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Intermediate Analysis

by: Alvena McDermott

Intermediate Analysis MATH 3333

Marketplace > University of Houston > Mathmatics > MATH 3333 > Intermediate Analysis
Alvena McDermott
GPA 3.69


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This 6 page Class Notes was uploaded by Alvena McDermott on Saturday September 19, 2015. The Class Notes belongs to MATH 3333 at University of Houston taught by Staff in Fall. Since its upload, it has received 15 views. For similar materials see /class/208423/math-3333-university-of-houston in Mathmatics at University of Houston.


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Date Created: 09/19/15
Math 3333 Intermediate Analysis David Blecher Notes on integration contd The following condition is very useful Theorem Riemann condition A bounded function f ab 7 R is integrable if and only if V6 gt 0 3 a partition P of ab such that Uf P 7 Lf P lt 6 Proof By de nition f is integrable if and only if Uf Lf 7 Suppose that V6 gt 0 3 a partition P of ab such that Uf P 7 Lf P lt 6 Using this and the de nition of Lf and Uf Uf S UMP lt LfP 6 S Lf 6 Since this is true for every 6 gt 0 we have Uf S Lf by Theorem 117 But Lf S Uf by Theorem 296 so Lf Uf 7 If Uf Lf and if 6 gt 0 is given choose by de nition of Lf and Uf partitions Q and R such that Lf Q gt Lf 7 g and UfR lt Uf Let P Q U R the re nement of both Q and R obtained by taking their union Then by the last equations and Theorem 294 we have UltfPgt UltfRgt lt Uf g 7 Lf g lt LltfQgt 37 3 LltfPgt 6 Looking at the left and right side of the last line we see that Uf P 7 Lf P lt 6 which is what was required D De nition A function f D 7 R is called uniformly continuous if V6 gt 0 36 gt 0 such that fz 7 lt 6 whenever my 6 D and x 7m lt 6 Can you spot the difference between the de nitions of f being continuous and f being uniformly continuous The only difference is that in uniformly continuity the 6 does not depend on the points in D Theorem If D is compact and f D 7 R is continuous then f is uniformly continuous Proof Suppose that f is not uniformly continuous Thus 3 6 gt 0 such that V6 gt 0 3 z and y in D such that x 7m lt 6 but 7 2 6 Let us take 6 i for n E N Thus 3 xn and yn in D such that la 7ynl lt i but 7 2 6 Since D is compact it is bounded so by Theorem 197 the sequence has a convergent subsequence unlwmwm say Suppose that this subsequence converges to a point z say Since D is closed by the Bonus problem we must have x E D Similarly the subsequence ynmynwyna has a convergent subsequence ym1ym2ym3 say converging to a point y E D By Theorem 194 the further subsequence zmlwmwma converges to z too Now zmk 7 ymk mik 7 0 as k 7 00 On the other hand by Fact 9 on sequences zmk 7 gm 7 z 7 y and so by the squeezing Fact 5 on sequences x 7 yl 0 Thus z y Now fxmk 7 and fymk 7 fy by Theorem 212b so by an argument similar to the we 7 tumor 7 we 7 mm 7 0 oh the other hand we 7 tumor 2 e so limk 7 2 6 by Fact 4 on sequences This is a contradiction D 1 Theorem If f 11 7 R is continuous then f is integrable Proof By the previous theorem f is uniformly continuous Thus given 6 gt 0 there is a number 6 gt 0 such that 7 lt bfa whenever my 6 11 and lx 7 yl lt 6 Choose a partition P zox1 zn of 11 such that An xk 7M4 lt 6 for every k 1 2 n Consider the interval xk1xk By the Min max theorem f has a maximum value Mk and a minimum value mk on this interval so there are numbers 5 and t in k1zk with fs Mkft mk Since ls 7 tl S An lt 6 we conclude that E Mk mk lf5 ftl lt bia Now n n n UfP7LfP ZAszk i ZAkak ZAzkMk7mk k1 k1 k1 and so UltfPgt7LltfPgt lt ZAMbe 7 b7agtb E k1 7a 6 Thus f satis es the Riemann condition7 Theorem above and so f is integrable D Example Calculate fol 2 dz using upper or lower sums Solution You probably saw this example worked in Calculus 1 If you did not please read 297 in the Text for more details if you need them Here we take the partition Pn 0 i 7 T l 1 So Azk i for all k 12 n If you draw a picture it is clear that the numbers Mk and mk Thus A 31 V UfPZMkAzkZ 2Zk2 k1 k1 k1 We know see section on Mathematical induction that the last sum equals W Thus 1 nn12n1 1 1 1 1 UPn7 71727 7 f gt n 6 6ltngtltngt73 as n 7 00 A similar argument shows that 1 1 1 1 L033 81 2 S g as n 7 00 It is easy to see from these and using also Fact 4 for sequences that Uf S and Lf 2 But Lf S Uf so Lf Uf Hence fol xzdm More properties of the Riemann integral Many of the properties of the Riemann integral follow from its de nition and from general properties of inf7s7 and sup7s7 which are applied to the mk and Mk Lets illustrate this idea by rst proving the following properties of in ma and suprema and then applying these properties to prove a well known fact about the integral If A is a set of real numbers and K gt 0 then infKy yEA Kinfyy A supKy y E A K supy y E A and inf7yy A isupyy A sup7yy EA iinfyy A These are pretty easy to prove Lets prove If 04 supA then 04 is an upper bound for A Thus ify E A then y S 04 and so Ky 3 K04 Thus K0 is an upper bound for Ky y E A Thus supKy y E A 3 K04 K supy y E A So the left side of is S the right side On the other hand for any y E A Ky S supKy y E A so that y S supKy y E A Since this is true for all y E A 04 S supKy y E A Hence K0 3 supKy y E A This proves that the right side of is S the left side and thus is proved A similar argument proves We stated in Section 12 I leave the proof as an exercise Fact 11 If f is integrable on ab and if K is a constant then Kf dx KL7 fdz Proof If K 0 then both sides are zero Next suppose that K gt 0 If P zox1 xn is a partition of ab let A z E zk1zk By we have infKf x E xk1xk Kinff x E xk1xk Kmk Thus LKfP Z infKfz x e zk1zkmk K 2 mkAzk KLfP k1 k1 Now take the suprernurn over the set P of all partitions P of ab We have using supLKf P P E P supK Lf P P E P KsupLf P P E P Thus LKf KLf A similar argument shows that UKf KUf Hence LKf KLU KUU UKf7 and so Kf is integrable and Kf dx Kf7 fdz Next suppose that K 71 If P zox1 zn is a partition of ab then by we have inf7fz x E zk1xk isupf x E xk1xk iMk Thus Lof P 2 n1pr x e xk1xkmk 7 MkAxk 7Uf P k1 k1 Taking the suprernurn over the set P of all partitions P of ab we have using L7f supL7fP P 6P sup7UfP P E P iinfUfP P 6P Uf Sirnilarly U7f 7Lf and so Lf Uf Mf Uf Thus if is integrable and if dx 7 fdx Finally if K is any negative number then combining the facts we have proved we get from ab77Kfdz 77Kfdx 77Kabfdz Kabfdz n We already proved earlier that Fact I2 If f ab a R is integrable and if m S f S M for all z 6 ab then mb7a bfdx Mbia From this it is easy to deduce that if f ab a R is integrable and if f 2 0 for all z 6 ab then fdx 2 0 Indeed simply take m 0 in Fact l2 Fact I3 de Kb 7 a if K is a constant Proof Apply Fact l2 with m M K and f K we get Kb7a S de S Kb7a which gives what we want D Fact I4 If f and g are integrable on ab and if fx 3 g for all z 6 ab then fdx S f gdx Proof If we take a partition P of ab and if mk are the in rnurns used in the de nition of Lf P and if m are the in rnurns used in the de nition of Lg P then mk inff x E xk1xk S infgx x E xk1xk Thus n n 17 L021 2 mm 3 2 mm LltgPgt Lltggt gdz k1 k1 Taking the suprernurn over all partitions P we deduce that Lf S gdz which is what we need since Lf fdx D Fact I5 f 9 dx fdx gdz if f and g are integrable on ab Proof This uses in rna and suprerna in the kinds of ways we have done in the last few Facts See textbook for details D Fact I6 If f is integrable on ab then so is lfzl and fdzl S lfldm Proof To see that lfl is integrable on ab we show that it satis es Riernann7s condition Let E gt 0 be given Since f satis es Riernann7s condition let 73 0z1 zn be a partition of ab with 7 Z 7 Z mkA k Z i lt 6 k1 k1 k n 1 1f57t lk717klthenlf5llftl Slf5ftl Mrmk Thus lf8l S Mrmk t and so suplftl xk1 S t S zk S Mk7mklftl Hence suplftl xk1 S t S xk7Mk7mk S lftl and so suplftl xk1 S t S zk 7 Mk 7 mk S inflftl xk1 S t S This gives suplftl xk1 S t S zk 7 inflftl 1 S t S zk S Mk 7 mk Multiplying this last equation by An M 7 1 and adding over all k we have that Z An suplftl xk1 S t S zk 7 Askinf t xk1 S t S zk 3 Z Mk 7 mkAzk k1 k1 But this says precisely that Ulfl773 MUCH S UMP LU lt 6 So lfl satis es Riernann7s condition We have 7lfxl S f S By Fact l4 and I1 we have that 7lfldzlb7lfldz lbfd lblfld That is if fdzl f lfldx D Fact I7 fdx fdx fob fdx if a S c S b and if f is integrable on ac and on 0J9 Proof We will rst show that f satis es Riernann7s condition on ab So suppose that E gt 0 is given Since f is integrable on ac and on 0J9 it satis es Riernann7s condition on those intervals So there are partitions 731 of ac and 732 of 0b such that Uf731 Lf731 lt 62 Uf732 7 1067732 lt 62 Let 73 731 U 732 a partition of ab It is easy to check that Lf73 Lf731 Lf732 draw a picture of the rectangles under the curve Sirnilarly U067 Uf7731 Uf732 Thus Uf73 7 Lf73 Uf731 7 Lf731 Uf732 7 Lf732 lt 62 62 7 6 Thus f satis es Riernann7s condition on ab and so is integrable on ab Also b fdz Uf73 Uf731 Uf732 fdz 6 b fdze Since 6 was arbitrary we have fdx S fdx fob fdx Sirnilarly b fdz 2 L027 7 Leta Lf7732 2 fdz b fdz 7 2e and so f fdzgfjfdzff fdz Thus f fdzf fdzff fdz D Fact 18 If f is monotone on ab then f is integrable on ab Proof This one is relatively easy to prove but since it is not so important we refer to the text for the proof D Fact 19 The rst fundarnental theorem of Calculus If f is integrable on ab de ne f fdt for z 6 ab If f is continuous at a point c 6 ab then F c fc Proof Recall that F c lirntnc To show that this limit exists and equals fc according to the 6 6 de nition of the limit if E gt 0 is given we have to show that there exists a 6 gt 0 such that 7 fc lt 6 whenever t 6 ab lt 7 cl lt 6 To this end let 6 gt 0 be given Since f is continuous at c 6 ab there exists a 6 gt 0 such that lft 7 fcl lt 6 whenever t 6 ab lt 7 cl lt For any such t we have Ft 7 Fc fdx 7 fdx f fdx and so M 1 fdz7f0 t7c T c t7c Assume that t gt c the case that t lt c is almost identical and left as an exercise Now fc is a constant and so fc i f fc dz and therefore fdx 7 fc i f f 7 fc dx Thus and also using also Fact l8 we have E E efo l1 1 fW 1 S t C C ic UWl7 Wm Since the z here is between c and t we have by Equation that lft 7 fcl lt 6 Thus t 7 0 Fact 110 The second fundamental theorem of Calculus If f ab 7 R is continuous and is differentiable on ab and if f is integrable on ab set f a f b 0 if they are not already de ned then f f dz fb e fa Proof Suppose that P x0z1 zn is a partition of ab By the MVT on k1xk there is a number tk E 4amp0 such that u 7 fxk1 f tkk 7 4 Thus n fb 7 fa u uid Z f tkxze 7 k71 k1 k1 On the other hand we have Lf7P E Z fthM kei S Uf7P7 1e1 and so LUCP S fb fa S Uf 7P Taking the suprernurn over partitions P we get 17 f d Lf supLf P partitions P S fb 7 fa Sirnilarly taking the in rnurn over partitions P we get b N 7 fa Um f dz Thus f f dz fb e fa D


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