Honours Analysis III Math 354 Prof. Dmitry Jacobson Notes Taken By: R. Gibson Fall 2010 1 Contents 1 Overview 3 1.1 p-adic Distance . 4 2 Introduction 5 2.1 Normed Linear Spaces . 5 2.2 Inner Product Spaces . 5 2.3 Metric Space Techniques . 10 3 Completeness 21 3.1 Basics & Definitions . 21 3.2 Completion & Density Revisited . 25 3.3 O.D.E. & Contraction . 27 4 Compactness 30 4.1 Basics & Definitions . 30 4.2 Arzela & Compacta . 33 5 Basic Point-Set Topology 40 5.1 Product Topology . 40 5.2 Conectedness . 42 5.3 Connected Components/ Path Components . 44 6 Banach Space Techniques 46 6.1 Linear Functionals . 46 6.2 Baire's Category Theorem, Banach-Steinhaus Theorem & The Open Mapping Theorem . 48 6.3 Convex Sets . 53 6.4 Kernel . 57 6.5 Conjugate Space . 58 6.6 Linear Functionals Revisited . 60 6.7 Bernstein Polynomials . 63 6.8 Inverse & Implicit Function Theorem In Rn ............................ 65 7 Linear Operators & The Operator Norm 68 7.1 The Hahn-Banach Theorem . 68 7.2 Examples . 70 2 Chapter 1 Overview Definition 1. Let X be a metric space. We define Distance d : X × X ! R to satisfy (i) 8x 2 X; d(x; x) = 0 (ii) 8x 6= y 2 X; d(x; y) > 0 (iii) 8x; y 2 X; d(x; y) = d(y; x) (iv) 8x; y; z 2 X; d(x; z) + d(z; y) ≥ d(x; y) Example 1. The following are some of the main examples of metric spaces for this course. • Spaces of sequences (finite or infinite) - with lp norm. • 1 X p f(x1; : : : ; xn);:::): jxkj < 1g k=1 This leads us to a proposition. Proposition 1. If x = (x1; x2;:::); y = (y1; y2;:::), then 1 !1=p X p d(x; y) = jxk − ykj k=1 is a distance. Example 2. If x = 0 and y = (1; 1=2; 1=3; 1=4;:::), then can we prove that r r 1 1 1 π2 d(x; y) = 1 + + + ··· + + ··· = 4 9 h2 6 ...probably not. Example 3. For p ≥ 1, an example would be continuous functions f; g on [a; b]. Proposition 2. In the space of continuous functions on [a; b], 1=p "Z b # d(f; g) = jf(x) − g(x)jpdx a defines a distance. We recall the Lp norm to be 1=p Z b ! p jjfjjp = jf(x)j dx a 3 We now examine polygons in the plane. There is a problem in the assignment which asks if the difference of areas defines a distance. So we must prove that d(P1;P2) = Area(P1 4 P2) which is the area of (P1 [ P2)n(P1 \ P2) Section 1.1 p-adic Distance Let p 2 f2; 3; 5; 7; 11; 13;:::g = P rimes. Let x; y 2 Q so a a x = 1 y = 2 b1 b2 Definition 2. The p-adic Norm is defined as follows c x = pk · d where k 2 Z and −k jjxjjp = p Definition 3. The p-adic Norm is defined to be dp(x; y) = jjx − yjjp Example 4. Working with p-adic distance. Suppose that 24 x = 49 then we have that 24 x = 7−2 =) jjxjj = 72 1 7 8 x = 31 =) jjxjj = 3−1 49 3 3 x = 23 =) jjxjj = 2−3 49 2 Proposition 3. jjx − yjjp defines a p-adic distance on Q. NB: If x is as above, then jjxjjEucl · jjxjj2 · jjxjj3 · jjxjj7 = 1 Neat, eh? 4 Chapter 2 Introduction Section 2.1 Normed Linear Spaces Linear means exactly what you would think it means. A good way to show this is 1 1 1 (an)n=1 + (bn)n=1 = (an + bn)n=1 (f + g)(x) = f(x) + g(x) 8 x 2 [a; b] Definition 4. Suppose that X is a linear space, then we say that the Norm of x 2 X is a map jj · jj : X ! R+ such that (i) jjxjj = 0 () x = 0 (ii) jjt · xjj = jtj · jjxjj (iii) jjx + yjj ≤ jjxjj + jjyjj It is a fact that d(x; y) = jjx − yjj defines a distance since d(y; x) = jjy − xjj = jj − (x − y)jj = jjx − yjj = d(x; y) d(x; z) = jjx − zjj = jj(x − y) + (y − z)jj ≤ jjx − yjj + jjy − zjj = d(x; y) + d(y; z) Section 2.2 Inner Product Spaces An Inner Product Space is a space together with a map (·; ·): X × X ! R such that (i) 0 ≤ (x; x) (ii) (x; y) = 0 () x = 0 (iii) (x; αy + βz) = α(x; y) + β(x; z); x; y; z 2 X; α; β 2 R (iv) (γx + δy; z) = γ(x; z) + δ(y; z) Example 5. The following are some examples of inner products 5 • Dot Product (x; y) = x · y = x1y1 + ··· + xnyn • Function Space Inner Product Z b f; g 2 C([a; b]) =) (f; g) = f(x)g(x)dx a and s Z b jjfjj = [f(x)]2dx a Proposition 4. jjxjj := p(x; x) always defines a norm. Proof. We will need a lemma. Lemma 5. The following identity holds. (x; y) ≤ jjxjj · jjyjj Proof. (x + ty; x + ty) = (x; x) + 2t(x; y) + t2(y; y) we know that jjx + tyjj2 ≥ 0 now, D = 4(x; y)2 − 4(x; x) · (y; y) ≤ 0 (x; y) ≤ p(x; x) · p(y; y) = jjxjj · jjyjj With this lemma, we can now prove the above proposition. So Proof. The triangle inequality states that jjx + yjj ≤ jjxjj + jjyjj and now we can square both sides to obtain (jjx + yjj)2 ≤ (jjxjj + jjyjj)2 and now (x + y; x + y) = (x; x) + 2(x; y) + (y; y) and jjxjj2 + 2jjxjj · jjyjj + jjyjj2 = (x; x) + 2jjxjj · jjyjj + (y; y) by the lemma, and we cancel to obtain =) (x; y) ≤ jjxjj · jjyjj which yields our result. How to guess whether jjxjj comes from (x; x) or not? The answer is (x + y; x + y) + (x − y; x − y) = (x; x) + 2(x; y) + (y; y) + (x; x) − 2(x; y) + (y; y) 6 then (∗) jjx + yjj2 + jjx − yjj2 = 2jjxjj2 + 2jjyjj2 The norm comes from (·; ·) () (∗) holds for all x; y 2 X. Proposition 6. Let (a1; : : : ; an;:::) be such that X 2 ai < 1 and let (b1; : : : ; bn;:::) be such that X 2 bj < 1 and let 1 !1=2 X 2 jjajj = ai i=1 Then 1 X aibi ≤ jjajj · jjbjj i=1 and so v v u n u n uX 2 uX 2 (a; b) = a1b1 + ··· + anbn ≤ t ai · t bj ! jjajj · jjbjj i=1 j=1 as n ! 1. Proof. To prove convergence, v n+m s un+m X X 2 u X 2 aibi ≤ ak · t bk k=n k=n k=n the two components on the right converge separately to 0 as n ! 1. This is because 1 X 2 ak < 1 k=1 2 and the same goes for the sum of bk. Thus, the whole thing converges to 0. 2 2 Summarizing, we proved that (C([a; b]);L -norm) is a metric space, and so is l : Next, we look at Lp-norm which is (1=p) "Z b # p jjfjjp = jf(x)j dx a and at lp-space, p 6= 2, we have 1 !(1=p) X p jjajjlp = jakj k=1 Thus, lp is the space of all a such that 1 X p jakj < 1 k=1 To prove this we need a lemma. 7 Lemma 7. Let a > 0; b > 0, and p; q ≥ 1 and such that 1 1 + = 1 p q NB: We say that p; q are Conjugate Exponents. Then ap bq (∗) a · b ≤ + p q Proof. Let x > 0, then f(x) = log (x) which is a concave function. 1 1 f 0(x) = f 00(x) = − < 0 x x2 and since f is concave, we can say that f(αx + βy) ≥ αf(x) + βf(y) where α + β = 1. Also, we know that log (a · b) = log a + log b 1 1 = log (ap) + log (bq) p q 1 1 (∗∗) ≤ log ap + bq p q Theorem 8 (Holder's Inequality). Let p < 1 n n !1=p n !1=q X X p X q jakbkj ≤ jakj jbkj k=1 k=1 k=1 where 1 1 + = 1 p q Remark. Both sides are homogeneous of degree 1 in (ak); (bk). So WLOG, we can rescale ak − s so that n X p jakj = 1 k=1 and n X q jbkj = 1 k=1 so we say that n n p q X X jakj jbkj ja b j ≤ + k k p q k=1 k=1 n ! n ! 1 X 1 X = ja jp + jb jq p k q k k=1 k=1 1 1 = + p q = 1 8 by our normalization of the sums. Now, 1 n !1=p n ! q X p X q 1=p 1=q RHS = jakj jbkj = 1 1 = 1 k=1 k=1 and so LHS ≤ RHS Now, if we let n ! 1, then we get Holder for infinite series. The proof is left to the reader as an excercise. Theorem 9 (Minkowski Inequality). (For Sequences With n < 1) Recall that for x = (x1; : : : ; xn), we have n !1=p X p jjxjjp = jxkj k=1 We want jjx + yjjp ≤ jjxjjp + jjyjjp So, we know that n p X p jjx + yjjp = jxk + ykj k=1 n X p ≤ (jxkj + jykj) k=1 n X p−1 p−1 = [(jxkj + jykj) jxjk + (jxkj + jykj) jyjk] k=1 and we apply Holder to each inner sum and let bk = (jxkj + jykj); ak = jxkj 1=q 1=p n ! X n p X p−1 q ≤ k = 1 jxkj [(jxkj + jykj) ] k=1 1=q 1=p n ! X n p X p−1 q + k = 1 jykj [(jxkj + jykj) ] k=1 And now, 1 1 1 1 p − 1 1 + = 1 =) 1 − = =) = p q p q p q =) (p − 1)q = p and so 1=p 1=p 1=q X n p X n p hX n qi RHS = k = 1 jxkj + k = 1 jykj · k = 1 (jxkj + jykj) and n X p RHS ≤ LHS = (jxkj + jykj) k=1 " n #1−1=q X p =) (jxkj + jykj) ≤ jjxjjp + jjyjjp k=1 9 and we can conclude that jjx + yjjp ≤ jjxjjp + jjyjjp Example 6.