Analysis I Class notes

Transcription

1 UIC Mth 313 Anlysis I Clss notes John Wood This is course bout the foundtions of rel numbers nd clculus. 1. Rtionl numbers 2. Some consequences of the xioms Completeness 5. Sequences nd limits 6. Series 7. Series of nonnegtive terms 8. Limits of functions nd continuity 9. Functions continuous on n intervl 10. Inverse functions 11. Differentition 12. Functions differentible on n intervl 13. Integrtion 14. Integrbility 15. Properties of integrls 16. The fundmentl theorem of clculus 17. Drboux nd Riemnn integrls 1. Rtionl numbers We will ssume the rtionl numbers re fmilir nd begin by reclling some nottion nd fcts. N is the set of positive integers, N = {1, 2, 3,...}. Z is the set of integers, Z = {0, +1, 1, +2,...}. Q is the set of rtionl numbers, Q = {m/n : m Z, n N} with m/n = p/q if nd only if mq = np. A rtionl number r Q is positive if r = m/n with both m, n N. The rtionls stisfy the condition, clled trichotomy, tht for ech r Q either r is positive, r = 0, or r is positive (so r is negtive) nd only one of these is true. The sets of positive integers nd of positive rtionl numbers re closed under ddition nd multipliction. We write r > 0 if r is positive nd r > s if r s is positive. Note tht if q nd n re = qm pn qn which is positive positive, then m/n > p/q if nd only if qm > pn, since m n p q if nd only if qm pn is positive. Let us ccept Q s fmilir set with ddition, multipliction, nd the order, >. Also we will use rgument by induction nd the sttement tht ny nonempty set of positive integers hs smllest element. Q stisfies the following set of xioms. 1

2 Axioms for n ordered field + b = b + ( + b) + c = + (b + c) 0 s.t. + 0 = ( ) s.t. + ( ) = 0 b = b (b)c = (bc) 1 s.t. 1 = 0 1 s.t. 1 = 1 (commuttive) (socitive) (identity) (inverse) 1 0 (b + c) = b + c Exctly one of > 0, = 0, or > 0 is true. If > 0 nd b > 0, then + b > 0 nd b > 0. (nontrivility) (distributive) (trichotomy) (closure) The rel numbers, R, re lso n ordered field nd there re other exmples. Anything we prove from these xioms will be true of ny ordered field. 2. Some consequences of the xioms (1) x 0 = 0 nd 0 x = 0. Proof. Ech equl sign in the following string is consequence of one of the xioms. x 0 = x 0+0 = x 0+(x 0+( (x 0))) = (x 0+x 0)+( (x 0)) = x (0+0)+( (x 0)) = x 0 + ( (x 0)) = 0. (2) ( x) y = (x y) nd x ( y) = (x y). Proof. Now, in ddition to the xioms we my use previously proved sttement. (x y) = (x y) + 0 = (x y) + x (y + ( y)) = (x y) + (x y + x ( y)) = ( (x y) + x y) + x ( y) = 0 + x ( y) = x ( y) + 0 = x ( y), nd ( x) y = y ( x) = (y x) = (x y). (3) ( x) = x. Proof. ( x) = ( x) + 0 = ( x) + (( x) + x) = ( ( x) + ( x)) + x = 0 + x = x + 0 = x. (4) ( x) ( y) = x y, in prticulr ( 1) ( 1) = 1. Proof. ( x) ( y) = (( x) y) = ( (x y)) = x y. (5) 1 > 0. Proof. Since 1 0, by the first order xiom either 1 > 0 or 1 > 0. If 1 > 0, then by the second order xiom, ( 1) ( 1) > 0. Using (4) this implies 1 > 0 which, with the ssumption tht 1 > 0, contrdicts the first order xiom. Therefore 1 0 nd hence 1 > 0. If x > 0 we sy x is positive, If x > 0 we sy x is negtive. The sttement y is greter thn x is defined to men y x > 0; this is written y > x or equivlently x < y. 2

3 (6) Exctly one of x < y, x = y, or x > y is true. Proof. These sttements re equivlent to the sttements y x is positive, y x = 0, or (y x) is positive. Hence (6) is equivlent the first order xiom. The xiom nd (6) re clled trichotomy. (7) If x < y nd z is in F, then x + z < y + z. Proof. We need to show tht (y + z) (x + z) is positive when y x is positive. This follows from clcultion, using the sum xioms, tht (y + z) (x + z) = y x. (8) If x < y nd y < z, then x < z. Proof. By our hypothesis y x nd z x re positive. Then by the second order xiom the sum (z y)+(y x) positive. Using the sum xioms we clculte (z y)+(y x) = z x. Hence z x is positive, so x < z. (9) If x < y nd 0 < z, then xz < yz. Proof. By hypothesis, y x is positive nd z is positive, so the product (y x)z is positive. Using the distributive xiom nd (2), we find (y +( x))z = (yz)+(( x)z) = (yz)+( (xz)). It follows tht xz < yz. (10) If 0 < x nd x < y, then x 2 < y 2. Proof. By (9) x 2 < xy. By (8) 0 < y, so by (9) xy < y 2. Then by (8) x 2 < y 2. (11) If 0 < x nd 0 < y, then x 2 < y 2 x < y. Proof. We use the method of contrdiction. We ssume the hypotheses: 0 < x, 0 < y, nd x 2 < y 2. If x < y were flse, then either x = y or y < x by (6) If we suppose x = y, then x 2 = y 2. This contrdicts the hypothesis. If we suppose y < x, then by (10), with the roles of x nd y interchnged, we hve y 2 < x 2 which gin contrdicts the hypothesis. Therefore x < y. (12) If 0 < x nd 0 < y, then x 2 < y 2 x < y. Proof. This follows from (10) nd (11). (13) If xy = 0 then x = 0 or y = 0. This will be n exercise. It cn be proved for field using the existence of multiplictive inverses. It cn lso be proved for n ordered commuttive ring (the integers or polynomils over field) in which multiplictive inverses my be missing Here is n lgebric version of Pythgors s geometric proof tht the digonl of squre is not commensurble with its side. Theorem. There is no rtionl number whose squre is 2. Proof. Otherwise there is positive rtionl number whose squre is 2. We could write 2 = m/n in such wy tht n is the smllest possible denomintor in N. Then 2n 2 = m 2. We hve m > n becuse m n implies m 2 n 2 < 2n 2, contrdiction. Also m < 2n becuse m 2n implies m 2 4n 2 > 2n 2, gin contrdiction. Hence m n < n. Finlly, (2n m) 2 = 4n 2 4mn + m 2 = 2m 2 4mn + 2n 2 = 2(m n) 2. 3

4 Hence 2 = 2n m where the numertor nd denomintor re positive, but m n < n, m n contrdicting the choice of n s the smllest possible denomintor. This shows tht 2 is not rtionl number. If we regrd m/n s only n pproximtion to 2, for exmple 3/2 is not so fr from 2 since (3/2) 2 = 2 + 1/4, the rgument bove replces 3/2 by the worse pproximtion 1/1. We my reverse this process to get better pproximtions. Let = 2n m, b = m n. Solving for m nd n: m = + 2b, n = + b. Beginning with the estimte 1/1, we get the sequence of rtionl numbers: 1/1, 3/2, 7/5,..., where /b is followed by (+2b)/(+b). The sequence of squres is 2 1, 2+1/4, 2 1/25,..., sequence lterntely below nd bove 2 nd getting closer. In fct, if 2 2 b 2 = e b 2 then 2b2 2 = e nd 2( + b) 2 ( + 2b) 2 = b + 2b 2 2 4b 4b 2 = 2 2b 2 = e. ( ) 2 + 2b So 2 = e + b ( + b). 2 Strting with = 1, b = 1 we hve e = 1. The sequence is s 1 = 1/1, s 2 = 3/2, s 3 = 7/5, s 4 = 17/12,... for which we hve s 1 < s 3 < s 5 < < 2 s 2 > s 4 > s 6 > > 2. nd s 2n s 2n 1 0 s n This gives n explicit wy to find rtionl pproximtions to 2. The reder should crry these computtions further. 4. Completeness The set of rel numbers R is n ordered field contining the rtionls. The rels re chrcterized by n dditionl property clled completeness. In n ordered field we sy x y if x = y or x < y. It follows from (6) nd (8) in 2 tht the reltion stisfies the xioms: (1) x x, (2) x y nd y x implies x = y, 4

5 (3) x y nd y z implies x z, nd (4) for ll x nd y either x y or y x. The first three xioms define prtil order nd, with the fourth, totl order. Completeness Axiom. If A nd B re nonempty subsets of R nd if x A nd y B x y, then there exists z R such tht x A x z nd y B z y. Definition. A nonempty closed bounded intervl is subset of R of the form where nd b re rel numbers with b. [, b] = {x : x b} Definition. A sequence of such intervls, I n = [ n, b n ], is nested if m < n I m I n, or equivlently if m n b n b m. Cntor Intersection Theorem. If I 1 I 2 I 3 is sequence of nonempty closed bounded intervls, then n=1 I n is nonempty. Proof. Let I n = [ n, b n ] nd let A = { n : n N}, B = {b n : n N}. Then m m+n b m+n b n, hence A nd B stisfy the hypothesis of the completeness xiom. Hence there exists z R with n z b n, so for ll n, z I n. Therefore z n=1 I n. Let λ n = b n n be the length of I n. We sy lim n λ n = 0 if ( ε > 0)( k)(n k ε < λ n < ε). Corollry. If lim λ n = 0, then n n=1 I n contins unique element. Proof. If w nd z re both in n=1 I n with w < z, let ε = z w > 0. Then ( k)(n k ε < b n n < ε). Also n w < z b n, so ε = z w z n b n n < ε, contrdiction. Definition. An element b R is n upper bound for nonempty set A if x A, x b. The nonempty subset A R is bounded bove if there exists n upper bound for A, tht is b R such tht x A x b. Definition. We sy z R is lest upper bound for A if (1) z is n upper bound for A nd (2) if b is ny upper bound for A, then z b. Lemm. If A hs lest upper bound, it is unique. Proof. If z nd z 1 re both lest upper bounds for A, then z z 1 since z 1 is n upper bound nd z is lest upper bound. Also z 1 z since z is n upper bound nd z 1 is lest upper bound. Therefore z = z 1. 5

6 Theorem. If nonempty set A is bounded bove, then A hs lest upper bound. Proof. Let B = {y : y is n upper bound for A}. Then A nd B re nonempty by hypothesis. If x A nd y B, then x y. By the completeness xiom there exists z R such tht x A x z nd y B z y. Therefore z is n upper bound for A nd, if y is ny upper bound for A, then y B nd hence z y. Hence z is the lest upper bound of A. Property of Archimedes. N is not bounded bove in R. Proof. Assume N is bounded bove nd let b be its lest upper bound. Then b 1 is not n upper bound for N, so there exists n N with b 1 < n. But then b < n + 1, so b is not n upper bound, contrdicting the ssumption tht N is bounded bove. Corollry. ε > 0, n N, 1 n < ε. Proof. Since ε > 0, we hve ε 1 > 0. By Archimedes property, there is n n > ε 1. Therefore εn > εε 1 = 1. Thus 1 n < ε. 5. Sequences nd limits Definition. A sequence of rel numbers is function s from N to R; s(n) is usully written s n nd the sequence, s 1, s 2, s 3,... is written {s n }. A sequence {s n } is bounded bove if the set {s n : n N} is bounded bove, tht is if b n s n b. A sequence {s n } is incresing if s 1 < s 2 < s 3 <, tht is n N, s n < s n+1, nd it is nondecresing if s 1 s 2 s 3, tht is n N, s n s n+1. In either cse, by induction m < n s m s n. Exmples. (i) s 1 = 1, s 2 = 2, s 3 = 3,.... (ii) s 1 = 0, s 2 = 3/2, s 3 = 2/3,.... (iii) s 1 = 1, s 2 = 3/2, s 3 = 7/4,.... Listing the first few term of sequence is often not helpful. To be cler: (i) is incresing but not bounded, (ii) s n = 1 + ( 1)n n, (iii) s n = n 1. (ii) is bounded bove, but not incresing the terms oscillte, (iii) is bounded bove nd incresing. 6

7 Definition. The sequence {s n } converges to l if ( ε > 0)( m N)(n > m ε < s n l < ε). The sequence converges if there is n l R such tht {s n } converges to l, tht is, ( l R)( ε > 0)( m N)(n > m ε < s n l < ε). We write lim s n = l to men tht the sequence {s n } converges to l. We sy lim s n n n exists if the sequence converges. Exmple. Using this definition we show tht the sequence {1/n} converges to 0. Given ε > 0, by the Corollry to Archimedes property, we know there is n integer m with 1/m < ε. If n m, then 0 < 1/n 1/m < ε, hence ε < 1/n 0 < ε. Hence the definition is stisfied. Uniqueness of limits. If lim s n = l nd lim s n = k, then k = l. Proof. Otherwise we my ssume k < l, relbeling if necessry. Let ε = (l k)/2 > 0. There is n m such tht n > m ε < s n l, hence ε = ε + l k < s n l + l k = s n k. Therefore lim s n = k is flse contrdicting the ssumption k < l. The following result gives convergence without specifying the limit. Theorem. If sequence is nondecresing nd bounded bove, then the sequence converges. Proof. Let A = {s n : n N}. By hypothesis A is bounded bove. Hence A hs lest upper bound, cll it l. Then ( n)(s n l). If ε > 0, then l ε is not n upper bound, so ( m)(l ε < s m ). Since the sequence is nondecresing, if n m then s m s n. Therefore, Hence n m ε < s n l < ε. Exmple. Consider the sequence n m l ε < s m s n l < l + ε. 3, 3 + 3, ,... defined recursively by s 1 = 3 nd s n+1 = 3 + s n. This sequence is bounded bove: s 1 = 3 < 3 nd, if s n < 3 then 3 + s n < 6 < 9, so s n+1 < 3. The sequence is incresing. 3 < 3 + 3, so s 1 < s 2. If s n < s n+1, then 3 + s n < 3 + s n+1, so s n+2 < s n+1. Hence the sequence converges. At this point we do not know the limit. Knowing tht it exists will led to proof tht the limit is (1 + 13)/2. 7

8 We will use the definition nd three bsic properties of bsolute vlues. { if 0, Definition. = otherwise. + b + b, b = b, c c. The first property, the tringle inequlity cn be proved by considering the cses where nd b hve the sme, or different, signs. The second property follows from the closure xiom for <. The lst property follows from the first by tking c = + b. Theorem. If lim n n = nd lim n b n = b, then: (1) lim n ( n + b n ) = + b. (2) The set { n : n N} is bounded bove nd below. (3) lim n n b n = b. 1 (4) If n n 0 nd 0, then lim = 1 n n. (5) If n n b n, then b. (6) If = lim n n = lim n b n nd n c n b n, then lim n c n =. Proof (in five prts nd n exercise). (1) We hve n + b n ( + b) = ( n ) + (b n b) n + b n b. Given ε > 0 m s.t. n m n < ε/2 nd b n b < ε/2. Hence ( n + b n ) ( + b) < ε. (2) Tking ε = 1, m s.t. n m n < 1. Therefore n + n < + 1 when m n. Then the set { n : n N} is bounded bove by nd below by u. (3) We hve u = mx{ 1,..., m, + 1} n b n b = n b n n b + n b b n b n b + n b. Let M > 0 be bound for n, b n, nd b. This mens n, n < M, b n < M, nd b < M. Now given ε > 0, m s.t. n m n < ε Then n b n b < M + ε M = ε. 2M 2M 8 ε 2M nd b n b < 1 n.

9 (4) We first prove tht {1/ n } is bounded. Let ε = /2 > 0. Then m s.t. n m n < /2. Hence n < /2, so tht /2 < n or, equivlently, 1/ n < 2/ for n m. Let { } 1 M = mx 1,..., 1 m, 2. Then 1/ n < M for ll n. Next 1 1 n = n n = 1 1 n n 1 n M. Now, given ε > 0, Hence 1/ n 1/ < ε. m s.t. n m n < ε M. (5) This proof is by contrdiction. Assume > b nd let ε = ( b)/2 > 0. Then m s.t. n m n < ε/2 nd b n b < ε/2. Then b + n b n = n + b b n < ( b)/2. Since b < 0 by ssumption nd n b n 0 by hypothesis, b + n b n < 0 nd its bsolute vlue is b n + b n < ( b)/2 by the line bove. This implies ( b)/2 < n b n 0, contrdiction, therefore b. (6) This result, known s the pinching theorem, is left s n exercise. The finite geometric series is the sum Lemm 1. (x 1)s n = x n Series s n = 1 + x + x x n 1. Proof. s 1 = 1, so the sttement is true when n = 1. If we ssume the sttement is true for n, then (x 1)s n+1 = (x 1)(s n +x n ) = (x 1)s n +(x 1)x n = x n 1+x n+1 x n = x n+1 1. So by induction the sttement is true for ll n. Lemm 2. If x 1, then s n n nd x n 1 + (x 1)n. Proof. If x 1, then ech x j 1, so from the definition s n n. The second sttement follows from Lemm 1. Lemm 3. If x > 1, then the sequence {x n } is not bounded bove. Proof. If x > 1, then b m s.t. m > b/(x 1). Then n m x n > b. Lemm 4. If x < 1, then lim n x n = 0. Proof. If x = 0, then ech x n = 0. Assume 0 < x < 1. Then 1/ x > 1. Given ε > 0, by Lemm 3, m s.t. n m 1/x n > 1/ε, hence x n < ε. 9

10 Theorem. If x < 1 then lim s n = 1 n 1 x. Proof. lim s 1 x n n = lim n n 1 x = 1 1 x This result is lso written infinite series n=0 x n = 1 1 x ; s n = ( ) 1 lim x n = 1 n 1 x. n 1 x i is clled prtil sum of the i=0 x n. In generl, given sequence { n }, s m = n=0 prtil sum of the series n. n=1 m n is clled the mth Definition. If the sequence {s m } converges to l, we sy the infinite series n converges to l. Theorem. If n = nd n=1 b n = b, then n=1 n=1 ( n + b n ) = + b nd n=1 n=1 c n = c. Proof. The first prt follows from prt 1 of the Theorem of 4 pplied to the sequence of prtil sums {s m } nd {t m }. For the second prt let {t m } be the constnt sequence, t m = c nd use prt 3 of the Theorem. Two widely known geometric series re the deciml expnsion which is = n 1 = /10 = 1 nd the sum = 1. If n 0 we sy n=0 7. Series of nonnegtive terms n is series of nonnegtive terms. In this cse the sequence of n=0 prtil sums {s m } is nondecresing. If {s m } is bounded bove, then the series converges by 5. 1 For exmple consider the series. Computing the first severl prtil sums n(n + 1) n=1 suggests s m = 1 1/(m + 1) which cn be proved by induction. Hence {s m } is incresing nd bounded bove by 1. Thus the series converges. In fct we know tht lim s m = 1, so the series converges to 1. n=1 10

11 1 Now consider the series. Explicit prtils sums re not reveling. Compring n2 n=1 prtils sums with the previous series, we hve m m < (m 1) m = m < 2. Agin this implies this sum converges to limit between 1 nd 2. The limit cn be shown to be π 2 /6. 1 For the series n! = , the nth term, 1 n! < 1 (n 1)n. Compring n=0 terms we hve ! n! (n 1)n < 3. Hence this series converges to number, e, between 2 nd 3. To prove from the definition tht series converges we need to know the limit l of the prtil sums. Often the limit is not known but, if it exists, is some new object of mthemticl interest bout which we would like to know more. In tht cse we need some theoreticl rgument to show convergence. The result used bove, clled the comprison theorem, cn be stted s follows: m Theorem. If n 0 nd if b s.t. m, n < b, then n converges. Definition. lim x f(x) = l mens n=1 n=1 8. Limits of functions nd continuity ( ε > 0)( δ > 0)(0 < x < δ f(x) l < ε). The set {x : f is defined t x} is clled the domin of f, written dom f. If x / dom f then f(x) l < ε is flse. Hence for lim f(x) to exist, there must be some δ > 0 with x {x : 0 < x < δ} dom f. The vlue of f t, or even whether f is defined t, does not mtter. Uniqueness of limits. If lim f(x) = l nd lim f(x) = k, then k = l. x x Proof. If k l, then let 2ε = l k > 0. Then ( δ > 0)(0 < x < δ f(x) 0l < ε, f(x) k < ε), hence 2ε = l k = l f(x) + f(x) k l f(x) + f(x) k < 2ε, contrdiction.. Hence k = l. 11

12 Theorem. Let lim x f(x) = l nd lim x g(x) = m. Then: (1) lim x (f(x) + g(x)) = l + m. Proof. Given ε > 0, ( δ 1 > 0)(0 < x < δ 1 f(x) l < ε/2) nd ( δ 2 > 0)(0 < x < δ 2 g(x) m < ε/2). Tke δ = min{δ 1, δ 2 }. Then 0 < x < δ implies f(x) + g(x) l m f(x) l + g(x) m < ε. (2) ( δ > 0)(0 < x < δ f(x) < l + 1). Proof. Let ε = 1 then ( δ > 0)(0 < x < δ f(x) l < 1). Hence f(x) = l + f(s) l l + f(x) l < l + 1. (3) lim x (f(x)g(x)) = lm. Proof. Tking the minimum of three vlues for δ we find δ > 0 such tht 0 < x < δ implies f(x) < l +1, nd g(x) m < ε/(2( l +1)), nd f(x) l < ε/(2( m +1)). Then f(x)g(x) lm = f(x)g(x) f(x)m + f(x)m lm f(x) g(x) m + f(x) l m ε < ( l + 1) 2( l + 1) + ε 2( m + 1) m < ε. (4) If l 0, ( δ > 0)(0 < x < δ f(x) > l /2). Proof. Since l /2 > 0, ( δ > 0)(0 < x < δ f(x) l < l /2). Then l f(x) l f(x) < l /2, so l /2 < f(x). 1 (5) If l 0, then lim x f(x) = 1 l. Proof. Given ε > 0, tke δ > 0 so tht 0 < x < δ implies f(x) > l /2 nd f(x) l < εl 2 /2. Then 1 f(x) 1 f(x) l l = < ε. l f(x) Definition. f is continuous t if lim x f(x) = f() Hence f is continuous t if ( ε > 0)( δ > o)( x < δ f(x) f() < ε). 12

13 Proposition. A constnt function, f(x) = c, nd the identity function, f(x) = x, re continuous t every R. Proof. For f(x) = c, given ε > 0, for ny nd ny x, f(x) f() = c c = 0 < ε. For f(x) = x we my tke δ = ε since then, if x < δ we hve x < ε. Corollry. Any polynomil function is continuous t ll R. Any rtio of polynomils is continuous t ny which is not root of the denomintor. Proposition. If lim s n = l nd f is continuous t l, then lim f(s n ) = f(l). n n Proof. Given ε > 0, ( δ > 0)( x < δ f(x) f(l) < ε) nd ( k)(n k s n l < δ). Hence n k implies f(s n ) f(l) < ε. Therefore lim f(s n ) = f(l). n Proposition. If f() = b, f is continuous t, nd g is continuous t b, then g f is continuous t. Proof. Since g is continuous t b, given ε > 0, δ > 0 such tht y b < δ g(y) g(b) < ε. Then since f is continuous t nd δ > 0, γ > 0 such tht x < γ f(x) f() < δ. Hence g(f(x)) g(f()) < ε provided x < γ nd therefore g f is continuous t. 9. Functions continuous on n intervl Intermedite Vlue Theorem. If f is continuous on [, b] nd f()f(b) < 0, then there exists z [, b] such tht f(z) = 0. Proof. Assume tht f() < 0 nd f(b) > 0. (In the opposite cse, replce f by f.) Let I 0 = [, b]. We construct recursively nested sequence of nonempty, bounded, closed intervls. If I n = [ n, b n ] hs been constructed with f( n ) < 0 nd f(b n ) > 0, let x = (b n n )/2 be its midpoint. Define I n+1 by n+1 = n, b n+1 = x if f(x) > 0, n+1 = x, b n+1 = b n if f(x) < 0. (In the cse f(x) = 0, we my tke z = x nd the proof is complete.) By the Cntor intersection theorem there exists z n=1 I n. We clim tht f(z) = 0. If f(z) > 0, then there is δ > 0 such tht 0 < x z < δ f(z) < f(x) f(z) < f(z), nd therefore f(x) > 0. Further there exists n such tht b n n = (b )/2 n < δ. Now n z b n, so 0 z n b n n δ, so f( n ) > 0 which contrdicts the construction of n. Similrly, f(z) < 0 leds to contrdiction. Therefore f(z) = 0. Exmple. Given c > 0, let f(x) = x 2 c nd tke b = mx{c, 2}. Then f(0) = c < 0 nd, in either cse, f(b) > 2 > 0. By the intermedite vlue theorem, there exists z > 0 with z 2 = c. Definition. A function f is bounded bove on [, b] is there exists k such tht x b f(x) k. 13

14 Boundedness Theorem. If f is continuous on [, b], then f is bounded bove on [, b]. Lemm. Suppose < c < b. If f is bounded bove on [, c] nd on [c, b], then f is bounded bove on [, b]. Proof. By hypothesis, ( k 1 )( x c f(x) k 1 ) nd ( k 2 )(c x b f(x) k 2 ). Then if x b, either x c nd f(x) k 1 or c x nd f(x) k 2. Therefore f(x) mx{k 1, k 2 }. Proof of theorem. Sy f is not bounded on I 0 = [, b]. We construct recursively nested sequence of nonempty, bounded, closed intervls on which f is unbounded. If I n = [ n, b n ] hs been constructed nd f is unbounded on I n, let x = (b n n )/2 be its midpoint. By the lemm, f is unbounded on t lest one of the intervls [, x] or [x, b]. Let I n+1 be one of these on which f is unbounded. By the Cntor intersection theorem there exists z n=1 I n. Since f is continuous t z, there exists δ > 0 such tht x z < δ f(x) f(z) + 1. But there exists n such tht b n n = (b )/2 n < δ nd n z b n. If n x b n, then x z b n n < δ so f(x) f(z) + 1. Hence f is bounded bove on I n, contrdicting the construction. Therefore f is bounded bove on [, b]. Extreme Vlue Theorem. m, M, c 1, c 2 such tht If f is continuous on [, b], then there exist numbers m f(x) M for x [, b], f(c 1 ) = m, f(c 2 ) = M, where c 1, c 2 [, b]. Proof. We prove this for M nd c 2 nd pply this result to f to get m nd c 1. By the previous theorem the set {f(x) : x [, b]} is bounded. Hence there is lest upper bound M. Suppose for ll x [, b], f(x) M. Then f(x) < M nd hence the function g(x) = 1 M f(x) for x [, b] is continuous. Agin by the previous theorem, there is bound k such tht 0 < 1 M f(x) k nd hence M f(x) 1 k or f(x) M 1 k < M. This contrdicts the fct tht M is the lest upper bound. Therefore there is c 2 with f(c 2 ) = M. 10. Inverse functions Let A nd B be subsets of R nd let f be function with domin A which tkes vlues in B, f(x) B. We write f : A B nd cll f function, or mp, from A to B. Definitions. The imge of f, im f = f(a) = {y : x A, f(x) = y}; 14

15 f is surjective or onto B if im f = B; f is injective or one-to-one if f(u) = f(x) u = x; The identity mp 1 A : A is defined by 1 A (x) = x for ll x A; A mp g : B A is n inverse of f if g f = 1 A nd f g = 1 B. Notice tht f(x) = y if nd only if g(y) = x. If f : A B hs n inverse, the inverse is unique, nd if g : B A is n inverse of f, then f is n inverse of g. Proposition. If (g f) = 1 A, then f is one-to-one nd g is onto. Proof. If f(u) = f(x), then u = (g f)(u) = g(f(u)) = g(f(x)) = (g f)(x) = x, so f is one-to-one. If x A, let y = f(x). Then g(y) = g(f(x)) = x, so g is onto. Proposition. f : A B hs n inverse if nd only if f is one-to-one nd onto. Proof. If f hs n inverse, g, then (g f) = 1 A, so f is one-to-one. Also f g = 1 B, so f is onto by the previous proposition. Conversely, given y B, if f is onto, x A with f(x) = y. Since f is one-to-one, this x is unique. Thus g is well-defined if we set g(y) = x. Our gol is to prove tht if f is continuous nd hs inverse, then the inverse is continuous. A more generl type of intervl thn the nonempty, closed, bounded intervls used so fr is given by the following Definition. An intervl is set I such tht, if, b I nd x b, then x I. Theorem. If f is defined nd continuous on n intervl I nd f is one-to-one, then f is either incresing or f is decresing on I. Proof. The proof is n ppliction of the intermedite vlue theorem. Let, b, c, d I, < b, c < d, nd suppose tht f() < f(b). (If not, replce f by f.) We must prove tht f(c) < f(d). We define fmily of intervls [x t, y t ] for 0 t 1 with x 0 =, y 0 = b nd x 1 = c, y 1 = d. Set x t = (1 t)+tc nd y t = (1 t)b+td. Note tht y t x t = (1 t)(b )+t(d c) > 0 for 0 t 1, so x t < y t. Define h : [0, 1] R by h(t) = f(x t ) f(y t ); h is continuous function of t since it is composition of continuous functions. Since f is one-to-one, h(t) 0. Now h(0) = f() f(b) < 0. Then, by the intermedite vlue theorem pplied to h on the intervl [0, 1], we must hve f(c) < f(d), hence f is incresing. Lemm. If f is continuous on n intervl I, then J = f(i) is n intervl. Proof. Given u, v J,, b I such tht f(u) =, f(b) = v. Since I is n intervl, [, b] I so f is defined on [, b]. If y [u, v], then f() y f(b), so by the intermedite vlue theorem x [, b] with f(x) = y. Therefore y J nd hence J is n intervl. Lemm. If f : I J nd g : J I re inverse functions nd f is incresing, then g is lso incresing. Proof. Given u, v J with u < v, let g(u) =, g(v) = b. If b then u = f() f(b) = v, contrdiction, hence < b nd g is incresing. 15

16 Theorem. If f : I J nd g : J I re inverse functions nd f is continuous, then g is lso continuous. Proof. Assume tht f is incresing (nd pply this result to f if f is decresing). Given y J we must show the lim v y g(v) = g(y). Given ε > 0 we need δ > 0 such tht or equivlently v y < δ g(v) g(y) < ε, y δ < v < y + δ g(y) ε < g(v) < g(y) + ε. Let y = f(x) nd v = f(u). Since x ε < x < x + ε nd f is incresing we hve f(x ε) < f(x) < f(x + ε). Let δ = min{f(x + ε) f(x), f(x) f(x ε)}. If y δ < v < y + δ, then f(x ε) = y ( f(x) f(x ε) ) y δ < v < y + δ y + ( f(x + ε) f(x) ) = f(x + ε). Since g is incresing nd x = g(y) we hve g(y) ε < g(v) < g(y) + ε. Exmples. The power functions f n (x) = x n, n > 0, re incresing on R if n is odd nd on [0, ) if n is even. The inverse functions re n x. The compositions x m/n re continuous. For nonrtionl exponents further limit rgument, or the definition x = exp( ln x) for x > 0 is needed. Let f(x) = 3 + x. The sequence defined in 5 by s 1 = 3 nd s n+1 = f(s n ) ws shown there to converge to rel number l [1, 3]. Since f is continuous on [0, ), f(l) = f(lim s n ) = lim f(s n ) = lim s n+1 = l. By the qudrtic formul, l = (1 + 13)/2. Definition. The derivtive of f t is 11. Differentition f f(x) f() () = lim. x x If this limit exists, f is sid to be differentible t. Exmples. If f(x) = c for some rel number c, then f () = 0 for ny. Recll the finite geometric series from 6. In replce x with x/ nd multiply by n to get (x 1)(1 + x + x x n 1 ) = x n 1, (x )( n 1 + n 2 x + + x n 1 ) = x n n. ( ) 16

17 Set n 1 ϕ (x) = n 1 i x i. i=0 Then ϕ is continuous t (nd for ll x). If x, then ϕ (x) = (x n n )/(x ). Hence x n n lim x x = lim ϕ (x) = n n 1. Thus if f(x) = x n, f () = n n 1. x Next divide eqution ( ) by n x n for, x 0 to get Thus for, x 0, (x )( 1 x n + 2 x n n x 1 ) = n x n. (x n n )/(x ) = (x) 1 ϕ 1(x 1 ) = ψ (x) nd ψ is continuous t x = 0. Thus if f(x) = x n, f () = n n 1. Finlly, replce x by x 1/n nd by 1/n in eqution ( ) to find x 1/n 1/n x = 1 ϕ 1/n(x 1/n ) nd deduce tht the derivtive of f(x) = x 1/n = n x is f () = 1 n 1 n 1. In the three cses bove we introduced functions which were equl to the difference quotient for x nd which were continuous t. Constntin Crthéodory gve n equivlent reformultion of the definition of the derivtive in terms of such function. His definition leds to shorter proofs nd mkes the use of continuity more explicit. Crthéodory s Theorem. f is differentible t if nd only if there is function ϕ (x) which is continuous t such tht f(x) = f() + (x )ϕ (x). If f is differentible, then f () = ϕ (). Proof. Assume such function ϕ (x) exists. Then (f(x) f())/(x ) = ϕ (x) for x nd lim(f(x) f())/(x ) = lim ϕ (x) = ϕ (). x x Conversely, ssume f is differentible t nd f () = l Define { (f(x) f())/(x ) if x, ϕ (x) = l if x =. Then lim ϕ (x) = lim(f(x) f())/(x ) = l, so ϕ (x) is continuous t. x x Crthéodory s definition is the subject of n rticle by Stephen Huhn in the Americn Mthemticl Monthly 98 (Jn. 1991) pp

18 Theorem. If f is differentible t then f is continuous t. Proof. By Crthéodory, f(x) = f() + (x )ϕ (x) where ϕ (x) is continuous t. Then by the results of 8, f is continuous t. Sum rule. If f nd g re differentible t, then f + g is differentible t nd (f + g) () = f () + g (). Proof. We hve f(x) = f() + (x )ϕ (x) nd g(x) = g() + (x )ψ (x), hence (f + g)(x) = (f + g)() + (x )(ϕ (x) + ψ (x)). By 8, ϕ + ψ is continuous t nd (ϕ + ψ )() = ϕ () + ψ (). Product rule. If f nd g re differentible t, then fg is differentible t nd Proof. As bove, (fg) () = f ()g() + f()g (). (fg)(x) = ( f() + (x )ϕ (x) )( g() + (x )ψ (x) ) = (fg)() + (x ) ( ϕ (x)g() + f()ψ (x) + (x )ϕ (x)ψ (x) ) where ϕ (x)g() + f()ψ (x) + (x )ϕ (x)ψ (x) is continuous t nd t x = is equl to f ()g() + f()g (). Corollry (Linerity). (kf + lg) = kf + lg for k, l R. Chin rule. If f is differentible t, f() = b, nd g is differentible t b, then the composition g f is differentible t nd (g f) () = g (b)f (). Proof. (g f)(x) = g(f(x)) = g(b) + (f(x) b)ψ b (f(x)) = g(b) + ( f() + (x )ϕ (x) b ) ψ b (f(x)) = g(b) + (x )ϕ (x)ψ b (f(x)). Since f is continuous t, ψ b f is continuous t so (ψ b f)ϕ is continuous t nd ψ b (f())ϕ () = g (b)f (). Proposition. Let f be differentible t nd f() 0. Then (1/f) () = f ()/(f()) 2. Proof. If g(y) = y 1, g is differentible t b = f() nd g (b) = b 2. Then (g f)(x) = 1/f(x) nd (g f) () = (f()) 2 f (). Quotient rule. Let f nd g be differentible t with g() 0. Then ( ) f () = f ()g() f()g (). g g() 2 18

19 ( ) f Proof. Write (x) s the product f(x) (1/g(x)) nd use the previous proposition g nd the product rule. Inverse Function Theorem. If f : I J nd g : J I re inverses nd f is differentible t I with f () 0, then g is differentible t b = f() nd g (b) = 1/f (g(b)). Proof. We hve f(x) = f()+(x )ϕ (x) where ϕ is continuous t nd f () = ϕ (). Since g is continuous t b nd ϕ is continuous t g(b), ϕ g is continuous t b nd ϕ (g(b)) = f () 0. Hence there exists ε > 0 such tht when y b < ε, ϕ (g(y)) 0. Let y = f(x) nd x = g(y). Then y = b + (g(y) g(b))ϕ (x), hence g(y) = g(b) + (y b)/ϕ (g(y)). Therefore g (b) = 1/f (g(b)). Theorem. If f () > 0, then there is δ > 0 such tht δ < x < f() < f(x) < x < + δ f(x) < f(). Proof. Agin f(x) = f()+(x )ϕ (x), ϕ is continuous t, nd ϕ () > 0. Therefore there is δ > 0 such tht x < δ ϕ (x) > 0. If < x < + δ then x > 0 nd f() < f() + (x )ϕ (x) = f(x), while if δ < x < then x < 0 nd f() > f() + (x )ϕ (x) = f(x). Definition. A criticl point for f is number c such tht f (c) = 0. The criticl point test for extreme vlues is the following: Corollry. Let f be defined on (, b), < c < b, nd f be differentible t c. If f(x) f(c) for ll x (, b) or if f(x) f(c) for ll x (, b), then f (c) = 0. Thus if f is defined on [, b] nd c [, b] is n extreme point of f, then c is either n end point of [, b], point where f is not differentible, or criticl point of f. Proof. By contrdiction, pplying the theorem to either f or f. The conclusion of the previous theorem is sometimes described by sying tht f is incresing t. This does not imply tht f is incresing on ny intervl. An exmple of function which is incresing t 0, but not in ny intervl (0, δ) or ( δ, 0) for ny δ > 0 is { 0, if x = 0; f(x) = x/2 + x 2 sin(1/x), otherwise. This function is differentible t 0 nd in fct on R, but f is not continuous t 0. 19

20 12. Functions differentible on n intervl Rolle s Theorem. If f is continuous on [, b] nd differentible on (, b), nd if f() = f(b), then ξ (, b) such tht f (ξ) = 0. Proof. Since f is continuous on [, b], f hs n extreme vlue on [, b] by ( 9). If mximum point or minimum point occurs t ξ (, b), then by the criticl point test, f (ξ) = 0. If both the mximum nd minimum vlues occur t the end points then, since f() = f(b), f is constnt on [, b] nd f (ξ) = 0 for ny ξ (, b). Men Vlue Theorem. If f is continuous on [, b] nd differentible on (, b), then ξ (, b) such tht (f(b) f())/(b ) = f (ξ). f(b) f() Proof. Let h(x) = f(x) (x ). Then h() = f() nd h(b) = f(), so h b stisfies the hypotheses of Rolle s theorem. Therefore ξ (, b) such tht h (ξ) = 0. But h (ξ) = f (ξ) (f(b) f())/(b ). Corollry 1. If f is defined on n intervl I nd f (x) = 0 for ll x I, then f is constnt on I. Proof. Tke, b I nd pply the men vlue theorem. Corollry 2. If f (x) = g (x) on I then there is constnt c such tht g(x) = f(x)+c. Proof. Apply Corollry 1 to f g. Corollry 3. If f (x) > 0 on I then f is strictly incresing on I. If f (x) < 0 on I then f is strictly decresing on I. Proof. Let, b I with < b. By the men vlue theorem, there exists ξ (, b) with (f(b) f())/(b ) = f (ξ) > 0, hence f(b) > f(). For the second prt, pply the first prt to f. Corollry 4. If f nd g re continuous on [, b], f (x) g (x) on (, b), nd f() g(), then f(x) g(x) for ll x [, b]. Proof. For x = the conclusion is prt of the hypothesis. For x (, b] there is ξ (, x) such tht f(x) g(x) (f g)(x) (f g)() = (f g) (ξ) (x ) 0. Corollry 5. If f () = 0 nd f () > 0, then δ > 0 such tht for x ( δ, + δ) nd x, f(x) < f(). Proof. The existence of f () implies tht there is δ > 0 such tht f (x) exists for x < δ nd, by 10, tht δ < x < 0 < f (x) nd < x < + δ f (x) < 0. By Corollry 3, f is incresing on ( δ, ) nd decresing on (, + δ), nd therefore f(x) < f() for 0 < x < δ. 20

21 Cuchy Men Vlue Theorem. Let f nd g be continuous on [, b] nd differentible on (, b). Then ξ (, b) such tht (f(b) f())g (ξ) = (g(b) g())f (ξ). Note. If g (ξ) 0 nd g(b) g() we hve f(b) f() g(b) g() = f (ξ) g (ξ). Proof. Let h(x) = f(x)(g(b) g()) g(x)(f(b) f()). Then h is continuous on [, b], differentible on (, b), nd h() = h(b), so Rolle s theorem implies the result. l Hôpitl s Rule. Assume δ > 0 such tht when 0 < x < δ, f (x) exists, g (x) exists, nd g (x) 0. If (1) lim f(x) = 0, lim g(x) = 0, nd x x (2) lim f (x)/g (x) = l. x Then lim f(x)/g(x) = l. x Proof. The first three ssumptions re implicit in condition (2). If f or g is not defined or is not continuous t we my redefine them t by setting f() = 0 nd g() = 0. Then f nd g re continuous on ( δ, + δ) by (1) nd none of the limits s x goes to re chnged. Further g(x) 0 for x < δ, since, if g(x) = 0, then by the MVT ξ such tht ξ < x < δ nd g (ξ) = 0, contrdiction. Given ε > 0, by (2), δ 1, 0 < δ 1 δ such tht ξ, 0 < ξ < δ 1 f (ξ) g (ξ) l < ε. If 0 < x < δ 1, by the Cuchy MVT, ξ such tht ξ < x < δ 1 nd f(x) g(x) = f (ξ) g (ξ), therefore f(x) g(x) l < ε. Hence lim x f(x)/g(x) = l. 21

22 13. Integrtion Recll from 4 tht if A is nonempty set tht is bounded bove, then A hs lest upper bound. This bound is lso clled the supremum of A or sup A. By wy of review, we discuss the gretest lower bound in detil. This requires essentilly the sme rgument s for the lest upper bound, but it ws not given in 4. Let B be nonempty set tht is bounded below. Let A R be the set of ll lower bounds for B, A = { : b B b}. Since we ssumed tht B is bounded below, A is not empty. By the completeness xiom, z R such tht: (1) A z nd (2) b B z b. By (2) nd the definition of A, z A nd by (1), z is gretest lower bound for B. If z 1 were nother gretest lower bound for B, then z, z 1 A. This implies z z 1 nd z 1 z nd therefore z 1 = z, so the gretest lower bound is unique. It is clled the infimum of A or inf A. Note tht inf A sup A. Definition. A prtition P of the intervl [, b] is finite set of points, P = {x 0, x 1,..., x n } where the x i R re subscripted so tht = x 0 < x 1 <... < x i 1 < x i <... < x n = b. Definition. The function f is bounded on [, b] if m, M R s.t. x [, b] m f(x) M. Thus the set of vlues of f on [, b] is bounded. If f is bounded on [, b] it will be bounded on ny subset of [, b]. Definition. If f is bounded on [, b], the upper nd lower sums of f with respect to prtition of [, b] re defined by U(f, P ) = L(f, P ) = n M i (x i x i 1 ), i=1 n m i (x i x i 1 ), i=1 Lemm 1. L(f, P ) U(f, P ). Proof. This follows from m i M i. m i = inf{f(x) : x [x i 1, x i ]}, where nd M i = sup{f(x) : x [x i 1, x i ]}, for 1 i n. Lemm 2. If the prtition Q contins one more point thn the prtition P, then L(f, P ) L(f, Q) U(f, Q) U(f, L). 22

23 Proof. Sy the extr point is u nd t i 1 < u < t i. Then if m i = inf{f(x) : x [x i 1, u]} m i nd m i = inf{f(x) : x [u, x i ]} m i we hve m i(u x i 1 ) + m i (x i u) m i (x i x i 1 ). This with Lemm 1 gives the result first inequlity. The second is Lemm 1. The third is nlogous to the first, replcing m by M nd reversing the inequlities. We sy the prtition Q is refinement of P if Q is obtined from P by inserting finite number of points. As finite sets, P Q. A finite number of pplictions of lemm 2 shows tht L(f, P ) L(f, Q) U(f, Q) U(f, P ). If P nd Q re ech prtitions of [, b] then P Q contins ny point in either P or Q nd is common refinement of both. Lemm 3. If P nd Q re prtitions of [, b], then L(f, P ) U(f, Q). Proof. L(f, P ) L(f, P Q) U(f, P Q) U(f, Q). Fixing n intervl [, b] nd function f bounded on [, b], Lemm 3 sys tht ny upper sum U(f, Q) is n upper bound for the set of ll lower sums, so sup{l(f, P ) : P prtition of [, b]} U(f, Q). Now sup{l(f, P ) is lower bound for the set of ll upper sums. This proves the Theorem 1. sup{l(f, P )} inf{u(f, P )}. Definition. Let f be bounded on [, b], then f is integrble on [, b] if sup{l(f, P )} = inf{u(f, P )}, nd this number is, by definition, the integrl of f over [, b], written The integrl defined this wy is clled the Drboux integrl. Exmples. () If f(x) = c, then f(x) dx = c(b ). { 0, x irrtionl, (b) If f(x) = then f is not integrble. 1, x rtionl; f(x) dx. Theorem 2. Let f be bounded on [, b], then f is integrble if nd only if for ll ε > 0 there is prtition P of [, b] such tht U(f, P ) L(f, P ) < ε. Proof. Let P be such prtition for given ε > 0. Since L(f, P ) sup{l(f, Q)} inf{u(f, Q)} U(f, P ), we hve sup{l(f, Q)} inf{u(f, Q)} < ε. Since this is true for ny ε > 0, sup{l(f, Q)} = inf{u(f, Q)}. If f is integrble, then sup{l(f, Q)} = inf{u(f, Q)}, so for ny ε > 0 there re prtitions P nd P such tht sup{l(f, Q)} L(f, P ) < ε/2 nd U(f, P ) sup{l(f, Q)} < ε/2. Hence U(f, P ) L(f, P ) < ε. Then, tking P = P P, we hve U(f, P ) L(f, P ) U(f, P ) L(f, P ) < ε. 23

24 14. Integrbility Theorem 1. If the function f is bounded nd monotone on the intervl [, b], then f is integrble on [, b]. Proof. Suppose f is nondecresing on [, b], tht is, if u v b then f(u) f(v). Given ε > 0 choose n > (f(b) f())(b )/ε nd let P n be the prtition of [, b] with division points x i = + i(b )/n for i = 0... n. If x i 1 x x i, then f(x i 1 ) f(x) f(x i ), hence m i = f(x i 1 ) nd M i = f(x i ). Then U(f, P n ) L(f, P n ) = Σ n i=1(f(x i ) f(x i 1 ))(b )/n = (f(b) f())(b )/n < ε. By Theorem 13.2, this implies tht f is integrble. Exmples. (1) The step function: f(x) = { 0, if x < 0; 1, if x 0. (2) The function f(t) = 1/t is decresing for t > 0 so f is integrble on [1, x] or [x, 1] for x > 0 nd we cn define log x = x 1 1 dt for 1 x or t 1 x 1 dt for 0 < x 1. t Theorem 2. If the function f is continuous on the intervl [, b], then f is integrble on [, b]. Proof. Since f is continuous on [, b], f is bounded nd, for ny prtition P, the upper nd lower sums re defined. Given ε > 0 we must show there is prtition P with U(f, P ) L(f, P ) < ε. If for ech intervl in the prtition we hve then M i m i < µ = ε/(b ), U(f, P ) L(f, P ) = Σ n i=1(m i m i )(x i x i 1 ) < µ(b ) = ε. Suppose [, b] is divided into two intervls, [, c] nd [c, b] nd tht P 1 is prtition of [, c] nd P 2 is prtition of [c, b]. If M i m i < µ for intervls in the prtition P 1 nd for intervls in the prtition P 2, then this inequlity will hold for intervls in the prtition P 1 P 2 of [, b]. If we ssume there is no prtition of [, b] for which this inequlity is true for ech intervl, then either there is no such prtition of [, c] or else there is no such prtition of [c, b]. Tking c = ( + b)/2 we get n intervl [ 1, b 1 ] equl to either [, c] or [c, b] for which there is no such prtition. Inductively we get nested sequence of closed intervls [, b] [ 1, b 1 ] [ 2, b 2 ] 24

25 such tht for ech [ k, b k ] there is no prtition for which the inequlity holds. Further the lengths of the intervls tend to zero since b k k = 2 k (b ). The intersection of nested sequence of closed intervls is nonempty; let c be point in the intersection. Since f is continuous t c there is δ such tht f(x) f(c) < µ/2 for ny x with x c < δ. If b k k < δ then c [ k, b k ] nd, if x [ k, b k ] then x c < δ, hence f(x) f(c) < µ/2. Since f is continuous on [ k, b k ], m = inf{f(u) : k u b k } = f(x 1 ) nd M = sup{f(u) : k u b k } = f(x 2 ) for some x 1, x 2 [ k, b k ]. Then M m = f(x 2 ) f(x 1 ) f(x 2 ) f(c) + f(c) f(x 1 ) < µ. If we tke the prtition P = { k, b k } of [ k, b k ], then P stisfies the inequlity M m < µ for the only intervl in P which contrdicts the wy [ k, b k ] ws chosen. Therefore there is prtition of [, b] with M i m i < µ for ech intervl nd f is integrble. 15. Properties of integrls Theorem 1. If < c < b, then f is integrble on [, b] if nd only if f is integrble on both [, c] nd [c, b]. Further, when these conditions hold, (1) f(x) dx = c f(x) dx + c f(x) dx. Proof. Suppose f is integrble on both [, c] nd [c, b]. If ε > 0 there re prtitions P 1 of [, c] nd P 2 of [c, b] such tht Then Q = P 1 P 2 is prtition of [, b] nd U(f, P i ) L(f, P i ) ε/2 for i = 1, 2. L(f, Q) = L(f, P 1 ) + L(f, P 2 ) U(f, Q) = U(f, P 1 ) + U(f, P 2 ), hence U(f, Q) L(f, Q) ε. Therefore f is integrble on [, b]. Also so L(f, Q) L(f, P 1 ) L(f, P 2 ) c c c f(x) dx + f(x) dx U(f, P 1 ) f(x) dx U(f, P 2 ), c f(x) dx U(f, Q). Since f(x) dx lso lies between L(f, Q) nd U(f, Q), f(x) dx c f(x) dx+ c ε. This is true for ny ε > 0, proving (1). 25 f(x) dx <

26 Now ssume f is integrble on [, b] nd let Q be prtition with (2) U(f, Q) L(f, Q) < ε. If c / Q, replce Q by Q {c}. By 13 Lemm 2, (2) still holds. Let P 1 = Q [, c] nd P 2 = Q [c, b]. Then ( U(f, P1 ) L(f, P 1 ) ) + ( U(f, P 2 ) L(f, P 2 ) ) = U(f, Q) L(f, Q) < ε. The two differences on the left re nonnegtive, hence ech is less thn ε. Therefore f is integrble on [, b] nd by the first prt of the proof, (1) holds. Theorem 2. If f is integrble on [, b] nd k R then kf is integrble on [, b] nd kf(x) dx = k f(x) dx. Proof. For k > 0, inf{kf(x) : u x v} = k inf{f(x) : u x v}, hence L(kf, P ) = kl(f, P ) nd similrly for the upper sum. The result follows. For k < 0, we use 0 = b ( k)f(x) + kf(x) dx = ( k)f(x) dx + kf(x) dx = ( k) f(x) dx + kf(x) dx by Theorem 1 nd the rgument bove with k > 0 tking the role of k. Theorem 3. If f nd g re integrble on [, b], then f + g is integrble on [, b] nd f + g dx = f dx + g dx. Proof. Let I be some intervl, [x i 1, x i ], in prtition of [, b]. For ny bounded function f, let m f = inf{f(x) : x I}. Then for x I, m f +m g f(x)+g(x), hence m f +m g m f+g. Similrly for the supremum, M f + M g M f+g. Now for ny ε > 0, let P be prtition of [, b] with both U(f, P ) L(f, P ) ε/2 nd U(g, P ) L(g, P ) ε/2. Then L(f, P ) + L(g, P ) L(f + g, P ) U(f + g, P ) U(f, P ) + U(g, P ). Hence U(f +g, P ) L(f +g, P ) < ε. Therefore f +g is integrble. Since both f +g dx nd b f dx+ g dx lie between these lower nd upper sums, f dx+ g dx f +g dx < ε. This is true for ll ε > 0 proving the equlity. Proposition. If m f(x) M for x b nd f is integrble, then m(b ) f(x) dx M(b ). Proof. Let P = {, b}, the prtition with just one intervl. Then m(b ) L(f, P ) f(x) dx U(f, P ) M(b ). 26

27 Corollry. If f nd g re integrble on [, b] nd f(x) g(x), then f(x) dx g(x) dx. 16. The fundmentl theorem of clculus In the cse b < define f(x) dx = f(x) dx whenever the second integrl exists. Then 15 Theorem 1 holds with ny permuttion of, b, c. In prticulr x b f(t) dt = b f(t) dt + x f(t) dt. If f is integrble on n intervl [u, v], nd, x [u, v], we define the b function F (x) = x f(t) dt. Fundmentl Theorem of Clculus. If f is integrble on [u, v], u < b < v, [, b] [u, v], nd f is continuous t b, then F is differentible t b nd F (b) = f(b). Proof. Since F (x) nd x f(t) dt differ by the constnt f(t) dt, it suffices to prove the b theorem in the specil cse = b; we let F (x) = x Since f is continuous t b by hypothesis, hence By the proposition of 15, if b < x, then b f(t) dt = x f(t) dt. ( ε > 0)( δ > 0)( t b < δ f(t) f(b) < ε), t b x b < δ f(b) ε < f(t) < f(b) + ε. (f(b) ε)(x b) x b f(t) dt (f(b) + ε)(x b), (1) ε F (x) f(b) ε for x b < δ. x b On the other hnd, if x < b, then (f(b) ε)(b x) x f(t) dt (f(b) + ε)(b x), Since f(t) dt = x f(t) dt = F (x) nd b x = (x b), this gin gives (1). x b Hence for ny ε > 0 there is δ > 0 such tht (1) holds for ny x with x b < δ, hence F (x) = lim x b F (x) x b = f(b). 27

28 The fundmentl theorem proves the existence of ntiderivtives functions F such tht F is given function f. Of course knowing lot of derivtives my llow one to guess such n F. But for f(x) = 1/x or 1/(1 + x 2 ) or e x2, guessing is not rel option unless you re lredy fmilir with n nswer. On the other hnd, finding n ntiderivtive for given f permits one to evlute definite integrl of f. Corollry. If G (x) = f(x) nd f is continuous on [, b], then f(x) dx = G(b) G(). Proof. Since f is continuous, it is integrble. Let F (x) = x f(t) dt. Then F (x) = f(x) = G (x), so (F G) (x) = 0. By Corollry 2 of 12 there is c R with F (x) = G(x) + c. Then F (b) = F (b) F () = G(b) + c G() c = G(b) G(). The following lterntive proof gives the result of the Corollry with slightly weker hypotheses. Theorem. If G (x) = f(x) snd f is integrble on [, b], then f(x) dx = G(b) G(). Proof. Given ε > 0 let P = {x 0,..., x n } be prrtition of [, b] with U(f, P ) L(f, P ) < ε. Then Therefore G(b) G() = = L(f, P ) = n G(x i ) G(x i 1 ) i=1 n f(ξ i )(x i x i 1 ) where ξ i (x i 1, x i ) i=1 n M i (x i x i 1 ) = U(f, P ) i=1 n m i (x i x i 1 ) i=1 f(η i )(x i x i 1 ) = G(b) G(). nd L(f, P ) G(b) G() U(f, P ) f(x) dx G(b) + G() < ε. nd 28

29 17. Drboux nd Riemnn integrls Let P = {x 0,..., x n } be prtition of [, b] nd let t = {t 1,..., t n } stisfy x i 1 t i x i. Definition. R(f, P, t ) = Σ n i=1f(t i )(x i x i 1 ) is clled Riemnn sum. Definition. mesh(p ) = mx{x 1 x 0,..., x n x n 1 }. Definition. f is Riemnn integrble on [, b] if there is number A such tht ( ε > 0)( δ > 0)( P, t )(mesh(p ) < δ A ε < R(f, P, t ) < A + ε). The Riemnn integrl of f is A. Theorem. A bounded function f is Drboux integrble if nd only if it is Riemnn integrble. The two integrls re equl. Proof. If f is Drboux integrble on [, b], then for ll ε > 0 there is prtition P with U(f, P ) L(f, P ) < ε nd L(f, P ) f U(f, P ). Sy f(x) B on [, b] nd P = {x 0,..., x m }. Tke δ = min{mesh(p ), ε/(bm)}. Let Q = {y 0,..., y n } be ny prtition of [, b] with mesh(q) < δ. If [y j 1, y j ] [x i 1, x i ] for ny i, then ( k, 1 k m 1)(y j 1 < x k < y j ). There re t most m 1 such subintervls in Q. For the rest we hve [y j 1, y j ] [x i 1, x i ] for some i depending on j. Then Hence Also So U(f, Q) U(f, P ) + (m 1)δB < U(f, P ) + ε, L(f, Q) L(f, P ) (m 1)δB > L(f, P ) ε. U(f, Q) L(f, Q) < 3ε. L(f, Q) R(f, Q, t ) U(f, Q). R(f, Q, t ) f < 3ε. Hence f is Riemnn integrble nd the integrls re equl. If f is Riemnn integrble there is number A such tht ( ε > 0)( δ > 0)( P, t )(mesh(p ) < δ A ε < R(f, P, t ) < A + ε) 29

30 Given ε > 0, fix such P = {x 0,..., x m }. Let M i = sup{f(x) : x i 1 x x i }. Since M i is lest upper bound, there is t i [x i 1, x i ] such tht f(t i ) > M i ε/m, hence x i x i 1 M i (x i x i 1 ) < f(t i )(x i x i 1 ) + ε/m. This nd similr rgument for m i gives U(f, P ) < R(f, P, t ) + ε < A + 2ε, L(f, P ) > R(f, P, s ) ε > A 2ε So U(f, P ) L(f, P ) < 4ε nd f is Drboux integrble. Now from the first prt of the proof we know the integrls re equl. Note tht we hve not ssumed f is continuous in this proof. We hve shown tht the set of integrble functions is the sme in the two theories nd tht the integrls coincide. 30