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Introduction to Normed Vector Spaces

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Introduction to Normed Vector Spaces Audrey Terras March 29, 2009 1 Why worry about in…nite dimensional normed vector spaces? We want to understand the integral from Lang’s perspective rather than that of your calculus book. Secondly we want to understand convergence of series of functions - something that proved problematic for Cauchy in the 1800s. These things are important for many applications in physics, engineering, statistics. We will be able to study vibrating things such as violin strings, drums, buildings, bridges, spheres, planets, stock values. Quantum physics, for example, involves Hilbert space, which is a type of normed vector space with a scalar product where all Cauchy sequences of vectors converge. The theory of such normed vector spaces was created at the same time as quantum mechanics - the 1920s and 1930s. So with this chapter of Lang you are moving ahead hundreds of years from Newton and Leibnitz, perhaps 70 years from Riemann. Fourier series involve orthogonal sets of vectors in an in…nite dimensional normed vector space: C[a; b] = f :[a; b] R f continuous : f ! j g The L2 norm of a continuous function f in C[a; b] is b 1=2 f = f(x) 2 dx : k k2 0 j j 1 Za @ A This is an analog of the usual idea of length of a vector f = (f(1); :::; (f(n)) Rn : 2 1=2 n f = f(j) 2 : k k2 0 j j 1 j=1 X @ A There are other natural norms for f C[a; b] such as: 2 b f = f(x) dx: k k1 j j Za f = max f(x) : k k1 a x b j j On …nite dimensional vector spaces such as Rn it does not matter what norm you use when you are trying to …gure out whether a sequence of vectors has a limit. However, in in…nite dimensional normed vector spaces convergence can disappear if a di¤erent norm is used. Not all norms are equivalent in in…nite dimensions. See Homework 9, problem 6. Note that C[a; b] is in…nite dimensional since the set 1; x; x2; x3; :::; xn; ::: is an in…nite set of linearly independent f g n j vectors. Prove this as follows. Suppose that we have a linear dependence relation cjx = 0; for all x in [a; b]. This j=0 X implies all the constants cj = 0: Why? Extra Credit. Prove this. In…nite dimensional vector spaces are thus more interesting than …nite dimensional ones. Each (inequivalent) norm leads to a di¤erent notion of convergence of sequences of vectors. 1 2 What is a Normed Vector Space? In what follows we de…ne normed vector space by 5 axioms. We will not put arrows on our vectors. We will try to keep vectors and scalars apart by using Greek letters for scalars. Our scalars will be real. Maybe by the end of next quarter we may allow complex scalars. It simpli…es Fourier series. De…nition 1 A vector space V is a set of vectors v V which is closed under addition and closed under multiplication 2 by scalars R. This means u + v V and v V and the following 5 axioms must hold for all u; v; w V and all 2 2 2 2 ; R: VS1.2 u + (v + w) = (u + v) + w VS2. 0 V s.t. 0 + v = v 9 2 VS3. v V v0 V s.t. v + v0 = 0: VS4. v8 + 2u = u9+ v2 VS5. 1v = v; ( v) = ( )v; ( + )v = v + v; (u + v) = u + v You may say we cheated by putting 4 axioms into VS5. De…nition 2 A vector space V is a normed vector space if there is a norm function mapping V to the non-negative real numbers, written v ; for v V; and satisfying the following 3 axioms: k k 2 N1: v 0 v V and v = 0 if and only if v = 0: k k 8 2 k k N2: v = v ; v V and R: Here = absolute value of : k k j j k k 8 2 8 2 j j N3: u + v u + v ; u; v V: Triangle Inequality: k k k k k k 8 2 De…nition 3 The distance between 2 vectors u; v in a normed vector space V is de…ned by d(u; v) = u v : k k Example 1. 3-Space. x1 3 R = x2 x1; x2; x3 R : 80 1 2 9 x3 < = @ A De…ne addition and multiplication by scalars as usual: : ; x1 y1 x1 + y1 x2 + y2 = x2 + y2 : 0 1 0 1 0 1 x3 y3 x3 + y3 @ A @ x1 A @ x1 A x2 = x2 ; R: 0 1 0 1 8 2 x3 x3 @ A @ A The usual norm is x1 2 2 2 x = x + x + x ifx = x2 : k k2 1 2 3 0 1 q x3 Other norms are possible: @ A x 2 = x1 + x2 + x3 or x = max x1 ; x3 ; x3 : k k j j j j j j k k1 fj j j j j jg The proof that these de…nitions make R3 a normed vector space is tedious. No doubt we will make it a homework problem. 2 Example 2. The space of continuous functions on an interval. C[a; b] = f :[a; b] R f continuous : f ! j g For f; g C[a; b]; de…ne (f + g)(x) = f(x) + g(x) for all x [a; b] and de…ne for R ( f)(x) = f(x) for all x [a; b]: We2 leave it as an Exercise (see Homework 9) to check the2 axioms for a vector space.2 The most non-trivial one is the2 one that says f + g and f are both continuous functions on [a; b]. Again there are many possible norms. We will look at 3: b 1=2 f = f(x) 2 dx : k k2 0 j j 1 Za @ A b f = f(x) dx: k k1 j j Za f = max f(x) : k k1 a x b j j Most of the axioms for norms are easy to check. Let’sdo it for the f 1 norm. N1. v 0 v V and v = 0 if and only if v = 0. k k k k 8 2 k k N2. v = v ; v V and R . N3. Trianglek k j Inequality.j k k 8 u2+ v u8 +2 v ; u; v V: Proof of N1. k k k k k k 8 2 Since f(x) 0 for all x we know that the integral is 0; because the integral preserves inequalities (Lang, Thm. 2.1, j j b p. 104 or earlier in these notes). Suppose that f(x) dx = 0: Since f is continuous by Theorem 2.4 of Lang, p. 104 or j j Za these notes near Figure 1, this implies f(x) = 0 for all x [a; b]: 2 Proof of N2. b b b Also for any R and f C[a; b]; we have f = f(x) dx = f(x) dx = f(x) dx = f : This 2 2 k k1 j j j j j j j j j j j j k k1 Za Za Za proves N2 for norms. Here we used the multiplicative property of absolute value as well as the linearity of the integral (i.e., scalars come out of the integral lecture notes p. 80). Proof of N3. Using the de…nition of the 1-norm, and the triangle inequality for real numbers as well as the fact that the integral preserves ; we see that b b f + g = f(x) + g(x) dx ( f(x) + g(x) ) dx: k k1 j j j j j j Za Za To …nish the proof, use the linearity of the integral to see that b b b ( f(x) + g(x) ) dx = f(x) dx + g(x) dx = f + g : j j j j j j j j k k1 k k1 Za Za Za Putting it all together gives f + g f + g : k k1 k k1 k k1 3 3 Scalar Products. You have seen the dot (or scalar or inner) product in R3: It is x1 y1 x2 y2 = x1y1 + x2y2 + x3y3: 0 1 0 1 x3 y3 @ A @ A It turns out there is a similar thing for C[a; b]: First let’sde…ne the scalar product on a vector space and see how to get a norm if in addition the scalar product is positive de…nite. De…nition 4 A (positive de…nite) scalar product < v; w > for vectors v; w in a vector space V is a real number < v; w > such that the following axioms hold: SP1: < v; w >=< w; v >; v; w V (symmetry) 8 2 SP2: < u; v + w >=< u; v > + < u; w >; u; v; w V 8 2 SP3: < v; w >= < v; w >; v; w V and R 8 2 8 SP4: < v; v > 0; v V and < v; v >= 0 v = 0 (positive de…nite) 8 2 () Axioms SP1,2,3 say < v; w > is linear in each variable holding the other variable …xed. Axiom SP4 says the scalar product is positive de…nite. We will always want to assume SP4 because we want to be able to get a norm out of the scalar product. De…nition 5 If V is a vector space with a (positive de…nite) scalar product < v; w > for v; w V; de…ne the associated 2 norm by v = p< v; v >; for all v V k k 2 Before proving that this really gives a norm, let’slook at some examples.
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