Piecewise Continuous φ


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1 Piecewise Continuous φ φ is piecewise continuous on [, b] if nd only if b in R nd φ : [, b] C There is finite set S [, b] such tht, for ll t [, b] S, φ is continuous t t: φ(t) = lim φ(u) u t u [,b] For ll t [, b), φ hs finite right limit t t. For ll t (, b], φ hs finite left limit t t. Rmsey Complex Integrtion
2 Piecewise Continuous, Slide 2 Lemm: φ is piecewise continuous on [, b] if nd only if R(φ) nd I(φ) re piecewise continuous on [, b]. Theorem from Advnced Clculus: f : [, b] R is Riemnn integrble if nd only if f is bounded nd the set of discontinuities of f hs outer mesure 0. Specil Problem (Chi): Let f be relvlued nd piecewise continuous on [, b]. Then 1 f is bounded 2 f is Riemnn integrble Rmsey Complex Integrtion
3 Piecewise Continuous, Slide 3 Specil Problem (Kodgis): Assume tht φ is piecewise continuous on [, b]. g : E C with E C g is continuous on n open set U E There is closed set F such tht rnge(φ) F U. Then g φ is piecewise continuous on [, b]. Exmple: Note tht the bsolute vlue function is continuous on ll of C. Consequently, if φ is piecewise continuous, so is φ. This is used implicitly in Section 3 of Chpter 6 on pge 66. Rmsey Complex Integrtion
4 Piecewise Continuous, Slide 4 Specil Problem (Chevlier): Suppose tht φ is piecewise continuous on [, b]. Suppose tht τ : [c, d] [, b] is onetoone nd continuous. Then φ τ is piecewise continuous on [c, d]. Hint: You cn prove tht τ is strictly incresing or strictly decresing. Tht leds to two cses. Rmsey Complex Integrtion
5 Integrls of piecewise continuous φ Def n: If φ is piecewise continuous on [, b] we define φ(t) dt by φ(t) dt := R(φ(t)) dt + i I(φ(t)) dt Specil Problem (Reckwerdt): Let φ nd λ be piecewise continuous on [, b]. Then φ + λ is piecewise continuous on [, b], nd (φ + λ)(t) dt = φ(t) dt + λ(t) dt Rmsey Complex Integrtion
6 Integrls of piecewise continuous φ, Slide 2 Specil Problem (Felicino): If φ is piecewise continuous on [, b] nd c is complex number, then cφ is piecewise continuous on [, b] nd (cφ)(t) dt = c φ(t) dt Theorem, pge 65: If φ 1 nd φ 2 re piecewise continuous on [, b], nd c 1 nd c 2 re complex numbers, then c 1 φ 1 + c 2 φ 2 is piecewise continuous on [, b] nd [c 1 φ 1 + c 2 φ 2 ](t) dt = c 1 φ 1 (t) dt + c 2 φ 2 (t) dt Rmsey Complex Integrtion
7 Integrls of piecewise continuous φ, Slide 3 Specil Problem (Rder) Suppose tht φ is piecewise continuous on [, b]. Suppose tht [c, d] [, b] with c < d. Then d d δ φ(t) dt = lim φ(t) dt c δ 0 c+δ Proof Hints: You my use for free tht, when 0 < δ < (d c)/2, c+δ c d δ d d φ(t) dt + φ(t) dt + φ(t) dt = φ(t) dt c+δ d δ c Somehow use the boundedness of φ to finish the rgument. Rmsey Complex Integrtion
8 Integrls of piecewise continuous φ, Silde 4 The previous problem is vlid more generlly. It remins vlid if R(φ) nd I(φ) re Riemnn integrble on [, b] R(φ) nd I(φ) re Lebesgue mesurble on [, b] with R(φ(t)) dt < nd I(φ(t)) dt < Rmsey Complex Integrtion
9 Integrls of piecewise continuous φ, Slide 5 Specil Problem (Hedley): Let φ nd ψ be piecewise continuous on [, b]. Suppose there is finite set W such tht, for ll t [, b] W, φ(t) = ψ(t) Then φ(t) dt = ψ(t) dt Rmsey Complex Integrtion
10 Integrls of piecewise continuous φ, Slide 6 Lemm: Let φ nd ψ be piecewise continuous on [, b]. Then φ ψ is piecewise continuous on [, b]. Lemm: Let φ : [, b] C. Let < c < b. Set λ equl to the restriction of φ to [, c] nd ψ equl to the restriction of φ to [c, b]. Then φ is piecewise continuous on [, b] if nd only if λ is piecewise continuous on [, c] nd ψ is piecewise continuous on [c, b]. If φ is piecewise continuous on [, b], then φ(t) dt = c λ(t) dt + ψ(t) dt c Rmsey Complex Integrtion
11 Differentiting φ : [, b] C Def n: Let < b nd φ : [, b] C. We sy tht φ is differentible t t 0 if nd only if t 0 [, b] nd there is complex number L such tht φ(u) φ(t 0 ) L = lim u t0 u t 0 u [,b] L is necessrily unique, nd we cll it φ (t 0 ). Specil Problem (Billington): Let < b nd φ : [, b] C. Then 1 φ is differentible t t 0 if nd only if R(φ) nd I(φ) re differentible t t 0. 2 If φ is differentible t t 0 then φ (t 0 ) = [R(φ)] (t 0 ) + i [I(φ)] (t 0 ) Note tht this problem sys tht, when φ (t) exists, [R(φ)] (t) = R[φ (t)] nd [I(φ)] (t) = I[φ (t)]. Rmsey Complex Integrtion
12 Piecewise C 1 Def n: Let < b nd φ : [, b] C. φ is piecewise C 1 if nd only if φ is continuous on [, b] There is finite set S such tht, for ll t [, b] S φ is differentible t t. For ll t [, b], if φ is differentible t t then φ is continuous t t. For ll t [, b), φ hs finite right limit t t. For ll t (, b], φ hs finite left limit t t. Rmsey Complex Integrtion
13 Piecewise C 1, Slide 2 Specil Problem (Holmes): Let < b, φ : [, b] : C, nd φ piecewise C 1. Suppose tht g : E C, with E C. Let F G E such tht F is closed nd G is open The rnge of φ is subset of F g is holomorphic on G g is C 1 on G (mening, if g = u + iv with u nd v rel, u x (z), u y (z), v x (z) nd v y (z) re continuous t every z G). Then g φ is piecewise C 1. Rmsey Complex Integrtion
14 Piecewise C 1, Slide 3 Specil Problem (Thompson): Let < b, φ : [, b] C, nd φ piecewise C 1. Let τ : [c, d] [, b] be onetoone with continuous derivtive (C 1 ). Then φ τ is piecewise C 1. Hint: Prove tht τ must be strictly decresing or strictly incresing. Tht leds to two cses. Rmsey Complex Integrtion
15 Piecewise C 1, Slide 4 Lemm: Let < b, φ : [, b] C. Then φ is piecewise C 1 if nd only if R(φ) nd I(φ) re piecewise C 1. Fundmentl Theorem of Clculus: Let < b, φ : [, b] C. Suppose tht φ is piecewise C 1. Then S is finite, where S is the set of t [, b] such tht φ is not differentible t t. Let h : [, b] C stisfy h(t) = φ (t) for t [, b] S. h on S cn be n rbitrry, independent ssignment of complex numbers. Then h is Riemnn integrble nd h(t) dt = φ(b) φ() Rmsey Complex Integrtion
16 Proof of the Fundmentl Theorem of Clculus Clim 1: h is continuous t t [, b] S. Proof: Let t / S. Becuse S is finite, there is some δ 0 > 0 such tht u t < δ 0 u / S Thus, if u t < δ 0 nd u [, b], we hve h(u) = φ (u). Becuse φ (t) exists (since t / S), φ is continuous t t by hypothesis. Becuse φ nd h gree on [, b] (t δ 0, t + δ 0 ) we hve h continuous t t. Rmsey Complex Integrtion
17 Proof Continued, Slide 2 Clim 2: For ll t [, b), h hs finite right limit t t. Proof: Let t [, b). If t / S, then h is continuous t t. Becuse t = b, this implies tht the right limit of h t t is h(t), which is finite. Suppose tht t S. Becuse S is finite, there is some δ 0 > 0 such tht (t δ 0, t + δ 0 ) S = {t}. Hence u (t, min{t + δ 0, b}) h(u) = φ (u) Note tht both t + δ 0 nd b re strictly bigger thn t. Thus h nd φ gree on n open intervl of positive length with left end t. Becuse φ hs finite right limit t t, so does h. Rmsey Complex Integrtion
18 Proof Continued, Slide 3 Clim 3: For ll t (, b], h hs finite left limit t t. Proof: Let t [, b). If t / S, then h is continuous t t. Becuse t =, this implies tht the left limit of h t t is h(t), which is finite. Suppose tht t S. Becuse S is finite, there is some δ 0 > 0 such tht (t δ 0, t + δ 0 ) S = {t}. Hence u (min{, t δ 0 }, t) h(u) = φ (u) Note tht both t δ 0 nd re strictly less thn t. Thus h nd φ gree on n open intervl of positive length with right end t. Becuse φ hs finite left limit t t, so does h. Rmsey Complex Integrtion
19 Proof Continued, Slide 4 By Clims 1, 2 nd 3, h is piecewise continuous on [, b] nd thus Riemnn integrble. Clim 4: Let c < d in [, b] with (c, d) S =. Then d c h(t) dt = φ(d) φ(c) Proof: Consider δ < (d c)/2. Then J = [c + δ, d δ] (c, d). Hence no point of S is in J nd thus h(u) = φ (u) for u J. Note tht φ is continuous on J becuse it is continuous on the open intervl (c, d) contining J. Becuse φ = [R(φ)] + i[i(φ)], tht mkes both [R(φ)] nd [I(φ)] continuous on J. Rmsey Complex Integrtion
20 Proof Continued, Slide 5 By the Fundmentl Theorem of Clculus, pplied to the rel nd imginry prts of φ, we hve d δ c+δ h(t) dt = = = d δ c+δ d δ c+δ d δ c+δ φ (t) dt R[φ (t)] dt + i [R(φ)] (t) dt + i d δ c+δ d δ c+δ I[φ (t)] dt = {[Re(φ)](d δ) [R(φ)](c + δ)} [I(φ)] (t) dt + i {[I(φ)](d δ) [I(φ)](c + δ)} = φ(d δ) φ(c + δ) Rmsey Complex Integrtion
21 Proof Continued, Slide 6 By Rder s Specil Problem, Therefore d c d c d δ h(t) dt = lim h(t) dt δ 0 c+δ h(t) dt = lim δ 0 {φ(d δ) φ(c + δ)} Becuse φ is C 1, φ is continuous. Therefore the left limit t d of φ is φ(d) nd the right limit t c of φ is φ(c). Hence, since difference in C preserves limits, d c h(t) dt = φ(d) φ(c) Rmsey Complex Integrtion
22 Proof Continued, Slide 7 Suppose S (, b) =. Apply Clim 4 with [c, d] = [, b] to conclude tht h(t) dt = φ(b) φ() Suppose tht S (, b) =. Let {s j } n j=1 enumerte S (, b), with s j < s j+1 for 1 j n 1. Let s 0 = nd s n+1 = b. For 0 j n, (s j, s j+1 ) S =. By Clim 4, sj+1 s j h(t) dt = φ(s j+1 ) φ(s j ) Rmsey Complex Integrtion
23 Proof Continued, Slide 8 Hence h(t) dt = = n j=0 sj+1 s j h(t) dt n [ φ(sj+1 ) φ(s j ) ] j=0 = φ(s n+1 ) φ(s 0 ) = φ(b) φ() Rmsey Complex Integrtion
24 Tringle Inequlity for Integrtion Theorem: If φ is piecewise continuous on [, b], then φ(t) dt φ(t) dt Note: By one of the specil problems, φ is piecewise continuous on [, b] nd hence Riemnn integrble. Rmsey Complex Integrtion
25 Arc Length (Totl Distnce Trveled) Lemm: Let < b, γ : [, b] C, nd γ be piecewise C 1. Suppose tht h 1 : [, b] C nd h 2 : [, b] C such tht, for t [, b] with γ differentible t t, h 1 (t) = h 2 (t) = γ (t) Then h 1 nd h 2 re piecewise continuous on [, b] h 1 nd h 2 re Riemnn integrble nd h 1 (t) dt = h 2 (t) dt Proof: As in the justcompleted proof of complex version of the FTC, both h 1 nd h 2 re piecewise continuous. By specil problem, h 1 nd h 2 re piecewise continuous. By nother specil problem, both re Riemnn integrble. Rmsey Complex Integrtion
26 Proof Continued, Slide 2 Let S be the set of t [, b] t which γ is not differentible. Clim 1: Suppose tht c < d with [c, d] [, b] nd (c, d) S =. Then d d h 1 (t) dt = h 2 (t) dt c c Proof: On (c, d), h j (t) = γ (t) = h 2 (t). Hence, for 0 < δ < (d c)/2, for j = 1 nd j = 2 we hve d δ c+δ h j (t) dt = Then for j = 1 nd j = 2 we hve d c h j (t) dt = lim δ 0 d δ c+δ d δ c+δ γ (t) dt γ (t) dt Rmsey Complex Integrtion
27 Proof Continued, Slide 3 Suppose tht S (, b) =. Let [c, d] = [, b] in Clim 1 to conclude tht h 1 (t) dt = h 2 (t) dt Now suppose tht S (, b) =. Enumerte S (, b) s {s j } n j=1. Let = s 0 nd b = s n+1. For 0 j n, (s j, s j+1 ) S =. By Clim 1, sj+1 s j h 1 (t) dt = sj+1 s j h 2 (t) dt Rmsey Complex Integrtion
28 Proof Concluded, Slide 4 Finlly, h 1 (t) dt = = = n sj+1 j=0 s j sj+1 n j=0 s j h 2 (t) dt h 1 (t) dt h 2 (t) dt Rmsey Complex Integrtion
29 The Definition of Arc Length (s Totl Distnce Trveled) Def n: Let < b nd γ : [, b] C be peicewise C 1. Let h : [, b] C such tht, for ll t [, b], γ is differentible t t, then h(t) = γ (t). Then the length of γ is L(γ) = h(t) dt In the book, Srson writes simply γ (t) dt. Rmsey Complex Integrtion
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