Several Important Theorems by Francis J. Narcowich November, 2014 1 The Projection Theorem Let H be a Hilbert space. When V is a finite dimensional subspace of H and f 2 H, we can always find a unique p 2 V such that kf − pk = minv2V kf − vk. This fact is the foundation of least-squares approximation. What happens when we allow V to be infinite dimensional? We will see that the minimization problem can be solved if and only if V is closed. Theorem 1.1 (The Projection Theorem). Let H be a Hilbert space and let V be a subspace of H. For every f 2 H there is a unique p 2 V such that kf − pk = minv2V kf − vk if and only if V is a closed subspace of H. To prove this, we need the following lemma. Lemma 1.2 (Polarization Identity). Let H be a Hilbert space. For every pair f; g 2 H, we have kf + gk2 + kf − gk2 = 2(kfk2 + kgk2): Proof. Adding the ± identities kf ±gk2 = kfk2 ±hf; gi±hg; fi+kgk2 yields the result. The polarization identity is an easy consequence of having an inner prod- uct. It is surprising that if a norm satisfies the polarization identity, then the norm comes from an inner product1. Proof. (Projection Theorem) Showing that the existence of minimizer im- plies that V is closed is left as an exercise. So we assume that V is closed. For f 2 H, let α := infv2V kv − fk. It is a little easier to work with this in 2 2 an equivalent form, α = infv2V kv − fk . Thus, for every " > 0 there is a 2 2 2 v" 2 V such that α ≤ kv" − fk < α + ". By choosing " = 1=n, where n 1 is a positive integer, we can find a sequence fvngn=1 in V such that 1 0 ≤ kv − fk2 − α2 < (1.1) n n 1Jordan, P. ; Von Neumann, J. On inner products in linear, metric spaces. Ann. of Math. (2) 36 (1935), no. 3, 719{723. 1 Of course, the same inequality holds for a possibly different integer m, 0 ≤ 2 2 1 kvm − fk − α < m . Adding the two yields this: 1 1 0 ≤ kv − fk2 + kv − fk2 − 2α2 < + : (1.2) n m n m By polarization identity and a simple manipulation, we have v + v kv − v k2 + 4kf − n m k2 = 2kf − v k2 + kf − v k2: n m 2 n m We can subtract 4α2 from both sides and use (1.2) to get v + v 2 2 kv −v k2+4(kf− n m k2−α2) = 2(kf−v k2+kf−v k2−2α2) < + : n m 2 n m n m 1 vn+vm 2 2 2 Because 2 (vn + vm) 2 V , kf − 2 k ≥ infv2V kv − fk = α . It follows that the second term on the left is nonnegative. Dropping it makes the left side smaller: 2 2 kv − v k2 < + : (1.3) n m n m 1 As n; m ! 1, we see that kvn − vmk ! 0. Thus fvngn=1 is a Cauchy sequence in H and is therefore convergent to a vector p 2 H. Since V is closed, p 2 V . Furthermore, taking limits in (1.1) implies that kp − fk = infv2V kv − fk. The uniqueness of p is left as an exercise. There are two important corollaries to this theorem; they follow from problem 3 of assignment 1 and Theorem 1.1.. We list them below. Corollary 1.3. Let V be a subspace of H. There exists an orthogonal projection P : H! V for which kf − P fk = minv2V kf − vk if and only if V is closed. Corollary 1.4. Let V be a closed subspace of H. Then, H = V ⊕ V ? and (V ?)? = V . 2 The Riesz Representation Theorem Let V be a Banach space. A bounded linear transformation Φ that maps V into R or C is called a linear functional on V . The linear functionals form a Banach space V ∗, called the dual space of V , with norm defined by jΦ(v)j kΦkV ∗ := sup : v6=0 kvkV 2 2.1 The linear functionals on Hilbert space Theorem 2.1 (The Riesz Representation Theorem). Let H be a Hilbert space and let Φ: H! C (or R) be a bounded linear functional on H. Then, there is a unique g 2 H such that, for all f 2 H, Φ(f) = hf; gi. Proof. The functional Φ is a bounded operator that maps H into the scalars. It follows from our discussion of bounded operators that the null space of Φ, N(Φ), is closed. If N(Φ) = H, then Φ(f) = 0 for all f 2 H, hence Φ = 0. Thus we may take g = 0. If N(Φ) 6= H, then, since N(Φ) is closed, we have that H = N(Φ) ⊕ N(Φ)?. In addition, since N(Φ) 6= H, there exists a nonzero vector h 2 N(Φ)?. Moreover, Φ(h) 6= 0, because h is not in the null space N(Φ). Next, note that for f 2 H, the vector w := Φ(h)f − Φ(f)h is in N(Φ). To see this, observe that Φ(w) = ΦΦ(h)f − Φ(f)h = Φ(h)Φ(f) − Φ(f)Φ(h) = 0: Because w = Φ(h)f − Φ(f)h 2 N(Φ), it is orthogonal to h 2 N(Φ)?, we have that 0 = hΦ(h)f − Φ(f)h; hi = Φ(h)hf; hi − Φ(f) hh; hi : | {z } khk2 Φ(h) Solving this equation for Φ(f) yields Φ(f) = hf; khk2 hi. The vector g := Φ(h) khk2 h then satisfies Φ(f) = hf; gi. To show uniqueness, suppose g1; g2 2 H satisfy Φ(f) = hf; g1i and Φ(f) = hf; g2i. Subtracting these two gives hf; g2 −g1i = 0 for all f 2 H. Letting f = g2 −g1 results in hg2 −g1; g2 −g1i = 0. Consequently, g2 = g1. 2.2 Adjoints of bounded linear operators We now turn the problem of showing that an adjoint for a bounded operator always exists. This is just a corollary of the Riesz Representation Theorem. Corollary 2.2. Let L : H!H be a bounded linear operator. Then there exists a bounded linear operator L∗ : H!H, called the adjoint of L, such that hLf; hi = hf; L∗hi, for all f; h 2 H. Proof. Fix h 2 H and define the linear functional Φh(f) = hLf; hi. Us- ing the boundedness of L and Schwarz's inequality, we have jΦh(f)j ≤ kLkkfkkhk = Kkfk, and so Φh is bounded. By Theorem 2.1, there is a 3 unique vector g in H for which Φh(f) = hf; gi. The vector g is uniquely determined by Φh; thus g = gh a function of h. We claim that gh is a linear function of h. Consider h = ah1 +bh2. Note that Φh(f) = hLf; ah1 +bh2i = ¯ a¯Φh1 (f) + bΦh2 (f). Since Φh1 (f) = hf; g1i and Φh2 (f) = hf; g2i, we see that ¯ Φh(f) = hf; ghi =a ¯Φh2 (f) + bΦh2 (f) = hf; agh1 + bgh2 i: It follows that gh = agh1 + bgh2 and that gh is a linear function of h. It is 2 also bounded. If f = gh, then Φh(gh) = kghk . From the bound jΦh(f)j ≤ 2 kLkkfkkhk, we have kghk ≤ jLkkghkkhk. Dividing by kghk then yields kghk ≤ kLkkhk. Thus the correspondence h ! gh is a bounded linear ∗ function on H. Denote this function by L . Since hLf; hi = hf; ghi, we have that hLf; hi = hf; L∗hi. Corollary 2.3. kL∗k = kLk. Proof. By problem 5 in Assignment 8, kLk = supf;h jhLf; hij, where khk = ∗ ∗ ∗ kfk = 1. On the other hand, kL k = supf;h jhL h; fij. Since hL h; fi = ∗ ∗ hf; L hi, we have that supf;h jhL h; fij = supf;h jhLf; hij. It immediately follows that kL∗k = kLk. Example 2.4. Let R = [0; 1]×[0; 1] and suppose that k is a Hilbert-Schmidt R 1 ∗ R 1 kernel. If Lu(x) = 0 k(x; y)u(y)dy, then L v(x) = 0 k(y; x)v(y)dy. Proof. We will use s; t as the integration variables and switch back, to avoid R 1 R 1 confusion. We begin with hLu; vi = 0 0 k(s; t)u(t)dt v(s)ds. By Fubini's theorem, we may switch the variables of integration to get this: Z 1 Z 1 Z 1 Z 1 k(s; t)u(t)dy v(s)ds = k(s; t)v(s)ds u(t)dt 0 0 0 0 Z 1 Z 1 = k(s; t)v(s)ds u(t)dt: 0 0 | {z } L∗v =hu; L∗vi The result follows by changing variables from t; s to x; y in the second equa- tion above . 3 The Fredholm Alternative Theorem 3.1 (The Fredholm Alternative). Let L : H!H be a bounded linear operator whose range, R(L), is closed. Then, the equation Lf = g 4 and be solved if and only if hg; vi = 0 for all v 2 N(L∗). Equivalently, R(L) = N(L∗)?. Proof. Let g 2 R(L), so that there is an h 2 H such that g = Lh. If v 2 N(L∗), then hg; vi = hLh; vi = hh; L∗vi = 0. Consequently, R(L) ⊆ N(L∗)?. Let f 2 N(L∗)?. Since R(L) is closed, the projection theorem, Theorem 1.1, and Corollary 1.3, imply that there exists an orthogonal pro- jection P onto R(L) such that P f 2 R(L) and f 0 = f −P f 2 R(L)?.