Analysis III Lecturer: Prof. Oleg Zaboronski, Typesetting: David Williams Term 1, 2015 Contents I Integration 2 1 The Integral for Step Functions 2 1.1 Definition of the Integral for Step Functions . 2 1.2 Properties of the Integral for Step Functions . 4 2 The Integral for Regulated Functions 6 2.1 Definition of the Integral for Regulated Functions . 6 2.2 Properties of the Integral for Regulated Functions . 8 3 The Indefinite Integral & Fundamental Theorem of Calculus 10 3.1 The Indefinite Integral . 10 3.2 The Fundamental Theorem of Calculus . 11 4 Practical Methods of Integration 12 4.1 Integration by Parts . 12 4.2 Integration by Substitution . 13 4.3 Constructing Convergent Sequences of Step Functions . 14 5 The Characterisation of Regulated Functions 15 5.1 Characterisation & Proof . 15 6 Improper & Riemann Integrals 17 6.1 Improper Integrals . 17 6.2 The Riemann Integral . 19 7 Uniform & Pointwise Convergence 21 7.1 Definitions, Properties and Examples . 21 7.2 Uniform Convergence & Integration . 25 7.3 Uniform Convergence & Differentiation . 27 8 Functional Series 28 8.1 The Weierstrass M-Test & Other Useful Results . 28 8.2 Taylor & Fourier Series . 29 II Norms 33 9 Normed Vector Spaces 33 9.1 Definition & Basic Properties of a Norm . 33 9.2 Equivalence & Banach Spaces . 35 10 Continuous Maps 40 10.1 Definition & Basic Properties of Linear Maps . 40 10.2 Spaces of Bounded Linear Maps . 42 11 Open and Closed Sets of Normed Spaces 43 11.1 Definition & Basic Properties of Open Sets . 43 11.2 Continuity, Closed Sets & Convergence . 44 11.3 Lipschitz Continuity, Contraction Mappings & Other Remarks . 45 12 The Contraction Mapping Theorem & Applications 46 12.1 Statement and Proof . 46 12.2 The Existence and Uniqueness of Solutions to ODEs . 47 12.3 The Jacobi Algorithm . 50 12.4 The Newton-Raphson Algorithm . 51 1 Part I Integration 1 The Integral for Step Functions 1.1 Definition of the Integral for Step Functions Motivation: \Integrals = Areas Under Curves" Consider constant functions of the form fc : x 7! f0 2 R for x 2 [a; b]. Let D be the area under the line at f0 from a to b. Then it makes sense to define Z b . fc(x) dx .= sign(f0) · Area(D) a = sign(f0) · (jf0j · (b − a)) = f0 · (b − a) Now this definition will be extended to piecewise constant or step functions. Definition 1: ' : [a; b] ! R is called a step function if there exists a finite set P ⊂ [a; b] with a; b 2 P such that ' is constant on the open sub-intervals of [a; b] n P . P is called a partition of [a; b]. More concretely, P = fa = p0; p1; p2; : : : ; pk−1; pk = bg, with p0 < p1 < p2 < : : : < pk−1 < pk, and ' = ci (an arbitrary constant), for i = 1; 2; : : : ; k − 1; k. (pi−1;pi) Notes: If φ is a step function then there are many partitions of [a; b] such that φ is constant on the open sub-intervals defined by the partition, call one P . There exists at least one other point greater than a in P (since both a and b belong to P by definition), call the first p. Then φ is also constant on the open intervals of [a; b] n (P [ fng), where a < n < p. Let Q be a partition of [a; b], with Q ⊃ P . Then Q is called a refinement of P (so (P [ fng) is a refinement of P ). The minimal common refinement of P and Q is denoted by P [ Q. If φ is a step function on [a; b], then any partition P = fa = p0; p1; p2; : : : ; pk−1; pk = bg, with p0 < p1 < p2 < : : : < pk−1 < pk, such that φ = ci (an arbitrary constant), for i = 1; 2; : : : ; k − 1; k, is called (pi−1;pi) compatible with, or adapted to, φ. Proposition 1: Let S[a; b] be the space of step functions on [a; b]. If f; g 2 S[a; b], then 8µ, λ 2 R, (λf + µg) 2 S[a; b] (so S[a; b] is closed under taking finite linear combinations of its elements). Proof: f 2 S[a; b], so 9P = fa = p0; p1; p2; : : : ; pk−1; pk = bg, with p0 < p1 < p2 < : : : < pk−1 < pk, which is compatible with f, and g 2 S[a; b] so 9Q = fa = q0; q1; q2; : : : ; ql−1; ql = bg, with q0 < q1 < q2 < : : : < ql−1 < ql, which is compatible with g. Consider P [ Q = fa = r0; r1; r2; : : : ; rm−1; rm = bg, with r0 < r1 < r2 < : : : < rm−1 < rm. Then this partition is compatible with (λf + µg). It remains to show that (λf + µg) = ci (an arbitrary constant), for i = 1; 2; : : : ; m − 1; m. (ri−1;ri) P ⊂ (P [ Q), so (ri−1; ri) ⊂ (pa−1; pa) for some a, and Q ⊂ (P [ Q), so (ri−1; ri) ⊂ (qb−1; qb) for some b. Then, since f is constant, f is constant, and since g is constant, (p ;p ) (r ;r ) (q ;q ) a−1 a i−1 i b−1 b g is constant, so (λf + µg) is constant. Since the choice of i was arbitrary, this proves (ri−1;ri) (ri−1;ri) the proposition. 2 Notes: This demonstrates that S[a; b] is a vector space over R. Define C[a; b] as the space of continuous functions on [a; b], and define B[a; b] as the space of bounded functions on [a; b]. These are also vector spaces. It was demonstrated in Analysis II that C[a; b] ⊂ B[a; b], and by definition S[a; b] ⊂ B[a; b]. k Definition 2: For φ 2 S[a; b], let P = fpigi=0 be a partition compatible with φ. Define (temporarily) b . R φi = φ . Then a : S[a; b] ! R is defined as (pi−1;pi) Z b k . X φ .= φi(pi − pi−1) a i=1 R b Remark: If φ ≡ φ0 2 R, then P = fa; bg, so a φ = φ0(b − a), which is precisely the area described in the motivation. R b Lemma 2: The value of a φ does not depend on the choice of partition compatible with [a; b]. Proof: For any φ 2 S[a; b], let P and Q be partitions compatible with [a; b]. Suppose there are k points R b in Q that are not in P . Denote them n1; n2; : : : ; nk. Temporarily let φ be the integral of φ from a a P to b defined by the partition P which is compatible with [a; b]. Now, consider P [ fn1g. Since Q is compatible with [a; b], we know that n1 2 [a; b], so 9i 2 N such that n1 2 (pi−1; pi) for some pi−1; pi 2 P . φ is constant on (pi−1; pi), and so is also constant on (pi−1; n1) and (n1; pi), so P [ fn1g is compatible with φ. By definition Z b Z b Z b Z b φ − φ = φi(n1 − pi−1) + φi(pi − n1) − φi(pi − pi−1) = 0, so φ = φ a P [fn1g a P a P [fn1g a P The proof is completed by induction, since there are a finite number of points. This is left as an exercise. 3 1.2 Properties of the Integral for Step Functions 1. Proposition 3 - Additivity: For any φ 2 S[a; b], 8v 2 (a; b), Z b Z v Z b φ = φ + φ a a v 2. Proposition 4 - Linearity: For any φ, 2 S[a; b], 8λ, µ 2 R, Z b Z b Z b λφ + µψ = λ φ + µ a a a 3. Proposition 5 - The Fundamental Theorem of Calculus: Let φ 2 S[a; b], and let P be a R x partition of [a; b] compatible with φ. Consider I : [a; b] ! R, I : x 7! a φ. Then (I) I is continuous on [a; b]. Sk Sk 0 (II) I is differentiable on i=1(pi−1; pi), and, 8x 2 i=1(pi−1; pi), I (x) = φ(x). Proof: k 1. Let P = fpigi=0 be a partition compatible with φ 2 S[a; b]. Then there are two cases: (i) If v 2 P , then by definition k v k Z b X X X Z v Z b φ = φi(pi − pi−1) = φi(pi − pi−1) + φi(pi − pi−1) = φ + φ a i=1 i=1 i=v+1 a v (ii) If v2 = P , then 9i 2 N, 1 ≤ i ≤ k, such that v 2 (pi−1; pi). Then fa = p0; : : : ; pi−1; vg, fv; pi; : : : ; pk = bg are partitions compatible with φ, so φ 2 S[a; v], φ 2 S[v; b], so (a;v) (v;b) their integrals are well-defined on those intervals. Then k Z b X φ = φj(pj − pj−1) a j=1 = φ1(p1 − p0) + φ2(p2 − p1) + ::: + φi(pi − pi−1) + ::: + φk(pk − pk−1) = φ1(p1 − p0) + ::: + φi(pi + v − v − pi−1) + ::: + φk(pk − pk−1) = φ1(p1 − p0) + ::: + φi(v − pi−1) + φi(pi − v) + ::: + φk(pk − pk−1) 2i−1 # " k 3 X X = 4 φj(pj − pj−1) + φi(v − pi−1) + φi(pi − v) + φj(pj − pj−1)5 j=1 j=i+1 Z v Z b = φ + φ a v 2. Let P be a partition compatible with φ 2 S[a; b], and let Q be a partition compatible with 2 S[a; b]. Then, as was proved in the previous section, P [ Q is compatible with both φ and .