LECTURE NOTE 1: NORMED VECTOR SPACE WEI XIONG 1. Vector space 1.1. vector space. Definition 1.1. Vector space Let X be a non-empty set and K be R or C. If there exists two mappings: (x; y) 2 X × X ! (x + y) 2 X and (α; x) 2 K × X ! αx 2 X called respectively addition and scalar multiplication, that together satisfy the following properties: • x + y = y + x for all x; y; z 2 X • x + (y + z) = (x + y) + z for all x; y; z 2 X • There exists a zero vector in θ s.t. x + θ = x; 8x 2 X • Given any x 2 X, there exists an element of X denoted by (−x) s.t. x + (−x) = θ • α(x + y) = αx + αy and (α + β)x = αx + βx • α(βx) = (αβ)x and 1x = x Observation 1.2. We can define metric on any non-empty set. Thus a metric space may not be avector space. 1.2. Quasi-norm and Norm on vector space. Suppose that (X; d) is a vector space with metric d s.t. • d(x + z; y + z) = d(x; y); 8x; y; z 2 X • d(xn; x) ! 0 ) d(αxn; αx) ! 0; 8α 2 K • αn ! α ) d(αnx; αx) ! 0; 8x 2 X Definition 1.3. Quasi-norm We define kk : X ! R : kxk = d(x; θ); 8x 2 X. Then it satisfies: • kxk ≥ 0, equality holds if and only if x = θ • kx + yk ≤ kxk + kyk • kxk = k − xk • k k k k limαn!0 αnx = 0; limkxnk!0 αxn = 0 Definition 1.4. F ∗ space and F space Given a vector space X with a quasi-norm, we define convergence w.r.t. kxn − xk ! 0. Then, the space is called a F ∗ space. Further, if the space is complete, it is called a Frechet space or F space. 1 Definition 1.5. Norm Let X be a vector space. A norm on X is any mapping k:k : x ! R that satisfies • kxk ≥ 0 equality if and only if x = 0 • kαxk = kαkkxk for all x 2 X; α 2 K • kx + yk ≤ kxk + kyk for all x; y 2 X Definition 1.6. B∗ space and Banach space Normed vector space is called a B∗ space and a Banach space if it is also complete. Definition 1.7. Equivalence of norms We say that k:k2 is stronger than k:k1 if kxnk2 ! 0 ) kxnk1 ! 0. Two norms are said to be equivalent if k:k1 is stronger than k:k2 and vice versa. Theorem 1.8. k:k2 is stronger than k:k1 if and only if there exists some constant c > 0 s.t. k:k1 ≤ ck:k2; 8x 2 X. k:k1; k:k2 are equivalent if and only if there exists constants c1; c2 > 0 s.t. c1kxk1 ≤ kxk2 ≤ c2kxk1; 8x 2 X. Proof. The second part is simply the corollary of the first part. So we focus on the proof of the first partof the theorem. ): kx1k ≤ ckxk2 ! 0. (: We prove by contradiction. Suppose that we cannot find c > 0 s.t. kk1 ≤ ckk2. Then, 8n 2 N, we can xn 1 find xn 2 X s.t. kxnk1 ≥ nkxnk2. Let yn =: where kynk1 = 1 for all n and 0 ≤ kynk2 ≤ . Thus, kxnk1 n kynk2 ! 0 but kynk1 does not converges to 0 which is a contradiction. □ Let X be a normed vector space of dimension n < 1. Then we can find a base fe1; :::; eng s.t. 8x 2 X, there n exists an unique expression, x = ξ1e1 +:::+ξnen. So, we can define a one-one mapping T x : x 2 X ! ξ 2 K . P k n k n ! We can also define a function p(ξ) = j=1 ξjej : K R. Theorem 1.9. A vector space X is of dimension n < 1. Then X and Kn are algebra isometric and topological isometric. Proof. We know that X and Y are algebra isometric if and only if dimX = dimY . The second part of theorem is a corollary of next lemmas. □ P j j n j j2 1=2 Let ξ =: [ j=1 ξj ] Lemma 1.10. 1 p(ξ) is uniformly continuous w.r.t. ξ. P j j j jk n ξj k j j ξ 6 2 p(ξ) = ξ j=1 jξj ej = ξ p( jξj ); ξ = θ 2 Proof. Let ξ; η 2 Kn. jp(ξ) − p(η)j ≤ p(ξ − η) triangle=inequality Xn ≤ jξj − ηjjkejk j=1 Xn Xn 2 1=2 2 1=2 ≤ [ jξj − ηjj ] [ kejk ] j=1 j=1 Xn 2 1=2 = jξ − ηj[ kejk ] j=1 = c0jξ − ηj P j j j jk n xij k 6 □ p(ξ) = ξ j=1 jξj ej = RHS if ξ = θ n We then define an (intermediate) norm onX: kkT : x 2 X ! T x = ξ 2 K ! jT xj 2 R. Lemma 1.11. Let X be a normed vector space of dimension n. Then its norm kk is equivalent to kkT . S n Proof. Note B(0; 1) X=B(0; 1) is open.(The ball induced by metric). So S1 =: fξ 2 K : jξj = 1g is compact. Thus p(ξ) attains its maximum c2 and minimum c1, i.e., c1 ≤ p(ξ) ≤ c2; 8ξ 2 S1. According to the second part of last lemma, we know that n c1jξj ≤ p(ξ) ≤ c2jξj; 8ξ 2 K We proceed to show that c1 > 0 by contradiction. ∗ ∗ ∗ 2 ∗ ∗ ∗ Suppose that c1 = 0. Then there exists ξ = (ξ1 ; ::; ξn) S1 s.t. p(ξ ) = 0. ξ1 e1 + ::: + ξn = θ if and only if ξ∗ = 0 which is a contradiction. □ Corollary: Let X be a vector space of finite dimension n. Then, all norms defined above are equivalent. Theorem 1.12. Finite normed vector space must be complete. Any sub-space of finite dimensional normed vector space is closed. (n) Proof. Let (xn) be a Cauchy sequence in finite normed vector space X and (ξ ) be the corresponding n (n) (m) (n) sequence in K . Then jξ − ξ j = jxn − xmjT ≤ c2kxn − xmk ! 0 as m,n tend to infinity. So, (ξ ) is a Cauchy sequence in Kn and thus converges to some ξ 2 Kn. Using the same argument, we show that −1 (n) kxn − T (ξ)k ≤ c1jξ − ξj ! 0. □ 2. Approximation problem Given a set of function fϕ1; :::; ϕng, we want to approximate some function f by the linear combination of them. The performance is evaluated by some loss function, e.g., the Fourier series. 3 ∗ n Given a B space X and a finite subset fe1; :::; eng of X, we want to find (λ1; :::; λn) 2 K s.t. for a fixed x 2 X, Xn Xn kx − λjejk = min kx − ajejk a2Kn j=1 j=1 P n The element j=1 λjej is called a best approximation. We proceed to show that there exists at least one best approximation and it is unique under certain assumption. WLOG, we assume that ei are linearly independent. Theorem 2.1. Existence of the best approximation Let X be a normed vector space. Then, 8x 2 X, we can find the optimal coefficients s.t. Xn Xn kx − λjejk = min kx − ajejk a2Kn j=1 j=1 Proof. We define the functions Xn n F (a) = kx − ajejka 2 K j=1 Xn P (a) = k ajejk j=1 F (a) is continuous and F (a) ≥ P (a) − kxk; 8a 2 Kn. We can verify that P (a) is a norm on Kn and thus n there exists c1 > 0 s.t. P (a) ≥ c1jaj; 8a 2 K according to the lemma 1.11. So, F (a) ! 1 as jaj ! 1. (Note kxk is fixed). Thus we may select M > 0 s.t. F (a) ≥ M; 8jaj > N and consider F (a) on the compact set B(0;N). So, F attains its minimum. □ Theorem 2.2. Existence of the best approximation Let M be a normed vector sub-space of X and M is of finite dimension. Then, for all x 2 X, the approximation of x exists. Proof. The problem is equivalent to that find the best approximation in spanfe1; :::; eng. □ If ei are linearly dependent, then the best approximation is not unique. So we suppose that ei are linearly independent. Definition 2.3. Strictly convex A normed vector space X is said to be strictly convex if 8x; y 2 X; x =6 y, we have kxk = kyk = 1 ) kαx + βyk < 1 holds 8α; β > 0; α + β = 1 Theorem 2.4. Let X be a strictly convex normed vector space. Given a set of linearly independent vectors fe1; :::; eng of X, then, 8x 2 X, its best approximation is unique. 4 Proof. Let M = span(e1; :::; en). Suppose that there exists y; z 2 M; y =6 z and kx − zk = kx − yk = d0 = d(x; M). Then, if d0 > 0 1 1 kx − (αy + βz)k = kα(x − y) + β(x − z)k d0 d0 x − y x − z = kα( ) + β( )k < 1 d0 d0 i.e., kx − (αy + βz)k < d0 which contradicts the fact that d0 is minimum. If d0 = 0, then the best approximation is exactly x itself. □ 3. Equivalent condition of B∗ space of finite dimension 3.1. Riesz Lemma. Lemma 3.1. F.Riesz Let X be a normed vector space. Let X0 ⊂ X and X0 =6 X be a closed sub-space. Then, 80 < ϵ < 1, 9y 2 X s.t. kyk = 1 and ky − xk ≥ 1 − ϵ, 8x 2 X0.