Hermitian inner products. Suppose V is vector space over C and (·; ·) is a Hermitian inner product on V . This means, by definition, that (·; ·): V × V ! C and that the following four conditions hold: (i) (v1 + v2; w) = (v1; w) + (v2; w) whenever v1; v2; w 2 V ; (ii) (cv; w) = c(v; w) whenever c 2 C and v; w 2 V ; (iii) (w; v) = (v; w) whenever v; w 2 V ; (iv) (v; v) is a positive real number for any v 2 V ∼ f0g. These conditions imply that (v) (v; w1 + w2) = (v; w1) + (v; w2) whenever v; w1; w2 2 V ; (vi) (v; cw) = c(v; w) whenever c 2 C and v; w 2 V ; (vii) (0; v) = 0 = (v; 0) for any v 2 V . In view of (iv) and (vii) we may set p jjvjj = (v; v) for v 2 V and note that (viii) jjvjj = 0 , v = 0. We call jjvjj the norm of v. Note that (ix) jjcvjj = jcjjjvjj whenever c 2 C and v 2 V . Suppose A : V × V ! R and B : V × V ! R are such that (1) (v; w) = A(v; w) + iB(v; w) whenever v; w 2 V . One easily verifies that (i) A and B are bilinear over R; (ii) A is symmetric and positive definite; (iii) B is antisymmetric; (iv) A(iv; iw) = A(v; w) whenever v; w 2 V ; 1 2 (v) B(v; w) = −A(iv; w) whenever v; w 2 V . Conversely, given A : V × V ! R which is bilinear over R and which is positive definite symmetric, letting B be as in (v) and let (·; ·) be as in (1) we find that (·; ·) is a Hermitian inner product on V . The interested reader might write down conditions on B which allow one to construct A and (·; ·) as well. Example 0.1. Let Xn n (z; w) = zjwj for z; w 2 C . j=1 The (·; ·) is easily seen to be a Hermitian inner product, called the standard (Hermitian) inner product, on Cn. Example 0.2. Suppose −∞ < a < b < 1 and H is the vector space of complex valued square integrable functions on [a; b]. You may object that I haven't told you what \square integrable" means. Now I will. Sort of. To say f :[a; b] ! R is square integrable means that f is Lebesgue measurable and that Z b jf(x)j2 dx < 1; a of course IR haven't told you what \Lebesgue measurable" means and I haven't told b you what a means, but I will in the very near future. For the time being just think of whatever notion of integration you're familiar with. Note that Z Z Z b b b f(x) dx = <f(x) dx + i =f(x) dx a a a whenever f 2 H. Let Z b (f; g) = f(x)g(x) dx whenever f; g 2 H. a You should object at this point that the integral may not exist. We will show shortly that it does. One easily verifies that (i)-(iii) of the properties of an inner product hold and that (iv) almost holds in the sense that for any f 2 F we have Z b (f; f) = jf(x)j2 dx ≥ 0 a with equality only if fx 2 [a; b]: f(x) = 0g has zero Lebesgue measure (whatever that means). In particular, if f is continuous and (f; f) = 0 then f(x) = 0 for all x 2 [a; b]. This Example is like Example One in that one can think of f 2 H as a an infinite-tuple with the continuous index x 2 [a; b]. Henceforth V is a Hermitian inner product space. The following simple Proposition is indispensable. Proposition 0.1. Suppose v; w 2 V . Then jjv + wjj2 = jjvjj2 + 2<(v; w) + jjwjj2: 3 Proof. We have jjv + wjj2 = (v + w; v + w) = (v; v) + (v; w) + (w; v) + (w; w) = (v; v) + (v; w) + (v; w) + (w; w) = jjvjj2 + 2<(v; w) + jjwjj2: □ Corollary 0.1 (The Parallelogram Law.). We have ( ) jjv + wjj2 + jjv − wjj2 = 2 jjvjj2 + jjwjj2 : Proof. Look at it. □ Here is an absolutely fundamental consequence of the Parallelogram Law. Theorem 0.1. Suppose V is complete with respect to jj · jj and C is a nonempty closed convex subset of V . Then there is a unique point c 2 C such that jjcjj ≤ jjvjj whenever v 2 C. Remark 0.1. Draw a picture. Proof. Let d = inffjjvjj : v 2 Cg and let C = fC \ B0(r): d < r < 1g: Note that C is a nonempty nested family of nonempty closed subsets of V . Suppose C 2 C, d < r < 1 and v; w 2 C. Because C is convex we have 1 2 \ 0 2 (v + w) C B (R) so 1 1 jjv + wjj2 = jj (v + w)jj2 ≥ d2: 4 2 Thus, by the Parallelogram Law, 1 1 ( ) 1 jjv − wjj2 = jjvjj2 + jjwjj2 − jjv + wjj2 ≤ r2 − d2: 4 2 4 It follows that inffdiam C \ B0(r): d < r < 1g = 0: By completeness there is a point c 2 V such that fcg = \C: □ Corollary 0.2. Suppose U is a closed linear subspace of V and v 2 V . Then there is a unique u 2 U such that jjv − ujj ≤ jjv − u0jj whenever u0 2 U. Remark 0.2. Draw a picture. Remark 0.3. We will show very shortly that any finite dimensional subspace of V is closed. 4 Proof. Let C = v − U and note that C is a nonempty closed convex subset of V . (Of course −U = U since U is a linear subspace of U, but this representation of C is more convenient for our purposes.) By virtue of the preceding Theorem there is a unique u 2 U such that jjv − ujj ≤ jjv − u0jj whenever u0 2 U. □ Theorem 0.2 (The Cauchy-Schwartz Inequality.). Suppose v; w 2 V . Then j(v; w)j ≤ jjvjjjjwjj with equality only if fv; wg is dependent. Proof. If w = 0 the assertion holds trivially so let us suppose w =6 0. For any c 2 C we have 0 ≤ jjv + cwjj2 = jjvjj2 + 2<(v; cw) + jjcwjj2 = jjvjj2 + 2<(c(v; w)) + jcj2jjwjj2: Letting (v; w) c = − jjwjj2 we find that j(v; w)j2 0 ≤ jjvjj2 − jjwjj2 with equality only if jjv + cwjj = 0 in which case v + cw = 0 so v = −cw. □ Corollary 0.3. Suppose a and b are sequences of complex numbers. Then ! ! X1 X1 1=2 X1 1=2 2 2 janbnj ≤ janj jbnj : n=0 n=0 n=0 Proof. For any nonnegative integer N apply the Cauchy-Schwartz inequality with (·; ·) equal the standard inner product on CN , v = (a0; : : : ; aN ) and w = (b0; : : : ; bN ) and then let N ! 1. □ Theorem 0.3 (The Triangle Inequality.). Suppose v; w 2 V . Then jjv + wjj ≤ jjvjj + jjwjj with equality only if either v is a nonnegative multiple of w or w is a nonnegative multiple of v. Proof. Using the Cauchy-Schwartz Inequality we find that jjv + wjj2 = jjvjj2 + 2<(v; w) + jjwjj2 ≤ jjvjj2 + 2jjvjjjjwjj + jjwjj2 = (jjvjj + jjwjj)2: Suppose equality holds. In case v = 0 then v = 0w so suppose v =6 0. Since j(v; w)j ≥ <(v; w) = jjvjjjjwjj we infer from the Cauchy-Schwartz Inequality that w = cv for some c 2 C. Thus j1 + cjjjvjj = jj(1 + c)v)jj = jjv + cwjj = jjvjj + jjcwjj = (1 + jcj)jjvjj 5 from which we infer that 1 + 2<c + jcj2 = j1 + cj2 = (1 + jcj)2 = 1 + 2jcj + jcj2 which implies that c is a nonnegative real number. □ Definition 0.1. Suppose U is a linear subspace of V . We let U ? = fv 2 V :(u; v) = 0 for all u 2 Ug and note that U ? is a linear subspace of V . It follows directly from (iv) that U \ U ? = f0g: Proposition 0.2. Suppose U is a linear subspace of V . Then U ⊂ U ?? and U ? is closed. Proof. The first assertion is an immediate consequence of the definition of U ?. The second follows because U ? is the intersection of the closed sets fv 2 V :(u; v) = 0g corresponding to u 2 U; These sets are closed because V 3 v 7! (u; v) is continuous by virtue of the Cauchy-Schwartz Inequality. □ Orthogonal projections. Henceforth U is closed linear subspace of V . Definition 0.2. Keeping in mind the foregoing, we define P : V ! U by requiring that jjv − P vjj ≤ jjv − u0jj whenever u0 2 U. That is, P v is the closest point in U to v. We call P orthogonal projection of V onto U. Note that P u = u whenever u 2 U. Thus rng P = U and P ◦ P = P: Keeping in mind that U ? is a closed linear subspace of V we let P ? be orthogonal projection of V onto U ?. Theorem 0.4. Suppose W is a linear subspace of V and Q : V ! W is such that jjw − Qvjj ≤ jjv − wjj whenever v 2 V and w 2 W . Then W is closed and Q is orthogonal projection of V onto W . 6 Proof. Supposew ~ 2 cl W and ϵ > 0. Choose w 2 W such that jjw~ − wjj ≤ ϵ. Then jjw~ − Qw~jj ≤ jjw~ − wjj ≤ ϵ. Owing to the arbitrariness of ϵ we infer that jjQw~ − wjj = 0 so w = Qw~ 2 W and cl W ⊂ W . □ Theorem 0.5. We have u = P v , v − u 2 U ? whenever u 2 U and v 2 V .