HILBERT SPACES FRANZ LUEF 1. Orthogonality Let M be a subspace of a Hilbert space H. Then the orthogonal complement of M is defined by M ? = fx 2 H : hx; yi = 0 for all y 2 Mg: The linearity and the continuity of the inner product allow us to show the following fact. Lemma 1.1. M ? is a closed subspace. ? ? Proof. Let x be an element of the closure of M . Hence there is a sequence (xn) in M ? such that xn ! x. Now, for any y 2 M we have hx; yi = hlimn xn; yi = limnhxn; yi = ? 0. Hence x 2 M . Some elementary properties are (1) M ⊂ N implies that N ? ⊂ M ?. (2) M ⊆ M ??. A theorem of utmost importance is the closest point theorem for closed convex sets in Hilbert spaces. Theorem 1.2. If M is a closed convex set and x is any point of a Hilbert space H, then there is a unique point x0 2 M which is closest to x: For all x 2 H we have that kx − x0k ≤ kx − yk for all y 2 M: 1 This is not true in Banach spaces. For example, c0 is a closed subspace in ` , but there is no closest sequence in c0 to the sequence (1; 1; 1; :::). In fact, the distance between c0 and (1; 1; 1; :::) is 1, and this is achieved by any bounded sequence (an) with an 2 [0; 2]. Important application: Let M be a closed subspace of H. In this situation, we are able to express the least distance property in a geometrical manner. Theorem 1.3 (Projection theorem). For a closed subspace M of H the point x0 2 M ? is the closest point to an element x 2 H if and only if x − x0 2 M . Corollary 1.4. H = M ⊕ M ?. Date: 10.09.2015. 1 Corollary 1.5. If M is a closed linear subspace of H, then M ?? = M. If S is a set in H, then S?? = span(S). ?? ? Proof. Let x 2 M . Then x = x0 + y0 where x0 2 M and y0 2 M . We have 2 0 = hx; y0i = hx0; y0i + hy0; y0i = ky0k . Hence y0 = 0 and consequently, x 2 M. Corollary 1.6. If S is a set in H, then S?? = span(S). In particular, span(S) is dense in H if and only if S? = 0. Proof. Note that span(S) is the smallest closed subspace containing S. Therefore, ? since S ⊆ span(S), we get by the preceding corollary that span(S) ⊆ S? and hence ? ?? ? S ⊆ span(S) = span(S). We look at the mapping PM that takes a x 2 H to its closest point x0 2 M. Theorem 1.7. If M is a closed subspace of H, then the map P : H!H defined by ? P x = x0 is a continuous projection with im(P ) = M; ker(P ) = M and ker(P ) ? im(P ). Proof. By definition of P , for any x 2 H we have x − P x 2 M ? and so P x 2 M. The mapping P is linear: We have x + y − (P x + P y) = x − P x + y − P y 2 M ? and P x + P y 2 M. Hence P (x + y) = P x + P y. We also have P (αx) = αP x. P is a projection: P 2 = P . Since P x 2 M we have P 2x = P x. P is onto M: Since for any x 2 M we have P x = x. Moreover, x 2 ker(P ) if and only if P x = 0 if and only if x = x − P x 2 M ?. P is continuous: Since kxk2 = kx − P xk2 + kP xk2 by Pythagoras. Hence kP xk ≤ kxk. Look at the following example: Let M be the subspace of H spanned by an orthonor- mal vector y. Then the orthogonal projection is given by P x = hx; yiy. Another example: Find the quadratic polynomial closest to a function f in L2[0; 1]. 2 The space of quadratic polynomials P2 is three-dimensional and f1; x; x g is a basis 2 and P2 is a closed subspace of L [0; 1]. Hence by our general result, there exists a p 2 P2 closest to f. In fact, p is characterized by the condition p − f ? q for all q 2 M. Equivalently, hp; xii = hf; xii for i = 0; 1; 2. The right hand sides can be computed, because f is a given function. Hence we get three linear equations in three unknowns, which can be solved. 2. Orthonormal Bases A set of elements fen : n 2 Ng in H is called orthonormal if hei; eji = δij. We call it a basis if spanfei : ı 2 Ng is dense in H. 2 2 (1) ` has a basis consisting of fei : i = 1; 2; ::g for ei = (0; 0; 0; :::; 0; 1; 0; :::). Since 2 2 spanei : i 2 N is dense in ` . Let x = (xi) in ` be orthogonal to all ei. Then for any i we have xi = hx; eii = 0 and hence x = 0. P If feng is a basis for a Hilbert space, then we want to know when k akek converges. Proposition 2.1. Let M be a closed subspace of H with a countable Hilbert basis en. P1 2 Then k=1 akek converges in M if and only if (ak) is in ` . Pn Proof. Let xn = k=1 akek. Then assuming n > m: 2 kxn − xmk = hxn − xm; xn − xmi n m X X = h akek; akeki k=1 k=1 n n X X = h akek; akeki k=m+1 k=m+1 n n X X = ajakhej; eki j=m+1 k=m+1 n X 2 = jajj : j=m+1 2 Suppose (ak) 2 ` . Then this last sum converges to 0, implying that (xn) is a Cauchy sequence in M, which must converge to a point x 2 M. Conversely, suppose that the sequence (xn) converges in M. Then it is a Cauchy P n 2 sequence, and kxn − xmk ! 0, implying that ( j = 1 jajj ) is a Cauchy sequence in C and must converge as n ! 1. Theorem 2.2. If a closed subspace M of H has a basis (ek), then we have for x 2 H: P1 2 2 (1) Bessel inequality: jhx; ekij ≤ kxk ; P k=1 (2) P x = khx; ekiek. Pn Proof. Let xn = k=1hx; ekiek. Then xn 2 M and letting ak = hx; eki we get 2 0 ≤ kx − xnk = hx − xn; x − xni = hx; xh−hxn; xi − hx; xni + hxn; xni n n 2 X 2 X = kxk − 2 jakj + ajakhej; eki k=1 j;k=1 n 2 X 2 = kxk − jakj : k=1 3 Pn 2 2 Hence, k=1 jhx; ekij ≤ kxk , which is an increasing sequence bounded above. Hence the sum on the left must converge and this implies the Bessel inequality. Pn By the previous proposition the series k=1hx; ekiek converges to a point y 2 M. Moreover for all m we have n X hx − y; emi = hx; emi − hx; ekihek; emi = 0: k=1 ? Hence x − y 2 M and so y is the closest point in M to x. An important consequence of the previous theorem is the following result. Theorem 2.3 (Parseval's identity). If (ek) is a basis for a Hilbert space H, then for P 2 P1 2 all x 2 X we have x = khx; ekiek and kxk = k=1 jakj . Proof. Since P x = x for M = H we get the basis expansion for all x 2 H, and the norm relation follows by the linearity and continuity of the innerproduct. Another consequence is that every separable Hilbert space is isomorphic to `2. 2 Theorem 2.4. Consider the map J : H! ` , x 7! (hx; eki). Then Bessel's inequality shows that the Fourier coefficients are in `2. Parseval's identity also implies that the innerproducts are preserved: hJx; Jyi = hx; yi and so J is an isometry. Hence it is P injective, the map J is also surjective: Since it maps khx; ekiek to (hx; eki), which is a well-defined element in `2. Department of Mathematics, Norwegian University of Science and Technology, 7491 Trondheim. E-mail address: [email protected] 4.