Anlysis III Lecturer: Prof. Oleg Zboronski, Typesetting: Dvid Willims Term, 25
Contents I Integrtion 2 The Integrl for Step Functions 2. Definition of the Integrl for Step Functions.......................... 2.2 Properties of the Integrl for Step Functions.......................... 4 2 The Integrl for Regulted Functions 6 2. Definition of the Integrl for Regulted Functions....................... 6 2.2 Properties of the Integrl for Regulted Functions...................... 8 3 The Indefinite Integrl & Fundmentl Theorem of Clculus 3. The Indefinite Integrl...................................... 3.2 The Fundmentl Theorem of Clculus............................. 4 Prcticl Methods of Integrtion 2 4. Integrtion by Prts....................................... 2 4.2 Integrtion by Substitution................................... 3 4.3 Constructing Convergent Sequences of Step Functions.................... 4 5 The Chrcteristion of Regulted Functions 5 5. Chrcteristion & Proof.................................... 5 6 Improper & Riemnn Integrls 7 6. Improper Integrls........................................ 7 6.2 The Riemnn Integrl...................................... 9 7 Uniform & Pointwise Convergence 2 7. Definitions, Properties nd Exmples.............................. 2 7.2 Uniform Convergence & Integrtion.............................. 25 7.3 Uniform Convergence & Differentition............................. 27 8 Functionl Series 28 8. The Weierstrss M-Test & Other Useful Results........................ 28 8.2 Tylor & Fourier Series..................................... 29 II Norms 33 9 Normed Vector Spces 33 9. Definition & Bsic Properties of Norm............................ 33 9.2 Equivlence & Bnch Spces.................................. 35 Continuous Mps 4. Definition & Bsic Properties of Liner Mps......................... 4.2 Spces of Bounded Liner Mps................................. 42 Open nd Closed Sets of Normed Spces 43. Definition & Bsic Properties of Open Sets.......................... 43.2 Continuity, Closed Sets & Convergence............................. 44.3 Lipschitz Continuity, Contrction Mppings & Other Remrks............... 45 2 The Contrction Mpping Theorem & Applictions 46 2. Sttement nd Proof....................................... 46 2.2 The Existence nd Uniqueness of Solutions to ODEs..................... 47 2.3 The Jcobi Algorithm...................................... 5 2.4 The Newton-Rphson Algorithm................................ 5
Prt I Integrtion The Integrl for Step Functions. Definition of the Integrl for Step Functions Motivtion: Integrls Ares Under Curves Consider constnt functions of the form f c : x f R for x [, b]. Let D be the re under the line t f from to b. Then it mkes sense to define f c (x) dx. sign(f ) Are(D) sign(f ) ( f (b )) f (b ) Now this definition will be extended to piecewise constnt or step functions. Definition : ϕ : [, b] R is clled step function if there exists finite set P [, b] with, b P such tht ϕ is constnt on the open sub-intervls of [, b] \ P. P is clled prtition of [, b]. More concretely, P { p, p, p 2,..., p k, p k b}, with p < p < p 2 <... < p k < p k, nd ϕ c i (n rbitrry constnt), for i, 2,..., k, k. (pi,p i) Notes: If φ is step function then there re mny prtitions of [, b] such tht φ is constnt on the open sub-intervls defined by the prtition, cll one P. There exists t lest one other point greter thn in P (since both nd b belong to P by definition), cll the first p. Then φ is lso constnt on the open intervls of [, b] \ (P {n}), where < n < p. Let Q be prtition of [, b], with Q P. Then Q is clled refinement of P (so (P {n}) is refinement of P ). The miniml common refinement of P nd Q is denoted by P Q. If φ is step function on [, b], then ny prtition P { p, p, p 2,..., p k, p k b}, with p < p < p 2 <... < p k < p k, such tht φ c i (n rbitrry constnt), for i, 2,..., k, k, is clled (pi,p i) comptible with, or dpted to, φ. Proposition : Let S[, b] be the spce of step functions on [, b]. If f, g S[, b], then µ, λ R, (λf + µg) S[, b] (so S[, b] is closed under tking finite liner combintions of its elements). Proof: f S[, b], so P { p, p, p 2,..., p k, p k b}, with p < p < p 2 <... < p k < p k, which is comptible with f, nd g S[, b] so Q { q, q, q 2,..., q l, q l b}, with q < q < q 2 <... < q l < q l, which is comptible with g. Consider P Q { r, r, r 2,..., r m, r m b}, with r < r < r 2 <... < r m < r m. Then this prtition is comptible with (λf + µg). It remins to show tht (λf + µg) c i (n rbitrry constnt), for i, 2,..., m, m. (ri,r i) P (P Q), so (r i, r i ) (p, p ) for some, nd Q (P Q), so (r i, r i ) (q b, q b ) for some b. Then, since f is constnt, f is constnt, nd since g is constnt, (p,p ) (ri,r i) (qb,q b ) g is constnt, so (λf + µg) is constnt. Since the choice of i ws rbitrry, this proves (ri,r i) the proposition. (ri,r i) 2
Notes: This demonstrtes tht S[, b] is vector spce over R. Define C[, b] s the spce of continuous functions on [, b], nd define B[, b] s the spce of bounded functions on [, b]. These re lso vector spces. It ws demonstrted in Anlysis II tht C[, b] B[, b], nd by definition S[, b] B[, b]. Definition 2: φ i. φ (pi,p i) For φ S[, b], let P {p i } k i be prtition comptible with φ. Define (temporrily). Then : S[, b] R is defined s φ. k φ i (p i p i ) i Remrk: If φ φ R, then P {, b}, so φ φ (b ), which is precisely the re described in the motivtion. Lemm 2: The vlue of φ does not depend on the choice of prtition comptible with [, b]. Proof: For ny φ S[, b], let P nd Q be prtitions comptible with [, b]. Suppose there re k points in Q tht re not in P. Denote them n, n 2,..., n k. Temporrily let b φ P be the integrl of φ from to b defined by the prtition P which is comptible with [, b]. Now, consider P {n }. Since Q is comptible with [, b], we know tht n [, b], so i N such tht n (p i, p i ) for some p i, p i P. φ is constnt on (p i, p i ), nd so is lso constnt on (p i, n ) nd (n, p i ), so P {n } is comptible with φ. By definition φ P {n} φ φ i (n p i ) + φ i (p i n ) φ i (p i p i ), so P The proof is completed by induction, since there re finite number of points. exercise. φ P {n} φ P This is left s n 3
.2 Properties of the Integrl for Step Functions. Proposition 3 - Additivity: For ny φ S[, b], v (, b), φ v φ + 2. Proposition 4 - Linerity: For ny φ, ψ S[, b], λ, µ R, λφ + µψ λ v φ φ + µ 3. Proposition 5 - The Fundmentl Theorem of Clculus: Let φ S[, b], nd let P be prtition of [, b] comptible with φ. Consider I : [, b] R, I : x x φ. Then (I) I is continuous on [, b]. (II) I is differentible on k i (p i, p i ), nd, x k i (p i, p i ), I (x) φ(x). ψ Proof:. Let P {p i } k i be prtition comptible with φ S[, b]. Then there re two cses: (i) If v P, then by definition φ k φ i (p i p i ) i v φ i (p i p i ) + i k iv+ φ i (p i p i ) (ii) If v / P, then i N, i k, such tht v (p i, p i ). Then { p,..., p i, v}, {v, p i,..., p k b} re prtitions comptible with φ, so φ S[, v], φ S[v, b], so (,v) (v,b) their integrls re well-defined on those intervls. Then φ k φ j (p j p j ) j v φ + φ (p p ) + φ 2 (p 2 p ) +... + φ i (p i p i ) +... + φ k (p k p k ) φ (p p ) +... + φ i (p i + v v p i ) +... + φ k (p k p k ) φ (p p ) +... + φ i (v p i ) + φ i (p i v) +... + φ k (p k p k ) ] [ i k φ j (p j p j ) + φ i (v p i ) + φ i (p i v) + φ j (p j p j ) j v φ + v φ ji+ 2. Let P be prtition comptible with φ S[, b], nd let Q be prtition comptible with ψ S[, b]. Then, s ws proved in the previous section, P Q is comptible with both φ nd ψ. Let P Q { x, x,..., x k, x k b}, with x < x <... < x k < x k b. Then φ φ i nd ψ ψ i, so (λφ + µψ) (λφ + µψ) i, so P Q is comptible (xi,x i) (xi,x i) (xi,x i) with λφ + µψ. Then v φ λφ + µψ k k (λφ + µψ) j (x j x j ) (λφ j + µψ j )(x j x j ) j j k λ φ j (x j x j ) + µ k ψ j (x j x j ) λ φ + µ j j ψ 4
3. Let P {p i } k i be prtition comptible with φ S[, b]. Choose x k j (p j, p i ), then i N, i k such tht x (p i, p i ). Then x pi x I(x) φ φ + φ I(p i ) + φ i (x p i ) φ i x + constnt p i so I is continuous on k j (p j, p i ) nd I (x) φ i φ(x). Furthermore, lim x pi I(x) I(p i ), so I is right continuous on [, b). Similrly, lim x pi I(x) I(p i ) + φ i (p i p i ) p i φ + φ i (p i p i ) p i on [, b]. φ + p i p i φ p i φ I(p i), so I is left continuous on (, b], so I is continuous Definition 3: Let f B[, b]. Then the supremum norm (or sup norm) of f, denoted by f, with : B[, b] R is defined s f. sup f(x) x [,b] Proposition 6: Let φ S[, b] be bounded between m, M R, so m φ(x) M x [, b]. Then m(b ) φ M(b ), in prticulr φ φ (b ). Proof: For φ S[, b], choose comptible prtition P. Now then k m(p i p i ) i φ k φ i (p i p i ) i k φ i (p i p i ) so, due to telescoping sums on the upper nd lower bounds, m(b ) i k M(p i p i ) i k φ i (p i p i ) M(b ), so m(b ) i φ M(b ). If φ is given, then, by definition, φ φ(x) φ, so using the sme rgument s bove, φ (b ) b φ φ (b ), so b φ φ (b ). Note: So fr φ hs been defined for b >. Define, for > b, φ. φ. Then the bound proved in the previous proposition becomes b φ φ b. It is simple exercise to demonstrte tht such n extension preserves the three properties of the integrl for step functions proved previously. 5
2 The Integrl for Regulted Functions 2. Definition of the Integrl for Regulted Functions Definition 4: A function f on [, b] is clled regulted function if, ε >, φ S[, b] such tht φ f < ε, or, lterntively, if, ε >, f(x) φ(x) < ε x [, b]. We denote the set of regulted functions on [, b] by R[, b]. Definition 4.: A sequence (φ n ) n S[, b] is sid to converge uniformly to function f : [, b] R if lim n f φ n. Lemm : f : [, b] R is regulted iff (φ n ) n S[, b] such tht, s n, (φ n ) converges uniformly to f. Proof: The proof is routine nd is left s n exercise. Definition 4.2: A function f : A R R is clled uniformly continuous on A if, ε >, δ such tht x, y A, x y < δ nd f(x) f(y) < ε. Note: In generl, uniform continuity implies continuity, but not the other wy round. For exmple, f : (, ] R, f : x x, is continuous, but not uniformly continuous. However, on closed intervls, the two re equivlent. Lemm: For function f : A R, if A [, b], continuity implies uniform continuity. Proof: Suppose f is continuous on [, b] but not uniformly continuous. Then ε > such tht δ >, x, y [, b] such tht x y < δ but f(x) f(y) ε. Tking δn n s n yields two sequences (x n ) n, (y n ) n [, b] with x n y n < δ n. Since (x n ) n is bounded, it converges by the Bolzno-Weierstrss Theorem, so (x nk ) k (x n ) n which converges in [, b], since [, b] is closed. So lim k x nk u [, b]. Choose (y nk ) k, which converges by similr rgument, nd choosing δ n k demonstrtes tht lim k x nk y nk, so lim k y nk u. But f is continuous, so lim k f(x nk ) f(u) lim k f(y nk ), which contrdicts the condition tht x y < δ but f(x) f(y) ε, so f is uniformly continuous on [, b]. Proposition 7: If function f : [, b] R is continuous on [, b], then it is lso regulted on [, b]. Proof: For ny ε >, let P {p i } k i be prtition of [, b] such tht p i p i < δ, where δ is such tht x y < δ f(x) f(y) < ε. Such δ exists becuse f is continuous nd so φ f(p i ) for i,..., k uniformly continuous on [, b]. Let φ S[, b] be such tht [pi,p i). Now, φ(b) f(b) x [, b), j such tht x [p j, p j ), then f(x) φ(x) f(x) f(pj ) < ε since x pj < δ. Also, f(b) φ(b), so, x [, b], f(x) φ(x) < ε, so f is regulted by definition. Exercise: Prove tht ny monotone function is regulted. 6
Proposition 8: For ny f, g R[, b] nd λ, µ R, λf + µg R[, b]. Proof: First prove tht f, g B[, b], λ R, f + g f + g nd λf λ f. The reminder of the proof is left s n exercise. Definition 5: Let f R[, b]. Then : R[, b] R is defined s f. lim n φ n where (φ n ) n S[, b] converges to f uniformly s n. Proposition 9: lim n φ n exists nd is independent of the choice of (φ n ) n S[, b], which converges to f R[, b] uniformly s n. Proof: Let f R[, b] nd (φ n ) n, (ψ n ) n S[, b] converge to f uniformly s n. (φ n ) converges to f uniformly, so for ny ε >, eventully there will be n n such tht φ n f < ε 2(b ), so eventully there will be n nd m, when sufficiently lrge, such tht φ n φ m (φ n f) + (f φ m ) φ n f + f φ m ε b. Then φ n φ m (φ n φ m ) φ n φ m (b ) < ε, so ( φ n) n is Cuchy, so the limit exists. Let ω n (φ n, ψ n ) n (φ, ψ, φ 2, ψ 2,...) S[, b]. Then ω n converges to f uniformly s n, the proof of this fct is routine nd is left s n exercise. lim n ω n exists, so the subsequences φ n nd ψ n must converge to the sme limit f, so φ n ψ n. 7
2.2 Properties of the Integrl for Regulted Functions Proposition : For ny f R[, b], u [, b], f u f + u Proof: f is regulted, so (φ n ) n S[, b] which converges uniformly to f. Then (φ n ) converges [,u] uniformly to f, since restrictions re regulted. The sme holds for (φ n ). So [,u] [u,b] f u f + u ( u f lim φ n + n u φ n ) lim n φ n f by the dditivity of integrls of step functions. Proposition : The mp I : R[, b] R, I : f f is liner. Proof: The proof is similr to combintion of the lst proof nd the proof for the linerity of the integrl for step functions, nd so is left s n exercise. Note: Not ll functions re regulted. Exmple: Consider f : [, ] R, with f : x { x Q x / Q f is lso denoted by Q - the indictor of Q. Let φ be ny step function on [, b], with comptible prtition P, nd let φ φ Then f φ (po,p ). f φ mx( φ, φ ) (p,p 2, so ) there cnnot exist sequence of step functions converging uniformly to f. Proposition 2: Suppose tht f R[, b], nd tht, x [, b], m f(x) M for some m, M R. Then m(b ) f M(b ) Proof: f R[, b], so, ε >, φ ε S[, b] such tht f φ ε < ε, so x [, b], f(x) ε φ ε (x) f(x) + ε, so m ε φ ε (x) M + ε, now using the result proved for step functions (m ε)(b ) φ ε (M ε)(b ) Choose ε n for n N, then (φ n) n converges uniformly to f. Tke the limit of the bove with ε n, then m(b ) f M(b ) Exercise: Prove tht, for ny f R[, b], f f (b ) 8
Exmple: f(x) x is uniformly continuous on R, since f(x) f(y) x y, so when x y < δ ε, f(x) f(y) < ε. Proposition 3: Let f α : [, ) R, f α : x x α for some α. Then f α is uniformly continuous for ny α but not uniformly continuous for α >. Proof: For α >, f α (x+δ) f α (x) (x+δ) α x α x α ((+ δ x )α ) x α (+α δ x ) αδxα s x, which contrdicts the condition for uniform continuity. For α <, α, f α (x + δ) f α (x) x α (( + δ x )α ) x α ( + αδ x ), (α ) <, so xα s x. For ny ε >, choose δ ε 2αδ, then xα is uniform continuous. The cse when α is trivil nd is left s n exercise. Proposition 4: If f : [, b] R is wekly incresing, then f R[, b]. Proof: For f(b) > f(), let d f(b) f() n. Let P be prtition with n + elements such tht p, p j sup x [,b] {x : f(x) f() + dj}, nd p n+ b. By construction, since p j > p j, then x (p j, p j ), f()+d(j ) f(x) f()+dj. Let ϕ n S[, b] such tht ϕ n (pj,p j) f()+dj nd ϕ n (p j ) f(p j ). Now, let x (p j, p j ). Then, using the previous inequlity, d f(x) ϕ n (x), so f ϕ n d f(b) f() n s n, so f is regulted. Exercise: Prove tht if f : [, b] R is wekly decresing, then f is regulted. This proves tht ll monotonic functions on bounded intervls re regulted. Exmple: Let Q [, ] {,, 2, 3, 2 3, 4, 3 4, 5,,...} {q, q 2, q 3,...}, nd let { x > q n (x > q n ) x q n Let f : [, ] R be given by f : x 2 n (x > q n ) n which exists x [, ]. Then f is non-decresing, nd is regulted, but jumps n infinitely mny times, so is impossible to drw. 9
3 The Indefinite Integrl & Fundmentl Theorem of Clculus 3. The Indefinite Integrl Definition 6: Let f R[, b]. Define F : [, b] R by F is clled the indefinite integrl of f. F : x x f Lemm: R[, b] B[, b] Proof: f R[, b], so φ S[, b] such tht φ f, so x [, b], φ(x) f(x) φ(x) +. Since step functions re bounded, φ B[, b], so f B[, b]. Proposition 5: F is continuous on [, b]. Proof: Let x, y [, b]. Then x F (x) F (y) f y f x y f f x y, so by the sequentil definition of continuity, f is continuous. Note: Let f : [, b] R. Then, if L > such tht, x, y [, b], f(x) f(y) L x y. Then f is clled Lipschitz continuous or Lipschitz. Lipschitz continuity implies continuity, but the reverse is not generlly true. Exmple: f(x) e x is Lipschitz on [, ], but g(x) x is not Lipschitz on [, ].
3.2 The Fundmentl Theorem of Clculus Theorem 6 - FTC: Let f R[, b] such tht f is continuous t c (, b). Then F (x) x f is differentible t c nd F (c) f(c). Proof: f is continuous t c, so, ε >, δ > such tht, x (c δ, c + δ), f(x) f(c) < ε, so f(c) ε < f(x) < f(c) + ε. Choose < h < δ. Then, by integrting the previous sttement h(f(c) ε) < c+h c f < h(f(c) + ε), so h(f(c) ε) < c+h f So, for ny h (, δ), h(f(c) ε) < F (c + h) F (c) < h(f(c) + ε), so ε < F (c + h) F (c) h c f(c) < ε f F (c + h) F (c) < h(f(c) + ε) So F is differentible t c, nd F (c) f(c). The proof for the cse where δ < h < is similr nd so is left s n exercise. Corollry 7: If f C[, b], then f is differentible on (, b) nd F f. Proof: f is continuous, so it is regulted. Applying FTC completes the proof. Theorem 8 - FTC2: Let f : [, b] R be continuous, nd let g : [, b] R be differentible, with g (x) f(x) x [, b]. Then f g(b) g() This is the Newton-Leibniz Formul, nd g is clled the ntiderivtive of f. This is sometimes written g g(b) g() Proof: Let δ : [, b] R, with δ(x) g(x) x f. Then δ is continuous on [, b]. it is lso differentible on (, b) by FTC. By the MVT, δ(b) δ() δ(ξ)(b ) for some ξ (, b). But δ (ξ) g (ξ) f(ξ) by FTC, so δ (ξ) f(ξ) f(ξ), so δ() δ(b), so g() f g(b) f, so f g(b) g().
4 Prcticl Methods of Integrtion 4. Integrtion by Prts Proposition 2: If f, g R[, b], then f g R[, b]. Proof: f + g f + g. Then (ϕ n ), (ψ n ) S[, b] such tht (ϕ n ) f, (ψ n ) g uniformly. f g ϕ n ψ n f g ϕ n g + ϕ n g ϕ n ψ n (f ϕ n )g + ϕ n (g ψ n ). Since sup x [,b] (x) b(x) supx [,b] (x) supx [,b] b(x), this is less thn or equl to g f ϕ n + ϕ n g ψ n. Now, g is regulted, nd so is bounded. f ϕ n, g ψ n s n, nd ϕ n ϕ n f + f ϕ n f + f. Also, ϕ n f s n, nd f is regulted, nd so is bounded. Tken together, this mens tht f g ϕ n ψ n s n, so (ϕ n ψ n ) converges uniformly to f g, so f g is regulted. If F nd G re differentible such tht F nd G re contin- Corollry 9 - Integrtion by Prts: uous, hence regulted, then F G F (b)g(b) F ()G() F G Proof: (F G) F G + F G. Then By FTC2 So (F G) F G + F G F G + F (b)g(b) F ()G() F G + F G F G + F G F G F (b)g(b) F ()G() F G F G Note: F (b)g(b) F ()G() cn be denoted by F G b. Exmple: Then The Gmm Function Γ : (, ) R is given by Γ : x t x e t dt. L lim ε,l ε t x e t dt Γ(x) lim ε,l L ε ( ) x tx e t dt lim x ε,l (tx e t ) L ε x Γ(x + ) x t x e t dt So Γ(x + ) xγ(x), nd Γ(), so Γ(n + ) n! L ε t x ( e t ) dt 2
4.2 Integrtion by Substitution Let f C[, b], nd let g : [c, d] [, b] be differen- Corollry 2 - Integrtion by Substitution: tible with Im g [, b]. Then d f(g(x)) g (x) dx g(d) c g(c) f(t) dt Proof: Let F (x) x f. Since f is continuous, F is differentible nd F (x) f(x) by FTC, then by FTC2 d g(d) g(c) f(t) dt F (g(d)) F (g(c)) Also, dx F (g(x)) F (g(x)) g (x) f(g(x)) g (x) which is the initil integrnd. F (g(x)) d c F (g(d)) F (g(c)), so Now, by FTC2, d f(g(x)) g (x) dx g(d) c g(c) f(t) dt Note: g does not need to be injective or surjective. Exmple: Let g(x) x 2. Then f(x 2 )(x 2 ) dx F (x 2 ) Exmple: Let G(, j). for some >, j R. Then Now, substituting g(x) x j, e x2 +2jx dx G(, j) e j2 L lim e x 2 +2jx dx L,L 2 L 2 e x2 +2jx dx e (x2 + 2j x) dx e ((x j )2 j2 2 ) dx e (x j )2 dx e j2 e (x j )2 dx e j2 e t2 dt e j2 e ( t) 2 dt Now, substituting h(t) t, So e j2 e ( t) 2 dt e j2 G(, j) e j2 e y2 π dy e j2 π 3
4.3 Constructing Convergent Sequences of Step Functions Theorem 22: Let f : [, b] R be regulted. Then lim n j b h f( j) f(x) dx, where j + b h j Proof: First, consider φ S[, b]. Let P be prtition comptible with φ, with P k. Then b φ h φ( j) k b n 2 φ s n j since errors only occur nd contribute to the difference t jumps on the step function, of which there re k, nd the mximum error this cn be is twice the mximum vlue of the step function cross the whole section. Then b φ lim n h φ( j) Now, for f R[, b], let R (n) f j b h j f( j), where j + b h j Since f is regulted, then, ε >, φ S[, b] such tht f φ < ε 3. Then R(n) f f R(n) f R (n) φ + R(n) φ φ + φ f R (n) f R (n) φ + R(n) φ φ + φ f Now, nd nd R (n) f R(n) φ R (n) φ φ f (b ) < ε (b ) 3 φ < ε (b ) eventully in n 3 φ f (f( j ) φ( j )) b n j f( j ) φ( j ) b n < ε (b ) 3 j since f(j ) φ( j ) < ε 3, so so R(n) f f < ε(b ) eventully in n lim n R(n) f f 4
5 The Chrcteristion of Regulted Functions 5. Chrcteristion & Proof Proposition 23: A function f R[, b] iff, x (, b), f(x±) both exist, nd both f(+) nd f(b ) exist, where f(x+) lim y x f(y) (the right limit), nd f(x ) lim y x f(y) (the left limit). Alterntively, this cn be stted s ll one-sided limits exist. Note: This chrcteristion only requires the existence of these limits, not equlity, f(x+) f(x ) is cceptble so long s they both exist. Proof : f is regulted, so, ε >, φ S[, b] such tht f φ < ε. Let x (, b). There re now two cses, φ is continuous t x, or φ is discontinuous t x. In either cse, δ > such tht φ φ, constnt. Let (x n ) n R be ny sequence such tht, n N, x n > x, (x,x+δ) nd lim n x n x. Eventully x n (x, x + δ). Then, eventully in n nd m, f(xn ) f(x m ) f(xn ) φ + φ f(x m ) f(xn ) φ + f(xm ) φ < 2ε, since f(xn ) φ, f(xm ) φ < ε, so f(x n ) n is Cuchy, nd so converges. If y n is ny other sequence with y n x s n, then (f(y n )) n converges to the sme limit. The proof of this fct is routine fter considering (x n, y n ) n, nd so is left s n exercise. It hs been shown tht for ny (x n ) n R such tht x n x s n, lim n f(x n ) f(x+) exists nd does not depend on x n. A similr rgument cn be used to prove this result for the left limit, nd not much modifiction is required for x, b, so these proofs re left s exercises. Proof : All one-sided limits exist. Now, fix ny ε >. Let A ε {γ [, b] : ψ S[, b] such tht (f ψ) [,γ] < ε} The proof shll follow from proving three sttements: (i) A ε : f(+) exists, so, ε >, δ > such tht, x (, + δ), f(x) f(+) < ε. Tke γ δ 2. Then define { f() x φ(x) f(+) x (, + δ 2 ) It is cler tht φ(x) S[, + δ 2 ]. Since f(x) f(+) < ε nd f() φ(), (f φ) [,+ δ < ε, so γ δ 2 A ε, so A ε. 2 ] (ii) sup A ε b: Assume tht c. sup A ε < b. Then, δ >, (c δ) A ε, so φ S[, c δ] such tht (f φ) [,c δ] < ε. But f hs limits f(c±) t c, so ε >, δ, δ 2 such tht, x (c, c + δ 2 ), f(x) f(c+) < ε nd, x (c δ, c), f(x) f(c ) < ε. Choose δ < δ. Now, let ψ S[, b] be given by f(c ) x < c ψ(x) f(c) x c f(c+) x > c nd let φ S[, b] be given by φ(x) { φ(x) x [, c δ] ψ(x) x (c δ, c + δ2 2 ] Then (c + δ2 2 ) A ε, with δ2 2 >, which contrdicts the fct tht c sup A ε, so b c. 5
(iii) b A ε : It hs been shown tht, δ >, φ S[, b δ] such tht (f φ) [,b δ] < ε. But f(b ) exists, so δ > such tht, x (b δ, b), f(x) f(b ) < ε. Choose δ < δ. Now, let ψ S[, b] be given by { f(b) x b ψ(x) f(b ) x < b nd let φ S[, b] be given by φ(x) Now, by definition, f φ < ε, so b A ε. { φ(x) x [, b δ] ψ(x) x (b δ, b] So γ b, so ϕ S[, b] such tht f ϕ < ε, so f R[, b]. Exercise: Prove tht if f R[, b], then the set of discontinuities of f is countble. Hint: Check tht, ε >, the set of jumps of size ε or greter is countble. 6
6 Improper & Riemnn Integrls 6. Improper Integrls Motivtion: It is possible to generlise the regulted integrl to unbounded functions nd/or unbounded domins. Exmple: x 2 dx 2 becuse, lthough x 2 is not regulted on (, ], ε >, x 2 R[ε, ]. Then ε x 2 dx 2x 2 ε 2 2ε 2 2 s ε Exmple: becuse, L >, x R[, L]. Then 2 L x 2 dx x L x 2 dx L s L Exmple: dx lim dx lim x ε ε x log ε x ε lim log ε ε which does not exist, therefore the integrl is not defined. Definition 7.: If, for ny ε >, f : (, b] R is regulted on [ + ε, b] nd l. lim f ε +ε exists, then f converges nd is equl to l. If the limit does not exist, then f diverges. Definition 7.2: If, L >, f R[, L] nd exists, then l. lim L L f converges nd is equl to l. If the limit does not exist, then f diverges. f 7
Note: Let f R[ L, L 2 ] for some L, L 2 >. Then f converges if, c R, c f nd f c converge. In this cse, c f + f c f. As n exercise, show tht if c f nd f exist, then c c f + f does not depend on the choice of c. c Exmple: Let f(x) x R[ L, L] L >. Then However, tke ny c R. Then so do not exist. Then L L L lim L c x dx, so x dx, nd L lim L L x dx c lim x dx L L x dx does not exist, since the limits must exist independently. Notes: Suppose f : (, b) R, nd, for ny ε, ε 2 >, f R[ + ε, b ε 2 ]. Then f exists if, for ny c (, b), c f nd c f both exist (converge). Then f converges, nd f c f + c f. f both exist (indepen- Suppose f is defined on [, b) (b, c]. Then c f exists (converges) if f nd c b dently), with c f f + c b f. Exmple: f(x) for x [, ) (, ]. x Note: Let f : (, b) R such tht, for ny ε, ε 2 >, f [ + ε, b ε 2 ]. Then c (, b) such tht l (c) lim ε c +ε f nd l 2 (c) lim ε2 ε2 exist iff, c (, b), l (c ) nd l 2 (c ) exist. The first formultion is often the more convenient in prctice however. If this formultion is true of f, then f converges, nd f l (c) + l 2 (c). As n exercise, check tht l (c) + l 2 (c) l (c ) l 2 (c ) to demonstrte tht this integrl is independent of the choice of c. c f Definition 7.3: Let f R[ L, L 2 ] for some L, L 2 >. Then f. L lim L c f + lim L 2 c L 2 f Definition 7.4: If f : [, b) (b, c] R, then c f. lim ε ε c f + lim f ε2 b+ε 2 Note: All integrls in the limit must mke sense s regulted or improper integrls. Note: It is possible to define the integrl of functions with domins such s f : j Z(j, j + ) R Situtions like this occur when studying gmm functions. 8
6.2 The Riemnn Integrl Definition: Let f : [, b] R. Then U f. L f. inf φ S[,b] sup φ S[,b] { { } φ : φ f ( upper sum ) } φ : φ f ( lower sum ) Definition: f : [, b] R is Riemnn integrble if U f L f. Then R f. U f L f Proposition: If f R[, b], then it is Riemnn integrble, nd R f f Proof: f R[, b], so, ε >, φ S[, b] such tht, x [, b], φ(x) ε f(x) φ(x) + ε. Since φ(x) ε, φ(x) + ε S[, b], Tking ε, φ ε(b ) L f U f L f U f f R f φ + ε(b ) Note: There re functions which re Riemnn integrble but not regulted. Exmple: Let f : [, ] R be given by { x 2 n for n,, 2,... f : x otherwise Then f is not regulted since, p >, x (, p) such tht f(x) nd y (, p) such tht f(y), so step function cn get no closer thn hlf to both sections of the function on ny intervl, so f is not defined. However, let Now, φ N f, so φ N L f U f { x 2 N f(x) x > 2 N φ N lim φ N N 2 N Then L f U f, so so the Riemnn integrl of f is defined. R f 9
Note: There re functions which re not Riemnn integrble. Exmple: Let Q (x) { x Q x R \ Q Then U Q L Q, so Q is not Riemnn integrble. However, it is Lebesgue integrble, with L Q 2
7 Uniform & Pointwise Convergence 7. Definitions, Properties nd Exmples Exmple: Let (f n ) n be sequence of functions on [-,], with f n : x x 2 n for n, 2,... Fix x nd consider lim f n(x) n x > x x <, so lim n f n(x) sign x. So the limit of (f n (x)) n s n exists x [, ], nd f n is continuous n N, but the limit is not continuous t. lim n lim f n(x) ± lim x ± x ± sign x lim x ± lim n f n(x) This function hs non-commuttive limits nd sequence of continuous functions converges to discontinuous function, both of which re undesirble properties of pointwise convergence. Definition 8: Let (f n ) n be sequence of functions, nd let A R. Then f n f pointwise s n if, x A, lim n f n (x) f(x). Note: Pointwise convergence is very loose, it does not preserve properties of the f n s such s continuity (s in the cse of x 2 n sign x). Pointwise convergence cn lso be very non-uniform (the sme exmple converges pointwise very slowly ner ). Note: Let f nd (f n ) n be functions on A R. If (f n ) f pointwise s n, then f cn fil to be continuous, even if ll f n s re. Alterntively, x A such tht lim lim f n (x) lim n x x x x lim f n(x) n Exmple (The Witch s Ht): Let f n : [, ] R, with f n () nd 2n 2 x x [, 2n ] f n (x) n 2n 2 (x 2n ) x ( 2n, n ] x > n Then f n C[, ]. Fix x >, then lim n f n (x). Now, s n, f n f on [, ], which is continuous. However, f, but f n n 2n 2, so 2 lim n f n lim f n n It is cler from exmples such s this tht pointwise convergence must be strengthened. Definition 9: Let f, f n : A R R for n, 2,..., then (f n ) n f converges uniformly s n if, ε >, N ε such tht, n > N ε, fn (x) f(x) < ε x A, or, equivlently, if limn f f n. 2
Note: For the reminder of this section, ssume, unless specified otherwise, tht f, f n : A R R for n, 2,.... Remrk: Uniform convergence pointwise convergence, but pointwise convergence uniform convergence. Definition 9.: A sequence (f n ) n (of functions on A R) is clled uniformly Cuchy if, ε >, N ε such tht, n, m > N ε, fn (x) f m (x) < ε x A. Note: For fixed x A, if (f n (x)) n is Cuchy, then it is pointwise convergent (to common limit f). Lemm: A uniformly Cuchy sequence converges uniformly. Proof: Let (f n ) n, (f m ) m be uniformly Cuchy sequences converging to f. Then, ε >, N ε such tht, n, m > N ε, f m (x) ε 2 f n(x) f m (x) + ε 2 x A. Tke m, then f m f, so f(x) ε 2 f n(x) f(x) + ε 2 x A, so f n (x) f(x) < ε x A, so (f n ) n converges uniformly. Theorem 24: (i) f R[, b] Let (f n ) n R[, b]. Assume tht f n f uniformly. Then (ii) lim n f n lim n f n f Proof: (i) f n R[, b], so, ε >, φ n S[, b] such tht f n φ n < ε 2. Now, f φ n f f n + f n φ n f f n + f n φ n. Assume tht n is lrge enough such tht f f n < ε 2, which is gurnteed since f n f, so f φ n < ε, so f R[, b]. (ii) It ws shown in (i) tht f is regulted, so f is well defined nd b f f n f f n (b ) f f n s n, so lim n f n lim f n n f Theorem 25: Suppose tht (f n ) n C[, b], nd f n f uniformly s n. Then f C[, b]. This theorem lso holds for some rbitrry A R in plce of [, b], nd shll be proven for this more generl cse. Proof: f n f uniformly, so ε >, N ε such tht, n > N ε, fn (x) f(x) < ε x A, so fn is continuous n N. Then δ ε such tht, x (x δ ε, x + δ ε) A, f n (x) f n (x ) < ε 3 x (x δ ε, x + δ ε) A. Set δ ε δ ε. Then, x (x δ ε, x + δ ε) A, f(x) f(x ) f(x) fn (x) + f n (x) f n (x ) + f n (x ) f(x ) fn (x) f n (x ) + fn (x) f n (x ) + fn (x ) f(x ). f n f uniformly s n, so the first nd lst of these terms re less thn ε 3, nd the middle term ws shown to be less thn ε 3 erlier, so f(x) f(x ) < 3ε 3 ε, so f is continuous t x. 22
Exmple (Cntor s Function or The Devil s Stircse): Let (f n ) n C[, ] be such tht f nd 2 f n(3x) x [, 3 ) f n+ (x) 2 x [ 3, 2 3 ] 2 + 2 f n(3x 2) x ( 2 3, ] Proposition: lim n f n exists nd is continuous. It is clled Cntor s function. Proof: Clerly f C[, ]. Now, suppose tht f n C[, ]. Then f n+ is clerly continuous on ( 3, 2 3 ). Now, f n is continuous on [, 3 ) by ssumption, so 2 f n(3 ) is continuous, so f n+ is continuous on [, 3 ). Also, f n is continuous on ( 2 3, ] by ssumption, so 2 + 2 f n(3 2) is continuous, so f n+ is continuous on ( 2 3, ]. Now the only points remining to investigte the continuity of re 3 nd 2 3. f n+ () 2 + 2 f n(), which is if f n (). f (), so f n () n N by induction. Concerning the continuity of 3, f n+( 3 ) 2, nd f n+( 3 +) 2, so f n+ is right-continuous t 3, nd f n+ ( 3 ) 2 f n(3( 3 )) 2 f n( ) 2 f n() 2 since f n is continuous t, so f n+ is left-continuous t 3, so f n+ is continuous t 3. The proof for 2 3 is similr nd is left s n exercise. Now, ( f n+ f n mx (f n+ f n ) x [, 3 ), (f n+ f n ) x [ 3, 2 3 ], (f n+ f n ) ( mx (f ) n+ f n ) x [, 3 ),, (f n+ f n ) x ( 2 3,] ( mx 2 f n(3 ) ) ( 2 f n (3 ), x [, 2 f n(3 2) ) 2 f n (3 2) x ( 2 3 ) x ( 2 3,] 3,] ) 2 f n f n So, for some c, d R, f n f n 2 f n f n 2 2 f n f n 2... The constnts re introduced to correct for f. Now, fix n > m, then, for some D R, c 2 n+ f f d 2 n+ f n f m f n f n+ + f n+ f n 2 +... + f m f m f n f n + f n f n 2 +... + f m+ f m ( D 2 n + 2 n +... + ) 2 m+ D n m 2 m+ 2 k k D 2 m+ k 2 k D 2 m s m Since f n f m cn be mde rbitrrily smll by tking n > m lrge enough, (f n ) is uniformly Cuchy. Therefore, f lim n f n exists nd is continuous, since uniformly Cuchy sequence converges uniformly nd f n is continuous n N. The limit is clled Cntor s function or the Devil s Stircse. 23
Now, f C[, ], so f R[, ]. Tking the limit of f n (x), it is cler tht 2 f(3x) x [, 3 ) f(x) 2 x [ 3, 2 3 ] 2 + 2 f(3x 2) x ( 2 3, ] So f 3 2 3 f + f + 3 Let y 3x, z 3x 2. Then 2 3 f 3 f(3x) dx + 2 2 3 3 2 dx + 2 2 + f(3x 2) dx 2 3 f 6 + 6 f(y) dy + 6 + 6 f(z) dz 3 + 3 f, so 2 3 f 3, so f 2 Remrk: Assume tht f is constnt on E [, ]. Then f is constnt on ( 2, 2 3 ) 3 E ( 2 3 + 3E), so E 3 + 3 E + 3 E, so 3 E 3, so E, so f is loclly constnt on some E [, ] of length, but f : [, ] [, ] continuously. Note: It is possible to mp [, ] R onto (surjectively) [, ] [, ] R 2 continuously. If γ : [, ] [, ] [, ] is such mp, then, x [, ] [, ], t [, ] such tht γ(t) x. A surjective mp is reltively simple to construct, such s τ : [, ] [, ] [, ], with τ :.x x 2 x 3... (.x x 3 x 5...,.x 2 x 4 x 6...), but τ is not continuous. A continuous mp stisfying these criteri is clled spce filling curve, nd exmples were constructed by Hilbert nd Peno in the 89s. 24
7.2 Uniform Convergence & Integrtion Note: Throughout this section, let D [, b] [c, d] R 2, nd (x, t) D. Definition : A function f : D R is continuous t (x, t ) D if, ε >, δ ε (x, t ) > such tht, (x, t ) D with x x < δ ε (x, t ) nd t t < δ ε (x, t ), f(x, t) f(x, t ) < ε. An equivlent definition, which shll be introduced lter, uses Eucliden distnce. Definition.: A function f : D R is uniformly continuous on D if, ε >, δ ε such tht, (x, t), (x, t ) D with x x < δ ε nd t t < δ ε, f(x, t) f(x, t ) < ε. Lemm 26.2: If f is continuous on D, then it is lso uniformly continuous if D is closed. Proof: The proof is left s n exercise. Hint: Fix t [c, d], nd consider f(, t) : [, b] R with f : x f(x, t). Exercise: Prove tht, if f is continuous on D, then f(, t) is continuous on [, b] (so f(, t) C[, b]). Lemm 26.: If mp I is such tht I : t [c, d] f(x, t) dx then, if f C(D), I C[c, d] (I is often clled the integrl depending on prmeter). Proof: Let (t n ) n [c, d] with lim n t n t, nd let f n : [, b] R with f n : x f(x, t n ). Now, f is continuous on D, so f is uniformly continuous on D since D is closed, so, ε >, δ ε> such tht, if x x < δ ε nd t t < δ ε, then f(x, t) f(x, t ) < ε. Since (tn ), (t n+m ) t s n, then, eventully in n, t n t n+m < δ ε, so f n (x) f n+m (x) f(x, t n ) f(x, t n+m ) < ε eventully in n, so (f n ) is uniformly Cuchy, so (f n ) converges uniformly to f. Now, lim I(t n) lim n n f(x, t n ) dx lim n f n (x) dx lim f n(x) dx n f(x, t ) I(t ) So I is continuous t t, nd since t [c, d] ws rbitrry, I is continuous on [c, d]. Proposition 27: If f, f t re continuous on [, b] [c, d], then, t (c, d), t f(x, t) dx f (x, t) dx t Or, lterntively, t (c, d), F (t) f(x, t) dx nd G(t) both exist on (c, d), F is differentible on (c, d), nd F G. f (x, t) dx t 25
Proof: Fix t (c, d). Then f(, t), f t (, t) C[, b], so F nd G exist. Now, [c, d ] [c, d] such tht t (c, d ). But f t is continuous on [, b] [c, d], nd so is uniformly continuous on [, b] [c, d]. Then F (t + h) F (t) G(t) h f(x, t + h) f(x, t) f (x, t) dx h t MVT f f (x, τ) (x, t) dx τ t for some τ (t, t + h). Now, f τ is uniformly continuous on [, b] [c, d], so ε >, δ ε > such tht h with h < δ ε, f τ (x, τ) f t (x, t) < ε, so f τ (x, τ) f t (x, t) dx < ε ε(b ) so so F (t) G(t). lim F (t + h) F (t) h h G(t) Theorem 28 - Fubini s Theorem: If f : D R is continuous, then ( b ) d ( d ) b f(x, y) dy dx f(x, y) dx dy c c Proof: Let F (t) t for t [, b]. Now, F (), nd ( ) d f(x, y) dy dx c F (t) FTC d c f(t, y) dy d c d c ( ) t f(x, y) dx dy f(t, y) dy F is continuous on [, b] nd differentible on (, b), nd F (t), so F (b) F (), so ( b ) d ( d ) b f(x, y) dy dx f(x, y) dx dy c c Note: f must be continuous on the whole of D, or counterexmples such s the following cn rise: Let D [, 2] [, ], nd let f(x, y) : D R with xy(x 2 y 2 ) f : (x, y) (x 2 +y 2 ) (x, y) (, ) 3 (x, y) (, ) Then 2 ( ) f(x, y) dy dx 5 2 ( ) 2 f(x, y) dx dy 26
7.3 Uniform Convergence & Differentition Note: Suppose tht (f n ) n is sequence of functions on [, b], nd tht (f n ) f uniformly s n. Then, even if every f n is differentible on [, b], f is not necessrily differentible on [, b]. And even if f is smooth, (f n) does not necessrily converge to f. Exmple: Suppose tht f n (x) n cos nx on R. This function is smooth, nd f n(x) sin nx. Let f : x. Then f n f f n n cos n n s n, so (f n) f uniformly s n. All f n s re smooth, nd f is smooth, since f. But (f n) s n, in fct, (f n) does not converge t ll, except for t certin points such s x kπ for k N. Exmple: Let f n : R R, with f n : x x 2 + n. Then (f n(x)) x pointwise s n. Let f(x) x. Then fn (x) 2 f(x) 2 x 2 + n x2 n. fn (x) f(x) f n(x) 2 f(x) 2 f n(x)+f(x) n n n s n, so (f n ) f uniformly s n. Now, ll f n s re smooth, nd (f n ) f uniformly s n, but f is not differentible t. Note: If function f : [, b] R is continuously differentible, it is clled C function, or, lterntively, f C [, b] is written, where C [, b] is the spce of continuously differentible functions on [, b]. Theorem 29: Let (f n ) be sequence of C functions on [, b], with (f n ) f pointwise s n on [, b], nd suppose tht (f n) converges uniformly. Then f C [, b] nd lim n (f n) f. Proof: Let g lim n (f n). Now, n N, f n C[, b], so g C[, b] since (f n) g uniformly, so so, since x g x lim (f n) lim n n x f n f(x) f() + FTC lim n [f n(x) f n ()] f(x) f() x x g is continuous, so by the FTC, f g lim n (f n ), so g ( ) lim (f n) f FTC lim (f n) n n 27
8 Functionl Series 8. The Weierstrss M-Test & Other Useful Results Note: Let (f n ) n be sequence of functions on A R. Then thei functionl series k f k converges pointwise (or uniformly) on A if the sequence of prtil sums S n ( n k f k) n converges pointwise (or uniformly). Theorem 24 : Let (f n ) n R[, b]. Suppose tht S n n k f k converges uniformly. Then k f k R[, b] nd f k k k f k Proof: If (f k ) R[, b], then S n n k f k R[, b] n N, so k f k k f k. Theorem 25 : Suppose tht (f n ) n C[, b] such tht S n n k f k converges uniformly. Then k f k C[, b], nd, x [, b], lim x x k f k (x) lim f k (x) x x k Proof: S n C[, b] n N, so lim x x k f k(x) k lim x x f k (x). Theorem 29 : If (f n ) n C [, b] nd S n n k f k converges uniformly on [, b], then d dx f k (x) k k d dx f k(x) Proof: The proof is left s n exercise. Theorem 3 - Weierstrss M-Test: Let (f n ) n be sequence of functions on A R. Then, if (M k ) k R such tht, x A, fn (x) < Mn nd k M k converges, then n f n converges uniformly. Proof: k M k converges, so ( n k M k) n is Cuchy, so, ε n > N εn such tht n, m > N εn, with n > m, n k M k m k M k < εn. Then n km+ M k < ε n, so, since M K, n km+ M k < ε n. Therefore m f k (x) f k (x) f k (x) f k (x) M k < ε n k k km+ km+ km+ This holds x A, so the sequence of prtil sums of f k is uniformly Cuchy, so k f k converges uniformly to its pointwise limit. Corollry 3.: If (f n ) n is such tht k f k converges, then k f k converges uniformly. 28
8.2 Tylor & Fourier Series Motivtion: Tylor polynomils re of the form [ ] S n k! f (k) ()(x ) k nd were studied in Anlysis II. Fourier series re of the form 2 + k ( k cos kx + b k sin kx), x [, 2π] k Applictions of functionl series include solving PDEs, ODEs nd time series. Definition: For ny f R[, π], the series with coefficients given by k π b k π is clled the Fourier series generted by f. 2 + ( k cos kx + b k sin kx) k f(x) cos kx dx, k f(x) sin kx dx, k > Note: It is not immeditely obvious for which clsses of functions this series is convergent, or if it converges to f. Theorem 3: Let ( k ) k, (b k ) k R, with k k nd k b k convergent. Then 2 + k ( k cos kx + b k sin kx) converges uniformly on R. The pointwise limit f : R R is continuous. f is 2π-periodic (f(x + 2π) f(x)). Moreover, k π b k π so the terms of the series cn be recovered. f(x) cos kx dx, k f(x) sin kx dx, k > Proof (Convergence): x R, k cos kx + b k sin kx k cos kx + b k sin kx k + b k. M k. Then k M k k k + b k k k + k b k, so k M k converges, so the Fourier series converges by the M-test. Proof (Continuity): Let f(x). 2 + k ( k cos kx + b k sin kx). Then f C(R). Proof (Periodicity): Let S N (y). 2 + n k ( k cos kx + b k sin kx). Then f(x + 2π) f(x) lim n S n (x+2π) lim n S n (x). Both of these limits exist, so f(x+2π) f(x) lim n S n (x + 2π) S n (x), since cos (k(x + 2π)) cos kx nd sin (k(x + 2π)) sin kx. 29
Lemm: m Z e imx dx 2π(m ) where (m ) is the indictor function for m. Proof: For m Z \ {}: For m : So e imx dx cos mx + i sin mx dx e imx dx 2 sin πm cos mx dx + i sin mx dx m e dx e imx dx 2π(m ) dx 2π Proof (Recovery): so Now, nd f(x) dx f(x) cos mx dx 2 2 2 + ( k cos kx + b k sin kx) dx k [ ] π dx + k cos kx dx + b k sin kx dx π π sin kx cos mx dx k f(x) dx π f(x) cos x dx [ ] [ π ] π cos mx dx+ k cos kx cos mx dx + b k sin kx cos mx dx 4i k k 2i (eikx e ikx ) 2 (eimx e imx ) dx e i(k m)x e i(m k)x + e i(k+m)x e i(m+k)x dx (2π(k m ) 2π(m k )) 4i nd then so cos kx cos mx dx 4 2 (eikx + e ikx ) 2 (eimx e imx ) dx e i(k m)x + e i(m k)x + e i(k+m)x + e i(m+k)x dx 2π(((k m ) + (m k )) π(k m) 4 f(x) cos mx dx m π ( k π(k m)) π m k f(x) cos mx dx The proof for b k is similr nd so is left s n exercise. 3
Note: This proof relies on n ppliction of Theorem 24, where the sum is multiplied by bounded function. The proof tht the result holds is left s n exercise. Corollry 3.: If f : [, π] R is C 2 function, ( or f C[, π]), nd f(π) f(). Then the Fourier series generted by f converges to f uniformly. Note: It must be estblished tht (i) The Fourier series converges uniformly. (ii) The limit is f. However, (ii) is firly clunky rgument using existing tools, nd so shll be omitted nd left s n exercise, whilst (i) shll be proved. Proof: b k π So πk πk 2 f(x) sin kx dx f(x) cos kx dx πk πk f (x) cos kx dx πk 2 f (x) sin kx dx b k πk 2 f (x) sin kx dx πk 2 ( [f(x) cos kx ] π ( [f (x) sin kx ] π f(x) sin kx dx πk 2 f sin k 2π 2 k 2 f. c b k 2 f (x) cos kx dx ) f (x) sin kx dx nd c b is independent of k. A similr rgument estblishes tht k c k for k >, where c 2 is lso independent of k. k k converges, so the Fourier series converges by the M-test with M 2 k c+c b k 2 for k >. ) Exmple (Temperture Equilibrtion on Non-Uniformly Heted Ring): Consider the unit circle S. Let ny point on it be described by the ngle ϕ [, π] it mkes with the x-xis, nd let the initil temperture on the ring be given by f(ϕ) C 2 [, π] such tht f() f(π). Then the temperture T t time t t the point t ϕ, for t >, ϕ [, π], is given by the Fourier het eqution T t (t, ϕ) 2 T (t, ϕ) ϕ2 Since the point t π is the sme s the point t, there re some boundry conditions to consider lso: T (t, ) T (t, π) T T (t, ) (t, π) ϕ ϕ The initil condition is given by Throughout this exmple, T (, ϕ) f(ϕ) T. T t, T. T ϕ, T. 2 T ϕ 2 3
Now, (cos kϕ) k sin kϕ, nd (cos kϕ) k 2 cos kϕ, so A(t) cos kϕ seems suitble nstz to mke for solution. Substituting this into the het eqution yields A cos kϕ + k 2 A cos kϕ, so A(t) + k 2 A(t), which is n ODE. The boundry conditions re in fct stisfied by either this solution, or one using the nstz B(t) sin kϕ, but neither of these stisfy the initil conditions. Since finite liner combintion of ny of these functions will hve the sme shortcoming, it seems sensible to tke the next nstz s being n infinite liner combintion of them, such s T (t, ϕ) (t) 2 + ( k (t) cos kϕ + b k (t) sin kϕ) k Assume tht T, T, T nd T ll converge uniformly for t >, ϕ [, π]. Then substituting T into the het eqution nd differentiting pointwise yields So 2 + ( k cos kϕ + b k sin kϕ) + ( k k 2 cos kϕ + b k k 2 sin kϕ) k k 2 + [ ( k + k k 2 ) cos kϕ + ( b ] k + b k k 2 ) sin kϕ k which is the Fourier series generted by, so k + k 2 k b k + k 2 b k for k >. This yields infinitely mny ODEs, which require infinitely mny initil conditions. Expnding f(ϕ) C 2 [, π] by Fourier series yields f(ϕ) (f) 2 + k ( (f) k ) cos kϕ + b (f) k sin kϕ where n (f) k denotes the coefficient n k dependent on the function f. This gives the required mount of initil conditions. Then (t) (f) k (t) (f) t k e k2 b k (t) b (f) t k e k2 So T (t, ϕ) (f) 2 + k e k2 t ( (f) k ) cos kϕ + b (f) k sin kϕ Whilst this is solution, its uniqueness must lso be estblished, however tht shll not be done here. Exercise: Check tht T converges uniformly for t, nd tht T, T, T do so lso for t >. Note: Over time, the temperture becomes distributed evenly cross the whole ring, since lim t The pproch to this equilibrium is exponentil. (f) T (t, ϕ) 2 f(ϕ) dϕ Averge Temperture. 2π 32
Prt II Norms 9 Normed Vector Spces 9. Definition & Bsic Properties of Norm Definition : Let V be vector spce over R. Then function : V R, with : v V v which stisfies the following: (i) v V, v (positivity). Moreover, v v V (the bility to seprte points). (ii) λ R, v V, λv λ v (bsolute homogeneity). (iii) u, v V, u + v u + v (tringle inequlity). is clled norm on V. A pir (V, ) is clled normed vector spce, or normed spce. Remrks: (ii) nd (iii) use the liner structure of V. Positivity follows from (ii) nd (iii), since, v V, v bs. v v v v + v hom. v + v v + v 2 v, so v. Exmples:., the bsolute vlue, is norm on R. The tringle inequlity is stisfied since x + y x + y. The proofs of the other properties re left s n exercise. 2., the sup norm, is norm on B[, b]. (i) : f B[, b], if f, then sup x [,b] f(x), so, x [, b], f(x), so f on [, b], so cn seprte points ( f g f g, f g > f g). (ii) : λ R, f B[, b], λf sup x [,b] λf(x) supx [,b] λ f(x) λ supx [,b] f(x) λ f, so hs bsolute homogeneity. (iii) f, g B[, b], f + g sup x [,b] f(x) + g(x) sup x [,b] ( f(x) + g(x) ) sup x [,b] f(x) + sup x [,b] g(x) f + g, so f + g f + g, so stisfies the tringle equlity. Note: Norms generlise the notion of mgnitude of vectors in R n. Proposition 34:. 2. 3. Let x (x,..., x n ) R n ( vector). Then the following functions on R n re norms: x x 2 n i x i (the txicb or Mnhttn norm) i x i 2 (the Eucliden norm or Eucliden distnce) x mx i n x i 33
Proof:. Tringle inequlity: u + v u i + v i ( u i + v i ) i i i i u i + v i u + v 2. For ny u, v R n, u v u i v i u n v 2 cos θ u 2 v 2 i for some θ [, 2π]. This is known s the Cuchy-Schwrz inequlity for R n. Then So u + v 2 2 (u + v) (u + v) (u i + v i ) (u i + v i ) i u (u + v) + v (u + v) Cuchy u 2 u + v 2 + v 2 u + v 2 Schwrz u + v 2 2 u 2 u + v 2 + v 2 u + v 2 The cse where u + v is trivil to check, otherwise u + v 2 u 2 + v 2 3. u + v mx i n u i + v i mx i n ( u i + v i ) mx i n u i + mx i n v i u + v The verifiction of the other properties is left s n exercise. Remrk: All of these norms re instnces of the following fmily of norms: x p x i p i p, p, 2,... In prticulr, x lim p x p Tht this fmily of functions is fmily of norms follows from the Minkowski inequlity: x i + y i p i p x i p i p + y i p i p x, y R n Note : R n cn be thought of s the spce of functions on N n {, 2,..., n} by considering x : N n R, with x : j x j. 34
9.2 Equivlence & Bnch Spces Definition 2: Norms nd b on V re clled equivlent (or Lipschitz equivlent) if k, k 2 such tht, v V, k v b v k 2 v b. This equivlence is denoted by b. Remrk: The equivlence of norms is n equivlence reltion on the set of ll norms on V. Tht is,. (choose k k 2 ). 2. b b. 3. If b nd b c, then c. The proof is left s n exercise. Lemm 35:, 2 nd re ll equivlent on R n. Proof: Becuse equivlence is n equivlence reltion, it is enough to prove tht nd 2. Now, x R n, x mx i n x i x So with k, k 2 n. x i i mx x j n x j n i There is similr rgument for the second prt of the proof, lthough much more generl sttement will be proved shortly, nd so this shll be left s n exercise. Definition 3: Let (V, ) be normed spce. Then (i) A sequence (σ n ) n V converges to σ V if lim n σ n σ. (ii) A sequence (σ n ) n V is Cuchy if, ε >, N ε such tht, n, m > N ε, σ n σ m < ε. (iii) If every Cuchy sequence in V converges, then (V, ) is clled Bnch (or complete). A complete normed spce is lso clled Bnch spce. Remrks: Any Cuchy sequence on (R, ) converges, so (R, ) is Bnch. Consider (S[, b], ). Let (s n ) n S[, b] be uniform Cuchy sequence, so (s n ) is Cuchy with respect to. Then s n r R[, b] s n, so, in generl, (s n ) doesn t converge to point in S[, b] becuse not ll regulted functions re step functions, so (S[, b], ) is not Bnch. Note: All finite-dimensionl normed spces re complete. A proof for the cse when the spce is over R shll be given lter. Remrks:. If lim n σ n exists, then it is unique. Suppose tht σ n V nd σ n b V s n. Then b σ n + σ n b σn + σn b s n, so b, so b by seprtion of points. 2. (R[, b], ) is complete, becuse uniformly Cuchy sequence of regulted functions converges to regulted function. Also, (C[, b], ) nd (B[, b], ) re complete by results demonstrted erlier. 35
3. Suppose tht b. Then, if lim n σ n σ with respect to, then lim n σ n σ with respect to b. The proof is left s n (importnt) exercise. 4. If b, then (V, ) is Bnch iff (V, b ) is Bnch. This is essentilly consequence of the previous remrk. 5. (V, ) is Bnch iff ny series n σ n which converges bsolutely converges to n element in V. n σ n converges bsolutely if n σn converges. The proof of this is left s n exercise, lthough the direction is difficult. Theorem 36: All norms on R n re (Lipschitz) equivlent. Proof: Let (e,..., e n ) R n be the stndrd bsis of R n. Then, x R n, x n k x ke k. Now, x x k e k k n xk e k x k n n ek mx xj ek x ek j n k k k k Let k 2 n e k. Then k 2 such tht, x R n, x k 2 x. Now, let k { } x x J. inf, A. : x, B. { x : x x x x } Clerly B A by definition. If α A, then u, v R n such tht α u u u u v Then u v u u u so α B, so A B, so B A. Then J inf A inf B, nd clerly J. Suppose tht J. Then x, x 2, x 3,... R n with x k k (so lim k x k ) nd x k k. ) Now, it must be estblished tht there is convergent subsequence (x kp (x k ) k such tht ) p ( ) (x kp x (x (), x (2),..., x (n) ) R n s p with respect to. Let x k x () k, x(2) k,..., x(n) k. Since ( ) ( x k, x () k, so x () k is bounded, so by the Bolzno-Weierstrss theorem x () k k p ( ) )p x () k which converges to x () R. Applying this sme rgument to ech component of x k k ) ( ) yields (x kp such tht x (j) k p p converges to x )p (j) R for ny j n, so (x kp x ( p component-wise s p. Since x (j) k p x )p (j), then, ε >, N ε (j) such tht, n > N ε (j), x (j) k p x (j) x (j) < ε. Let Nε mx j n N ε (j). Then, k > N ε, x kp x mx j n k p x (j) < ε, ) xkp so lim p x, so (x kp x s p with respect to. p Now, x kp x kp x + x x kp x + x, so x x kp x > since xkp x s p, so x by the seprtion of points property of. Also, x x x kp + x kp xkp+ x xkp xkp+k2x xkp. (x kp ) (x k ), sox kp s p, nd it ws estblished tht x x kp s p, so xkp x + k2 xkp s p, so x, so x by the seprtion of points property of. However, this is contrdiction, so J >, so let k J. Then, from the definition of J, k x x x R n \ {}, nd if x, then x x, so k x x, so k x x x R n, so k x x k 2 x x R n, so on R n, so ll norms re equivlent on R n. 36