18.103 Fall 2013 Orthonormal Bases Consider an inner product space V with inner product hf; gi and norm kfk2 = hf; fi Proposition 1 (Continuity) If kun − uk ! 0 and kvn − vk ! 0 as n ! 1, then kunk ! kuk; hun; vni ! hu; vi: Proof. Note first that since kvn − vk ! 0, kvnk ≤ kvn − vk + kvk ≤ M < 1 for a constant M independent of n. Therefore, as n ! 1, jhun; vni − hu; vij = jhun − u; vni + hu; vn − vij ≤ Mkun − uk + kukkvn − vk ! 0 2 2 In particular, if un = vn, then kunk = hun; uni ! hu; ui = kuk . For u and v in V we say that u is perpendicular to v and write u ? v if hu; vi = 0. The Pythogorean theorem says that if u ? v, then ku + vk2 = kuk2 + kvk2 (1) Definition 1 'n is called an orthonormal sequence, n = 1; 2;::: , if h'n;'mi = 0 for n 6= m 2 and h'n;'ni = k'nk = 1. Suppose that 'n is an orthonormal sequence in an inner product space V . The following four consequences of the Pythagorean theorem (1) were proved in class (and are also in the text): N X If h = an'n, then n=1 N 2 X 2 khk = janj : (2) 1 1 N X If f 2 V and sN = hf; 'ni'n, then n=1 2 2 2 kfk = kf − sN k + ksN k (3) If VN = span f'1;'2;:::;'N g, then kf − sN k = min kf − gk (best approximation property) (4) g2VN If cn = hf; 'ni, then 1 2 X 2 kfk ≥ jcnj (Bessel's inequality): (5) n=1 Definition 2 A Hilbert space is defined as a complete inner product space (under the dis- tance d(u; v) = ku − vk). Theorem 1 Suppose that 'n is an orthonormal sequence in a Hilbert space H. Let 1 [ VN = span f'1;'2;:::;'N g;V = VN N=1 (V is the vector space of finite linear combinations of 'n.) The following are equivalent. a) V is dense in H (with respect to the distance d(f; g) = kf − gk), b) If f 2 H and hf; 'ni = 0 for all n, then f = 0. N X c) If f 2 H and sN = hf; 'ni'n, then ksN − fk ! 0 as N ! 1. n=1 d) If f 2 H, then 1 2 X 2 kfk = jhf; 'nij n=1 1 If the properties of the theorem hold, then f'ngn=1 is called an orthonormal basis or complete orthonormal system for H. (Note that the word \complete" used here does not mean the same thing as completeness of a metric space.) Proof. (a) =) (b). Let f satisfy hf; 'ni = 0, then by taking finite linear combinations, hf; vi = 0 for all v 2 V . Choose a sequence vj 2 V so that kvj − fk ! 0 as j ! 1. Then by Proposition 1 above 2 0 = hf; vji ! hf; fi =) kfk = 0 =) f = 0 2 N X (b) =) (c). Let f 2 H and denote cn = hf; 'ni, sN = cn'n. By Bessel's inequality 1 (5), 1 X 2 2 jcnj ≤ kfk < 1: 1 Hence, for M < N (using (2)) 2 N N 2 X X 2 ksN − sM k = cn'n = jcnj ! 0 as M; N ! 1: M+1 M+1 In other words, sN is a Cauchy sequence in H. By completeness of H, there is u 2 H such that ksN − uk ! 0 as N ! 1. Moreover, hf − sN ;'ni = 0 for all N ≥ n: Taking the limit as N ! 1 with n fixed yields hf − u; 'ni = 0 for all n: Therefore by (b), f − u = 0. (c) =) (d). Using (3) and (2), N 2 2 2 2 X 2 kfk = kf − sN k + ksN k = kf − sN k + jcnj ; (cn = hf; 'ni) 1 2 Take the limit as N ! 1. By (c), kf − sN k ! 0. Therefore, 1 2 X 2 kfk = jcnj 1 Finally, for (d) =) (a), N 2 2 X 2 kfk = kf − sN k + jcnj 1 2 2 Take the limit as N ! 1, then by (d) the rightmost term tends to kfk so that kf −sN k ! 0. Since sN 2 VN ⊂ V , V is dense in H. 3 Proposition 2 Let 'n be an orthonormal sequence in a Hilbert space H, and X 2 X 2 janj < 1; jbnj < 1 then 1 1 X X u = an'n; v = bn'n n=1 n=1 are convergent series in H norm and 1 X hu; vi = anbn (6) n=1 Proof. Let N N X X uN = an'n; vN = bn'n: 1 1 Then for M < N, N 2 X 2 kuN − uM k = janj ! 0 as M ! 1 M so that uN is a Cauchy sequence converging to some u 2 H. Similarly, vN ! v in H norm. Finally, N N N X X X huN ; vN i = haj'j; bk'ki = ajbkh'j;'ki = ajbj j;k=1 j;k=1 j=1 since h'j; fki = 0 for j 6= k and hfj; fji = 1. Taking the limit as N ! 1 and using the continuity property (1), huN ; vN i ! hu; vi, gives (6). 1 If H is a Hilbert space and f'ngn=1 is an orthonormal basis, then every element can be written 1 X f = an'n (series converges in norm) n=1 The mapping X fang 7! an'n n is a linear isometry from `2(N) to H that preserves the inner product. The inverse mapping is f 7! fang = fhf; 'nig 4 It is also useful to know that as soon as a linear mapping between Hilbert spaces is an isometry (preserves norms of vectors) it must also preserve the inner product. Indeed, the inner product function (of two variables u and v) can be written as a function of the norm function (of linear combinations of u and v). This is known as polarization: Polarization Formula. 2 2 2 2 hu; vi = a1ku + ivk + a2ku + vk + a3kuk + a4kvk (7) with a1 = i=2; a2 = 1=2; a3 = −(1 + i)=2; a4 = −(i + 1)=2 Proof. ku + ivk2 = hu + iv; u + ivi = kuk2 + hiv; ui + hu; ivi + kvk2 = kuk2 + i(hv; ui − hu; vi) + kvk2 Similarly, ku + vk2 = kuk2 + (hv; ui + hu; vi) + kvk2 Multiplying the first equation by i and adding to the second, we find that iku + ivk2 + ku + vk2 = (i + 1)kuk2 + 2hu; vi + (i + 1)kvk2 Solving for hu; vi yields (7). 5.