Theorem (4.11). Let M be a closed subspace of a Hilbert space H. For any x, y E, the parallelogram law applied to x/2 and y/2 gives.
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1 Math 641 Lecture # , 4.12, 4.13 Inner Products and Linear Functionals (Continued) Theorem (4.10). Every closed, convex set E in a Hilbert space H contains a unique element of smallest norm, i.e., there is exactly one x 0 E such that x 0 x for all x E. Proof. Let δ = inf{ x : x E}. For any x, y E, the parallelogram law applied to x/2 and y/2 gives 2 (1/4) x y 2 = (1/2) x 2 + (1/2) y 2 x + y 2. Since E is convex, the linear combination (1/2)(x + y) belongs to E. By the definition of δ, it follows that x y 2 2 x y 2 4δ 2. From this follows the uniqueness: if x = y = δ, then x y 2 0 x = y. Now for the existence: from the definition of δ, there is a sequence {y n } in E such that y n δ as n. Replace x and y by y n and y m in x y 2 2 x y 2 4δ 2 to get for all n and all m. y n y m 2 2 y n y m 2 4δ 2. As the left-hand side of this inequality goes to 0 as n and m, the sequence {y n } is Cauchy. Completeness of H implies there is x 0 H such that y n x 0, i.e., y n x 0 0 as n. Since y n E and E is closed, the limit x 0 of {y n } also belongs to E. Continuity of the norm function now implies that x 0 = lim n y n = δ. Thus E has a unique element of smallest norm. Theorem (4.11). Let M be a closed subspace of a Hilbert space H. (a) Every x H has a unique decomposition into a sum of P x M and Qx M. x = P x + Qx
2 (b) The elements P x and Qx are the nearest points to x in M and M. (c) The mappings P : H M and Q : H M are linear. (d) (Pythagorean Theorem) x 2 = P x 2 + Qx 2. Corollary. If M is a proper closed subspace of H, then there exists a nonzero y H such that y M. See the Appendix for a proof. Proof of Theorem (4.11). Parts (a) and (b). First the uniqueness of the decomposition is shown. Suppose that x + y = x + y for elements x, x M and y, y M. Then x x = y y. Since M and M are (closed) subspaces of H, it follows that x x M, y y M. However, M M = {0} because (x, x) = 0 implies x = 0 (Axiom 5 of an inner product). Thus x = x and y = y. Now on to the existence of the decomposition. For each x H, the translate x + M = {x + y : y M} is closed and convex. Define Qx to be the element of smallest norm in x + M; such an element is unique and exists by Theorem (4.10). Define P x = x Qx. Since Qx x + M, then P x = x Qx M, and so P maps H into M. Claim: Q maps H into M. This follows from showing that (Qx, y) = 0 for all y M. Assume y = 1, without loss of generality, and put z = Qx. The minimizing property of Qx shows that for all α C, (z, z) = z 2 z αy 2 = (z αy, z αy). Because y = 1, this inequality simplifies to The choice of α = (z, y) gives 0 α(y, z) α(z, y) + αα. 0 (z, y)(y, z) (z, y)(z, y) + (z, y)(z, y) = (z, y)(z, y) (z, y)(z, y) + (z, y)(z, y) = (z, y)(z, y) = (z, y) 2.
3 This implies that (z, y) = 0 for y arbitrary; this shows that Qx M. It remains to show that P x is the the element of M nearest to x. For y M, it follows since P x = x Qx with Qx M and P x y M, that x y 2 = Qx + (P x y) 2 = (Qx + (P x y), Qx + (P x y)) = (Qx, Qx) (Qx, P x y) + (P x y, Qx) + (P x y, P x y) = Qx P x y 2. Thus x y 2 is obviously minimized when y = P x. See the Appendix for proofs of parts (c) and (d). The Riesz Representation Theorem for Hilbert Spaces. If L is a continuous linear functional on H, then there exists a unique y H such that Lx = (x, y) for all x H. Proof. If Lx = 0 for all x H, then take y = 0. If Lx 0 for some x H, then set M = {x : Lx = 0}. The linearity of L implies that M is a subspace, and the continuity of L implies that the subspace M is closed. Since Lx 0 for some x H, the Corollary of the previous Theorem shows that M is not the trivial subspace of H: i.e., x = P x + Qx implies that Qx 0 since otherwise if Qx = 0, then x = P x meaning that x M, a contradiction. Hence there exists a z M with z = 1. For any such choice of z, the element belongs to M since Since z M, then (u, z) = 0. Thus for any x H, u = (Lx)z (Lz)x Lu = (Lx)(Lz) (Lz)(Lx) = 0. Lx = (Lx)(z, z) [(z, z) = 1] = ( (Lx)z, z ) = ( u + (Lz)x, z ) [u = (Lx)z (Lz)x] = (u, z) + ( (Lz)x, z ) = 0 + (Lz)(x, z) [(u, z) = 0] = (x, (Lz)z). Taking y = αz where α = Lz gives Lx = (x, y).
4 This is the only choice of y which works, because if (x, y) = (x, y ) for all x H, then 0 = (x, y y ) for all x H, implying in particular that Therefore, y y = 0. (y y, y y ) = 0.
5 Appendix Remark: The linear mappings (or operators) P : H H and Q : H H are called the orthogonal projections of H onto M and M ; they satisfy P 2 = P and Q 2 = Q. Proof of Corollary of Theorem (4.11). With M a proper subspace of H, there exists a nonzero x H \ M. If x M, then take y = x 0. If x is not orthogonal to M, then using the orthogonal projection Q from the Theorem, the element y = Qx M satisfies y M. Again by the Theorem, x = P x + Qx = P x + y; here y 0, for otherwise if y = 0, then x = P x M a contradiction. Proof of parts (c) and (d) of Theorem (4.11). Part (c). Apply the unique decomposition of Part (a) to x, y, and αx + βy for x, y H and α, β C: P (αx + βy) αp x βp y = αx + βy Q(αx + βy) α(x Qx) β(y Qy) = αx + βy Q(αx + βy) αx + αqx βy + βqy = αqx + βqy Q(αx + βy). The first term is in M and the last term is in M, meaning that both are zero. Thus P and Q are linear. Part (d). Since P x Qx, the equality x 2 = P x 2 + Qx 2 follows from Part (a) and the writing of the norms in terms of the inner products.
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