Orthonormal Bases Fall Consider an inner product space V with inner product f, g and norm

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1 8.03 Fall 203 Orthonormal Bases Consider an inner product space V with inner product f, g and norm f 2 = f, f Proposition (Continuity) If u n u 0 and v n v 0 as n, then u n u ; u n, v n u, v. Proof. Note first that since v n v 0, v n v n v + v M < for a constant M independent of n. Therefore, as n, u n, v n u, v = u n u, v n + u, v n v M u n u + u v n v 0 In particular, if u n = v n, then u n 2 = u n, u n u, u = u 2. For u and v in V we say that u is perpendicular to v and write u v if u, v = 0. The Pythogorean theorem says that if u v, then u + v 2 = u 2 + v 2 () Definition ϕ n is called an orthonormal sequence, n =, 2,..., if ϕ n, ϕ m = 0 for n m and ϕ n, ϕ n = ϕ n 2 =. Suppose that ϕ n is an orthonormal sequence in an inner product space V. The following four consequences of the Pythagorean theorem () were proved in class (and are also in the text): If h = a n ϕ n, then h 2 = a n 2. (2)

2 If f V and s N = f, ϕ n ϕ n, then If V N = span {ϕ, ϕ 2,..., ϕ N }, then f 2 = f s N 2 + s N 2 (3) f s N = min g V N f g (best approximation property) (4) If c n = f, ϕ n, then f 2 c n 2 (Bessel s inequality). (5) Definition 2 A Hilbert space is defined as a complete inner product space (under the distance d(u, v) = u v ). Theorem Suppose that ϕ n is an orthonormal sequence in a Hilbert space H. Let V N = span {ϕ, ϕ 2,..., ϕ N }, V = (V is the vector space of finite linear combinations of ϕ n.) The following are equivalent. N= a) V is dense in H (with respect to the distance d(f, g) = f g ), b) If f H and f, ϕ n = 0 for all n, then f = 0. c) If f H and s N = f, ϕ n ϕ n, then s N f 0 as N. d) If f H, then f 2 = f, ϕ n 2 If the properties of the theorem hold, then {ϕ n } 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 f, ϕ n = 0, then by taking finite linear combinations, f, v = 0 for all v V. Choose a sequence v j V so that v j f 0 as j. Then by Proposition above V N 0 = f, v j f, f = f 2 = 0 = f = 0 2

3 (5), (b) = (c). Let f H and denote c n = f, ϕ n, s N = Hence, for M < N (using (2)) s N s M 2 2 = c n ϕ n = M+ c n 2 f 2 <. c n ϕ n. By Bessel s inequality c n 2 0 as M, N. M+ In other words, s N is a Cauchy sequence in H. By completeness of H, there is u H such that s N u 0 as N. Moreover, f s N, ϕ n = 0 for all N n. Taking the limit as N with n fixed yields Therefore by (b), f u = 0. (c) = (d). Using (3) and (2), f u, ϕ n = 0 for all n. f 2 = f s N 2 + s N 2 = f s N 2 + c n 2, (c n = f, ϕ n ) Take the limit as N. By (c), f s N 2 0. Therefore, f 2 = c n 2 Finally, for (d) = (a), f 2 = f s N 2 + c n 2 Take the limit as N, then by (d) the rightmost term tends to f 2 so that f s N 2 0. Since s N V N V, V is dense in H. 3

4 Proposition 2 Let ϕ n be an orthonormal sequence in a Hilbert space H, and an 2 <, bn 2 < then u = are convergent series in H norm and Proof. Let Then for M < N, u N = a n ϕ n, v = u, v = b n ϕ n a n b n (6) a n ϕ n ; v N = u N u M 2 = b n ϕ n. a n 2 0 as M M so that u N is a Cauchy sequence converging to some u H. Similarly, v N v in H norm. Finally, u N, v N = a j ϕ j, b k ϕ k = a j b k ϕ j, ϕ k = a j b j j,k= since ϕ j, f k = 0 for j k and f j, f j =. Taking the limit as N and using the continuity property (), u N, v N u, v, gives (6). If H is a Hilbert space and {ϕ n } is an orthonormal basis, then every element can be written f = a n ϕ n (series converges in norm) The mapping {a n } n j,k= a n ϕ n is a linear isometry from l 2 (N) to H that preserves the inner product. The inverse mapping is f {a n } = { f, ϕ n } j= 4

5 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. u, v = a u + iv 2 + a 2 u + v 2 + a 3 u 2 + a 4 v 2 (7) with a = i/2, a 2 = /2, a 3 = ( + i)/2, a 4 = (i + )/2 Proof. u + iv 2 = u + iv, u + iv = u 2 + iv, u + u, iv + v 2 = u 2 + i( v, u u, v ) + v 2 Similarly, u + v 2 = u 2 + ( v, u + u, v ) + v 2 Multiplying the first equation by i and adding to the second, we find that Solving for u, v yields (7). i u + iv 2 + u + v 2 = (i + ) u u, v + (i + ) v 2 5