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209: Honors Analysis in Rn Differentiation and Function Spaces 209: Honors Analysis in Rn Differentiation and function spaces 1 Banach Space, Hilbert Space A vector (linear) space E over a field K is a set endowed with two operations + : E × E → E and · : K × E → E that satisfy the familiar axioms [(1) : f + g = g + f], [(2) : ∃ 0, f + 0 = f], [(3) : (f + g) + h = f + (g + h)], [(4) : ∃ − f , f + (−f) = 0], [(5) : 1 · f = f], [(6) : ∀λ, µ ∈ K, (λµ) · f = λ · (µ · f)], [(6) : λ · (f + g) = λ · f + λ · g]. Definition 1 Let E be a real or complex vector space. A function k·k : E → R+ is a norm on E if kfk = 0 ⇔ f = 0 kλfk = |λ|kfk, ∀ λ ∈ K (K = R or K = C kf + gk ≤ kfk + kgk. A vector space endowed with a norm is called a normed vector space. The norm defines a topology. An set U is open if for any f ∈ U there exists r > 0 so that B(f, r) ⊂ U where B(f, r) = {g | kf − gk < r}. The norm defines a distance d(f, g) = kf − gk. Definition 2 We say that E is a Banach space if it is a normed vector space that is complete, i.e. every Cauchy sequence converges. Basic examples of Banach spaces are: Rn, C(K), Lp(dµ), for µ a complete measure on a measurable space (X, Σ), with special case `p(Z), 1 ≤ p ≤ ∞. For K ⊂ Rn a compact set C(K) = {f : K → R | f continuous} with norm kfkC = supx∈K |f(x)|. Exercise 1 C(K) is a real Banach space. Tha analogous space of complex valued functions is a complex Banach space. Lp(dµ) is a real or complex Banach space. 1 Exercise 2 Let T : X → Y be a linear operator between two normed linear spaces. Linear means that T (f + g) = T f + T g and T (λf) = λT f. The following are equivalent: 1.T is continuous. 2.T is continuous at 0. 3. ∃C ≥ 0 such that kT (x)k ≤ Ckxk ho lds ∀ x ∈ X. Exercise 3 If T : X → Y is linear and continuous then kT xk kT xk sup = sup = sup kT xk x6==0 kxk {x;kxk≤1} kxk {x;kxk=1} holds. The common value is denoted kT k. Exercise 4 If T : X → Y is linear and continuous and if S : Y → Z is linear and continuous, then ST : X → Z defined by ST (x) = S(T (x)) is linear and continuous. n m Exercise 5 Let E = R , F = R , ej = (δjk)j,k=1,...n and fα = (δαβ)α,β=1,...,m the canonical bases. Let T : E → F be a linear operator. Prove that it is continuous. Associate to T the matrix (tαi)α=1,...m,i=1,...n defined by Pm β=1 tβifβ = T (ei). Prove that the matrix associated to the composition of two operators TS is the product of the corresponding matrices. Proposition 1 Let L(E, F ) = {T : E → F | T linear and continuous}. Then T 7→ kT k is a norm on L(E, F ) and L(E, F ) is a Banach space. Theorem 1 (Hahn-Banach, real). Let T : S → R be a linear map, defined on a linear subspace of the real vector space X. Assume that there exists a real valued function p : X → R+ such that p(x + y) ≤ p(x) + p(y) and p(tx) = tp(x) for all x, y in X and all t ≥ 0. Assume that T (x) ≤ p(x) holds for all x ∈ S. Then, there exists a linear functional F : X → R that satisfies F (x) = T (x) for all x ∈ S and F (x) ≤ p(x) for all x ∈ X. Theorem 2 (Hahn-Banach, complex). Let X be a complex vector space, S a linear subspace, p a nonnegative real valued function that satsifies p(x + y) ≤ p(x)+p(y) and p(zx) = |z|p(x) for all x, y in X and all z ∈ C. Let T : S → C be a linear functional satisfying |T (x)| ≤ p(x) for all x ∈ S. Then there exists a linear functional F defined on X such that F (x) = T (x) for allx ∈ S and |F (x)| ≤ p(x) for all x ∈ X 2 Proofs: Please see Kolmogorov and Fomin, Chapter 4, section 14. Definition 3 A scalar product on a real (complex) vector space H is a bi- linear (sesquilinear) map H × H → C, denoted (f, g), that is, a map with the properties (f, g) = (g, f), (λf1 + µf2, g) = λ(f1, g) + µ(f2, g), λ µ ∈ C (f, f) ≥ 0, (f, f) = 0 ⇔ f = 0. The scalar product (hermitian product, dot product) defines a norm kfk = p(f, f). The Schwarz inequality holds |(f, g)| ≤ kfkkgk. Definition 4 We say that H is a Hilbert space if it is a real or complex vector space endowed with a scalar product such that H is complete, i.e. any Cauchy sequence converges. A basic example of a Hilbert space is L2(dµ) = {f | R |f|2dµ < ∞} where µ is a complete measure on a measurable space (X, Σ). The scalar product is Z (f, g) = f(x)g(x)dµ X In any Hilbert space we have the parallelogram identity kf + gk2 + kf − gk2 = 2(kfk2 + kgk2). A set C ⊂ H is convex if, whenever f, g ∈ C then (1 − t)f + tg ∈ C holds for all t ∈ [0, 1]. Proposition 2 Let C ⊂ H be a closed convex subset of a Hilbert space. Then the minimum norm is attained: ∃c ∈ C such that kck = inf kfk f∈C 3 Proof. Let d = inff∈C kfk. Let fn ∈ C be such that limn→∞ kfnk = d. By the parallelogram identity " 2# 2 1 2 2 fn + fn+p kfn − fn+pk = 4 kfnk + kfn+pk − 2 2 Suppose that > 0 is given, and we choose N such that, for n ≥ N we have 2 2 4 1 kfnk ≤ d + 4 . Then, because of C’s convexity, we have 2 (fn + fn+p) ∈ C, 1 2 2 hence, by the definition of d, k 2 (fn + fn+p)k ≥ d . It follows then that kfn − fn+pk ≤ holds for all n ≥ N, p ≥ 0. Thus, the sequence fn is Cauchy. Because H is Hilbert, the sequence converges, fn → c, and because C is closed and fn ∈ C, the limit c ∈ C. Because the norm is continuous we obtain kck = d. If S ⊂ H is any set, we define S⊥ = {f ∈ H | (f, s) = 0, ∀s ∈ S} Exercise 6 S⊥ is a closed, linear subspace of H. (That means that S is a closed set and that it is a vectorial subspace of H). Proposition 3 Let M ⊂ H be a closed linear sunbspace. Then for every h ∈ H there exists a unique m ∈ M such that h − m ∈ M ⊥. The identity khk2 = kmk2 + kh − mk2 holds. The map P : H → M that assigns P h = m is linear, continuous and onto M. P is idempotent, i.e. composition with itself does not change it, P 2 = P . P is selfadjoint, i.e. (P f, g) = (f, P g) holds. The map P is called the orthogonal projector onto M. The proof follows from the previous proposition by observing that h + M is closed and convex. Then, let q ∈ x + M be of minimum norm. We can write q = x − m with m ∈ M (because M is a linear space). Assume by contradiction that there exists n ∈ M such that (q, n) 6= 0. Without loss of generality we may assume that (q, n) > 0. (Indeed, this is achieved by setting n0 = zn with z = (q, n).) Then a simple calculation shows that kqk2 > kq − tnk2 can be achieved by taking 0 < t small enough. This contradicts the minimality. 4 Theorem 3 (Riesz) Let L : H → C be a continuous linear map from the Hilbert space H. Then there exists a unique f ∈ H such that Lh = (h, f) holds for all h ∈ H. Proof. Consider kerL = {h | Lh = 0} and assume it is not the entire H. (If it is then f = 0 will do the job). Then let g ∈ (kerL)⊥. The existence of g 6= 0 is guaranteed by the orthogonal projector theorem, by taking h∈ / kerL and setting g = h − P h. Then, for any element f ∈ H we note that h = L(g)f −(Lf)g ∈ kerL, and thus (h, g) = 0 holds. This implies −2 that Lf = (f, ge) holds for all f with ge = [kgk L(g)]g. A family {ea}a∈A of elements in a Hilbert space is said to be orthonormal if (ea, eb) = δab where δab = 1 if a = b and δab = 0 if a 6= b. We define for any f ∈ H, fb(a) = (f, ea). Let F ⊂ A be a finite set. Consider M to be the linear subspace generated by {ea}a∈F . Then the orthogonal projector onto M is P given by P f = a∈F fb(a)ea. Note that one can think of P f as a solution to a variational problem: Find the element m of M that best approximates f, i.e. that minimizes the distance kf − mk. An orthonormal set {ea}A is said to be complete if it is maximal (under inclusion). Theorem 4 TFAE 1. {ea}A is maximal orthonormal 2. {ea}A is orthonormal and the linear space it generates is dense. 2 P 2 3. kfk = a∈A |fb(a)| The sum in 3 is the integral with respect to counting measure H0.
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