Fourier series. Preliminary material on inner products. Suppose V is vector space over C and (, )


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1 Fourier series. Preliminry mteril on inner products. Suppose V is vector spce over C nd (, ) is Hermitin inner product on V. This mens, by definition, tht (, ) : V V C nd tht the following four conditions hold: (i) (v + v, w) = (v, w) + (v, w) whenever v, v, w V ; (ii) (cv, w) = c(v, w) whenever c C nd v, w V ; (iii) (w, v) = (v, w) whenever v, w V ; (iv) (v, v) is positive rel number for ny v V {0}. These conditions imply tht (v) (v, w + w ) = (v, w ) + (v, w ) whenever v, w, w V ; (vi) (v, cw) = c(v, w) whenever c C nd v, w V ; (vii) (0, v) = 0 = (v, 0) for ny v V. In view of (iv) nd (vii) we my set v = (v, v) for v V nd note tht (viii) v = 0 v = 0. We cll v the norm of v. Note tht (ix) cv = c v whenever c C nd v V. Suppose A : V V R nd B : V V R re such tht () (v, w) = A(v, w) + ib(v, w) whenever v, w V. One esily verifies tht (i) A nd B re biliner over R; (ii) A is symmetric nd positive definite; (iii) B is ntisymmetric; (iv) A(iv, iw) = A(v, w) whenever v, w V ;
2 (v) B(v, w) = A(iv, w) whenever v, w V. Conversely, given A : V V R which is biliner over R nd which is positive definite symmetric, letting B be s in (v) nd let (, ) be s in () we find tht (, ) is Hermitin inner product on V. The interested reder might write down conditions on B which llow one to construct A nd (, ) s well. Exmple One. Let (z, w) = n z j w j for z, w C n. j= The (, ) is esily seen to be Hermitin inner product, clled the stndrd (Hermitin) inner product, on C n. Exmple Two. Suppose < < b < nd F is the vector spce of complex vlued Riemnn integrble functions on [, b]. Note tht Let (f, g) = f(x) dx = Rf(x) dx + i If(x) dx. f(x)g(x) dx whenever f, g F. One esily verifies tht (i)(iii) of the properties of n inner product hold nd tht (iv) lmost holds in the sense tht for ny f F we hve (f, f) = f(x) dx 0 with equlity only if {x [, b] : f(x) = 0} hs zero Jordn content. In prticulr, if f is continuous nd (f, f) = 0 then f(x) = 0 for ll x [, b]. This Exmple is like Exmple One in tht one cn think of f F s n infinitetuple with the continous index x [, b]. The following simple Proposition is indispensble. Suppose v, w V. Then v + w = v + R(v, w) + w. Proof. We hve v + w = (v + w, v + w) = (v, v) + (v, w) + (w, v) + (w, w) = (v, v) + (v, w) + (v, w) + (w, w) = v + R(v, w) + w. The CuchySchwrtz Inequlity. Suppose v, w P. Then (v, w) v w with eqlity only if {v, w} is dependent. Proof. If w = 0 the ssertion holds trivilly so let us suppose w 0. For ny c C we hve 0 v + cw = v + R(v, cw) + cw = v + R(c(v, w)) + c w. Letting (v, w) c = w
3 we find tht 0 v (v, w) w with equlity only if v + cw = 0 in which cse v + cw = 0 so v = cw. Corollry. Suppose nd b re sequences of complex numbers. Then ( ) / ( ) / n b n n b n. n=0 n=0 Proof. For ny nonnegtive integer N pply the CuchySchwrtz inequlity with (, ) equl the stndrd inner product on C N, v = ( 0,..., N ) nd w = (b 0,..., b N ) nd then let N. n=0 The Tringle Inequlity. Suppose v, w P. Then v + w v + w with equlity only if either v is nonnegtive multiple of w or w is nonnegtive multiple of v. Proof. Using the CuchySchwrtz Inequlity we find tht v + w = v + R(v, w) + w v + v w + w = ( v + w ). Suppose equlity holds. In cse v = 0 then v = 0w so suppose v 0. Since (v, w) R(v, w) = v w we infer from the CuchySchwrtz Inequlity tht w = cv for some c C. Thus from which we infer tht + c v = ( + c)v) = v + cw = v + cw = ( + c ) v which implies tht c is nonnegtive rel number. + Rc + c = + c = ( + c ) = + c + c Suppose U is liner subspce of V. We let U = {v V : (u, v) = 0 for ll u U} nd note tht U is liner subspce of V. It follows directly from (iv) tht U U = {0}. We sy subset A of V is orthonorml if whenever v, w A we hve (v, w) = { if v = w; 0 if v w. Orthogonl projections. We sy liner subspce of V is nice if for ech v V there is u U such tht () v u v ũ whenever ũ U; 3
4 tht is, no point of U is closer to v thn u. Suppose U is nice liner subspce of V. Suppose v V nd u U. Then () holds if nd only if (3) v u U. Proof. We hve (4) v w = (v u) + (u w) = v u + R(v u, w) + u w for w V. If (3) holds nd ũ U we set w = ũ in (4) nd infer tht v ũ = v u + u ũ v u so () holds. Suppose () holds nd ũ U. For nd t R nd c C we set w = u tcũ U in (4) nd obtin so v u v w = v u + tr(v u, cũ) + t cũ 0 tr(v u, cũ) + t cũ so R(v u, cũ) = 0. Letting c = ±i we infer tht (v u, ũ) = 0. Suppose v V nd, for j =,, we hve u j U nd v u j v ũ whenever ũ U. Then u = u. Proof. From the preceding Proposition we obtin (v u, ũ) = 0 nd (v u, ũ) = 0 whenever ũ U. Subtrcting we obtin Now let ũ = u u. (u u, ũ) = 0 for ll ũ U. In view of the preceding Proposition we my define the mp P : V U, cllled orthogonl projection of V onto U, by requiring tht v P v U whenever v V. We define by requiring tht P : V U P v = v P v whenver v V. Theorem. We hve 4
5 (i) P is liner; (ii) P P = P. (iii) (P v, w) = (v, P w) whenever v, w P. (iv) U is nice nd P is orthogonl projection of V onto U. (v) v = P v + P v for ny v V. Proof. All this follows from (3). Suppose v, w V nd c C. Then (v + cw) (P v + cp w) = (v P v) + c(w P w) U so P (v + cw) = P v + cp w so (i) holds. Tht P u = u for u U is immedite nd this implies (ii). We hve (P v, w) = (P v, P w + P w) = (P v, P w) = (P w, P v) = (P w, P v + P v) = (P w, v) = (v, P w) so (iii) holds. If w U we hve so (iv) holds. Finlly (v P v, w) = (P v, w) = 0 v = P v + P v = P v + R(P v, P v) + P v = P v + P v. The GrmSchmidt Process. Suppose P is orthogonl projection on the nice liner subspce U of V, ũ V U, Ũ = {u + cũ : c C} nd P v = P v + Then Ũ is nice nd P is orthogonl projection on Ũ. Proof. Esy exercise for the reder. (v, ũ) ũ ũ whenever v V. Remrk. If U = {0} then P = 0 so P (v) = (v, ũ) ũ ũ nd P is orthogonl projection on the line {cũ : c C}. Suppose U is finite dimensionl liner subspce of V. Then U is nice. Moreover, if B is finite orthonorml subset of U nd the number of elements in B equls the dimension of U then P v = (v, u)u nd P v = (v, u) whenever v V. u B u B Proof. Stndrd liner lgebr together with the GrmSchmidt process my be used to produce B s bove nd to show tht B is bsis for U. Let Lv = u B(v, u)u for v V. Suppose v V nd ũ B. It is evident tht (v Lv, ũ) = 0 which, s B is bsis for U, implies tht v Lv U ; thus P = L. 5
6 Finlly, if v V we hve = (u, v). u B Lv = ( (v, u)u, ũ)ũ) u B ũ B(v, = (v, u)(v, ũ)(u, ũ) u B, ũ B The rel stuff. Given rel number P, we sy complex vlued function f on R is P periodic if We let f(x + P ) = f(x) for ll x R. P be the vector spce of complex vlued periodic functions on R which re integrble over ny bounded intervl. For f, g P we let (f, g) = f(x)g(x) dx. We hve lredy shown tht (, ) is, essentilly, Hermitin inner product on P. For ech n Z we let E n (x) = e inx, C n for x R; evidently, E n P. Suppose A C {0}. Proof. Since e Ax dx = eax A b = eab e A. A d e Ax dx A = eax, for x R the Proposition follows from the Fundmentl Theorem of Clculus. Corollry (E m, E n ) = { if m = n, 0 else. Remrk. Thus the the set { E n : n Z} is orthonorml with respect to (, ). For ech N N we let T N 6
7 be the liner subspce of P spnned by {E n : n N} nd we cll the members of T N trigonometric polynomils of degree N. For ech f P we define ˆf : Z C, the Fourier trnsform of f, by letting ˆf(n) = (f, E n ) for n Z. One of our gols is to reconstruct f from its Fourier trnsform. As first step in this direction, for ech nonnegtive integer N nd ech f P we set P N f = ˆf(n)E n. From our erlier work we find tht if f P then P N f is orthogonl projection of f onto T N. Bessel s Inequlity. For ny f P we hve ˆf(n) f. n Z Remrk. Plncherel s Theorem, which comes lter, will give the opposite inequlity. Proof. This follows directly from our previous work with orthogonl projections. For ny f P nd ny R we hve Proof. Let n be tht integer such tht + + Mking the substitution x = w n we hve n+π f(x) dx = f(x) dx = f(x) dx. n π < n + π. n Mking the substitution x = w (n + ) we hve + n+π f(x) dx = n We complete the proof by dding these equtions. f(w n) dw = f(w (n + )) dw = n n f(w) dw. f(w) dw. Using this result we now show tht the Fourier trnsform behves nicely with respect to trnsltions. Whenever h R nd f P we define τ h f : R C by setting Evidently, τ h f P. We hve () τ h is liner for ech h R; τ h f(x) = f(x h) for x R; 7
8 () τ h τ h = τ h +h whenever h, h R; (3) (τ h f, τ h g) = (f, g) whenever f, g P nd h R. Proof. Exercise. Suppose f P nd h R. Then τ h f(n) = e inh ˆf(n). Proof. Exercise. Corollry. Suppose f P nd h R. Then P N (τ h f) = e inh P N f. Proof. Just unwrp the definition of P N. by setting For f P we define Af : R C Af(x) = f( x) whenever x R; Evidently, Af P. We sy f P is even if Af = f nd we sy f is odd if Af = f. We set f e = (f + Af) nd f o = (f Af). Evidently, f e is even, f o is odd nd f = f e + f o. Suppose f P. Then Âf(n) = ˆf(n) for n Z. Proof. Strightforwrd exercise. Suppose f, g P. For ech x R we set f g(x) = { f(x y)g(y) dy if f(x y)g(y) dy < 0 else nd we cll f g the convolution of f nd g. It is not hrd to show tht f g P. Suppose f, g P. Then f g = ˆfĝ. Proof. Exercise. For ech nonegtive integer N we define the Dirichlet kernel D N by letting D N = 8 E n.
9 (i) (ii) D N is even, Let N be nonegtive integer. Then 0 D N (x) = (iii) P N f = D N f for ny f P. Proof. Suppose N nd x R {0}. Then sin(n+ )x sin x if x 0 N + if x = 0; D N (x) dx = = D N (x) dx =. D N (x) = N n= N 0 e inx = e inx (eix ) N+ e ix = e i(n+ )x e i(n+ )x e i x e i x = sin(n + )x sin x, nd it is evident tht P N (0) = N+ so (i) holds. Tht D N is even follows directly from (i). We hve D N (x) dx = N n= N which, together with the fct tht D N is even implies (ii). To prove (3), suppose f P nd x R nd observe tht P N f(x) = = = (f, E n )E n ( = D N f(x). (E n, E 0 ) = ( ) π f(t)e int dt e inx e in(x t) )f(t) dt The Riemnn Lebesgue Lemm. Suppose < < b < nd f is integrble on (, b). Then lim t f(x) sin tx dx = 0. Proof. It follows directly from preceding Proposition tht the ssertion holds if f is step function. Let η > 0. Choose step function s such tht f s η. 9
10 Then f(x) sin tx dx = η + [f(x) s(x)] sin tx dx + f(x) s(x) dx + s(x) sin tx dx ; s(x) sin tx dx s(x) sin tx dx for ny t R. Let t nd note tht η is rbitrry to complete the proof. Corollry. Suppose f P. Then lim ˆf(n) = 0. n Lemm. Suppose 0 < x < π/. Then sin x + π x. Proof. Suppose 0 < b < π/. If 0 < x b then, by the Men Vlue Theorem there is ξ ( x, x ) such tht sin x = sin x sin 0 = cos ξ(x 0) cos b x. If b < x < π then Now let b = rctn π. sin x sin b sin b π x. Lemm. Then Suppose g P nd Proof. Let h : R C be such tht h(x) = so By the RiemnnLebesgue Lemm, P N g(0) = h(x) D N (x)g(x) dx = g(x) x dx <. lim P N g(0) = 0. N g(x) sin x/ if x 0 nd h(0) = 0. By virtue of the preceding Lemm, + π g(x) x h(x) dx <. if 0 < x < π ( h(x) sin N + ) x dx 0 s N. The Fourier Inversion Formul. Suppose f P, R, L +, L C nd f(x) L dx + x +π 0 f(x) L+ dx <. x
11 Then Very importnt remrk. lim P Nf() = L + L +. N For exmple, if f is differentible t the hypothesis holds with L ± = f(). Proof. Let A = L + L +. Let s P be such tht { L A if x < 0, s(x) = L + A if 0 x < π nd note the s is odd. Let c P be such tht c(x) = A for x R nd let We hve g = τ f s c. g(w) 0 w dw = f(w + ) L w dw + f(w + ) L + 0 w dw = f(x) L +π x dx f(x) L + x dx <. By the preceding Lemm we hve lim P N g(0) = 0. N Suppose N is positive integer. Since s is odd nd D N is even we hve tht D N s is odd so Moreover, But It follows tht P N s(0) = P N τ f(0) = D N (x)s(x) dx = 0. P N c(0) = A D N (x) dx = A. lim P N τ f(0) = A. N τ f(n) = e in ˆf(n) = PN f(). Suppose m is positive integer, f P nd f is m times continuously differentible. Then Proof. We use integrtion by prts to obtin ˆf (n) = = f (m) (n) = (in) m ˆf(n). f (x) e inx dx e inx d(f(x)) = e inx f(x) x=π x= = in f(x) e inx dx = in ˆf(n). f(x) d(e inx )
12 Thus the Proposition holds if m = nd follows for rbitrry m by induction. Remrk. A good wy to look t this is to note tht f = lim h 0 h (τ hf f) nd then to recll tht τ h f = e inh f. Corollry. Suppose m is positive integer, f P nd f is m times continuously differentible, N is nonnegtive integer nd x R. Then Proof. Note tht f(x) P N f(x) f (m). π N m ( ) / ( ) / n b n n b n n Z n= n Z whenever nd b re complex vlued functions on Z; tht nd tht, by Bessel s Inequlity, n=n+ n m dx N x m = N m ; n >N f (m) ˆ (n) f (m). Suppose x R. From the Fourier Inversion Formul we hve tht f(x) P N f(x) = lim M N<n M n m N< n <O ˆf(n)E n (s). Moreover, Suppose O nd N re positive integers nd O < N. ˆf(n)E n (x) N< n <O = ˆ (in) m f (m) (n)e n (x) N< n <O / f (m) ˆ (n) N m f (m). n >N / Corollry. Plncherel s (or is it Prsevl s?) Theorem. Suppose f P nd f = f(x) dx <. ˆf(n). n Z
13 Proof. Tht the right hnd side does not exceed the left hnd side is Bessel s Inequlity. To prove the opposite inequlity one combines n pproximtion rgument with the Fourier Inversion Formul. Theorem. Suppose f, g P, M is positive integer f (x) = g(x) for x in ny of the intervls nd the limits exist for j =,..., M. Then the limits exist for j =,..., M nd where < < < < M π, 0 = M, ( 0, ), (, ),..., ( M, M ) lim g(x), x j lim f(x), x j (f, E n ) = (g, E n ) + in lim g(x) x j lim f(x) x j M J m E n ( m ) j= J j = lim x j f(x) lim x j f(x), j =,..., M. Proof. For ny j =,..., M we hve j f(x)e inx dx j j ( ) e inx = f(x)d dx j in ( ) e inx j ( ) j e = f(x) f inx (x) dx in j j in ( = lim f(x)e n ( j ) lim f(x)e n ( j ) + in x j x j Now sum over j =,..., M. Exmple. Then ˆf(0) = 0 nd Let f P be such tht f(x) = x if x < π. j ˆf(n) = in ( )n if n 0. To see this let M = nd let = π. Then J = so if n 0 we hve (f, E n ) = in where g P is such tht g(x) = for < x < π. Corollry. ( (g, E n ) E n (π) n= n = π 6. ) j f (x)e inx dx = in ( )n+ Proof. Let f be the member of P such tht f(x) = x for x [, π) nd pply Prsevl s Theorem. 3 ).
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