is close to 0 and t to 2π so 2π − t + x ≤ δ. In any case, by assuming that u(0) = u(2π) and using the uniform continuity of a continuous function on [0, 2π], given � > 0 we can choose δ so small that (14.21) |u(x) − u(t)| ≤ �/2 on Γ(x). On the complement of Γ(x) we have (14.17) and since u is bounded we get the estimate (14.22) |Vn(x)−u(x)| ≤ �/2 Γ(x) Sn(x−t)+(n+1)−1C�(δ) ≤ �/2+(n+1)−1C�(δ). Here the fact that Sn is non-negative and has integral one has been used again to estimate the integral of Sn(x − t) over Γ(x) by …