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.� � � � � � 59 LECTURE NOTES FOR 18.102, SPRING 2009 Lecture 10. Tuesday, Mar 10 All of this is easy to find in the various reference notes and/or books so I will keep these notes very brief. (1) Bessel’s inequality If in a preHilbert space H, ei, i = 1, . . . , N are orthonormal – so (ei, ej ) = δij then for any element u ∈ H, set Nv = (u, ei)ei then i=1 N(10.1) � 2 2 2�v� = |(u, ei)| ≤ �u�H H , i=1 (u − v) ⊥ ei, i = 1, . . . , N. The last statement follows immediately by computing (u, ej ) = (v, ej ) and similarly �v�2 can be computed directly. Then the inequality, which is Bessel’s inequality, follows from Cauchy’s inequality since from the last statement (10.2) �v�2 = (v, v) = (v, u) + (v, v − u) = (v, u) = |(v, u)| ≤ �v��u� shows �v� ≤ �u�. (2) Orthonormal bases: Since in the inequality in (10.1) the right side is independent of N it follows that if {ei}∞is a countable orthonormal set then i=1 ∞(10.3) |(u, ei)|2 ≤ �u�2 H . i=1 From this it follows that the sequence n(10.4) vn = (u, ei)ei i=1 is Cauchy since if m > n, ∞(10.5) �vn − vm�2 = �(u, ej )|2 ≤ �(u, ej )|2 n<j≤mj=n+1 and the right side is small if n is large, independent of m. Lemma 5. If H is a Hilbert space – so now we assume completeness – and {ei}∞is an orthonormal sequence then for each u ∈ H, i=1 ∞(10.6) v = (u, ej )ej ∈ H j=1 converges and (u − v) ⊥ ej for all j. Proof. The limit exists since the sequence is Cauchy and the space is com-plete. The orthogonality follows from the fact that (u − vn, ej ) = 0 as soon as n ≥ j and (10.7) (u − v, ej ) = lim (u − vn, ej ) = 0 n→∞� � � � 60 LECTURE NOTES FOR 18.102, SPRING 2009 by continuity of the inner product (which follows from Cauchy’s inequality). Now, we say an orthonormal sequence is complete, or is and orthonormal basis of H if u ⊥ ej = 0 for all j implies u = 0. Then we see:-Proposition 15. If {ej �∞ is an orthonormal basis in a Hilbert space Hj=1 then ∞(10.8) u = (u, ej )ej ∀ u ∈ H. j=1 Proof. From the lemma the series converges to v and (u − v) ⊥ ej for all j so by the assumed completeness, u = v which is (10.8). � (3) Gram-Schmidt Theorem 6. Every separable Hilbert space has an orthonormal basis. Proof. Take a countable dense subset – which can be arranged as a se-quence {vj } and the existence of which is the definition of separability – and orthonormalize it. Thus if v1 =� 0 set ei = v1/�v1�. Proceeding by induction we can suppose to have found for a given integer n elements ei, i = 1, . . . , m, where m ≤ n, which are orthonormal and such that the linear span (10.9) sp(e1, . . . , em) = sp(v1, . . . , vn). To show the inductive step observe that if vn+1 is in the span(s) in (10.9) then the same ei work for n + 1. So it follows that n� w(10.10) w = vn+1 − (vn+1, ej )ej = 0 so em+1 = j=1 ��w� makes sense. Adding em+1 gives the equality of the spans for n + 1. Thus we may continue indefinitely. There are only two possibilities, either we get a finite set of ei’s or an infinite sequence. In either case this must be an orthonormal basis. That is we claim (10.11) H � u ⊥ ej ∀ j = ⇒ u = 0. This uses the density of the vn’s. That is, there must exist a sequence wj where each wj is a vn, such that wj → u in H. Now, each each vn, and hence each wj , is a finite linear combination of ek’s so, by Bessel’s inequality (10.12) �wj �2 = |(wj , ek)|2 = |(u − wj , ek)|2 ≤ �u − wj �2 k k where (u, ej ) = 0 for all j has been used. Thus �wj � → 0 and u = 0. � (4) Isomorphism to l2 A finite dimensional Hilbert space is isomorphic to Cn with its standard inner product. Similarly from the result above Proposition 16. Any infinite-dimensional separable Hilbert space (over the complex numbers) is isomorphic to l2 , that is there exists a linear map (10.13) T : H −→ L2� 61 LECTURE NOTES FOR 18.102, SPRING 2009 which is 1-1, onto and satisfies (T u, T v)l2 = (u, v)H and �T u�l2 = �u�H for all u, v ∈ H. Proof. Choose an orthonormal basis – which exists by the discussion above and set (10.14) T u = {(u, ej )�∞j=1. This maps H into l2 by Bessel’s inequality. Moreover, it is linear since the entries in the sequence are linear in u. It is 1-1 since T u = 0 implies (u, ej ) = 0 for all j implies u = 0 by the assumed completeness of the orthonormal basis. It is surjective since if {cj }∞thenj=1 ∞(10.15) u = cj ej j=1 converges in H. This is the same argument as above – the sequence of partial sums is Cauchy by Bessel’s inequality. Again by continuity of the inner product, T u = {cj } so T is surjective. The equality of the norms follows from equality of the inner products and the latter follows by computation for finite linear combinations of the ej and then in general by continuity. �� � � � � � � � � � 62 LECTURE NOTES FOR 18.102, SPRING 2009 Problem set 5, Due 11AM Tuesday 17 Mar. You should be thinking about using Lebesgue’s dominated convergence at several points below. Problem 5.1 Let f : R −→ C be an element of L1(R). Define (10.16) fL(x) = f(x) x ∈ [−L, L] 0 otherwise. Show that fL ∈ L1(R) and that |fL − f | → 0 as L → ∞. Problem 5.2 Consider a real-valued function f : R −→ R which is locally inte-grable in the sense that (10.17) gL(x) = f(x) x ∈ [−L, L] 0 x ∈ R \ [−L, L] is Lebesgue integrable of each L ∈ N. (1) Show that for each fixed L the function ⎧ ⎪⎨ ⎪⎩ gL(x) if gL(x) ∈ [−N, N] N if gL(x) > N −N if gL(x) < −N (N )(10.18) gL (x) = is Lebesgue integrable. (N)� |g − gL| → 0 as N → ∞. (3) Show that there is a sequence, hn, of step functions such that (2) Show that L (10.19) hn(x) f(x) a.e. in R.→ (4) …