NOTES WEEK 15 DAY 2 SCOT ADAMS

NOTES WEEK 15 DAY 2 SCOT ADAMS We will need the Fà di Bruno result, proved erlier. It sttes: Let X nd Y nd Z be finite dimensionl vector spces, let f : X Y nd g : Y Z. Let p P X nd let k P N 0. Assume p P domrd k fs nd fppq P domrd k gs. Then p P domrd k pg fqs. Let X nd Y be finite dimensionl vector spces nd let f : X Y. Recll tht, for ny S Ď X, by f is infinitely differentible on S, we men: @k P N 0, S Ď domrd k fs. By f is diffeomorphism, we men tht ll of the following conditions hold: f is injective, domrfs is open in X, imrfs is open in Y, f is infinitely differentible on domrf s nd f 1 is infinitely differentible on imrfs. For ny p P X, by f is diffeomorphism ner p, we men there exists n open nbd U of p in X such tht U Ď domrfs nd such tht f U is diffeomorphsim. Let X nd Y be finite dimensionl vector spces, let f, g : X Y nd let p P X. Assume tht there exists nbd U of p in X such tht: @x P U, fpxq gpxq. Then, for ll k P N 0, we hve D k pf D k pg. Let X nd Y be finite dimensionl vector spces nd let f : X Y. Then: @p P X, D z pf T p q D p f. Also, @k P N 0, @open subset U of X, D k pf Uq pd k fq U. Also, @k, l P N 0, domrd k D l fs domrd k`l fs. Until the end of the proof of the Inverse Function Theorem, we dopt the following nottion: Let n P N. let W : R n. Let z : 0 W. Let W : LpW, W q. Let I : id W P W. For ll T P T, recll: rt s P R nˆn denotes the mtrix of T. Define DET : W Ñ R by DETpT q detrt s. Define } } P N pw q by }T } rt s 2. Let W ˆ : t T P W DETpT q 0 u. Dte: My 4, 2017 Printout dte: My 11, 2017. 1

2 SCOT ADAMS Define R : W ˆ Ñ W ˆ by RpT q T 1. We proved, during the lst clss, tht R is infinitely differentible on W ˆ. We will lso use the next two results, ll tken from the lst clss: LEMMA 0.1. Let U nd V be open subsets of W, φ : U Ñ V homeomorphism. Assume domrdφs U, imrdφs Ď W ˆ. Then Dpφ 1 q R pdφq pφ 1 q. Proof. Proved in the lst clss. LEMMA 0.2. Let f : W W nd let U be n open bll in W. Assume, for ll x P U, tht }pd x fq I} ď 1{2. Then both of the following re true: (1) f U : U Ñ W is n.t. nd (2) @x P U, D x f P W ˆ. Proof. Proved in the lst clss. Finlly, we will need one result, proved in Week 14, Dy 2: THEOREM 0.3. Let φ : W W be n.t. Let U : domrφs nd let V : imrφs. Assume U is open in W. Then V is open in W nd φ : U Ñ V is homeomorphism. Proof. Proved Week 14 Dy 2 (Theorem 0.6). We cn now prove the Inverse Function Theorem: THEOREM 0.4. Let f : W W nd let p P W. Assume tht f is infinitely differentible on nbd of p. Assume tht D p f P W ˆ. Then f is diffeomorphism ner p. Proof. Let q : fppq. Define α, β : W Ñ W by αpwq p ` w nd βpwq q ` w. Then α, β : W Ñ W re diffeomorphisms Then, by Fà di Bruno (twice), pβ 1 q f α is infinitely differentible on nbd of z. Let f 0 : fp T. Then f 0 pzq fp T pzq z. Also, f 0 pβ 1 q f α, nd so f 0 is infinitely differentible on nbd of z. As β f 0 pα 1 q f, by Fà di Bruno (twice), it suffices to show: f 0 is diffeomorphism ner z. Let L : pd z f q 1 0. Then L 1 D z f 0. Since D z f 0 D z pf T p q D p f P W ˆ,

NOTES WEEK 15 DAY 2 3 it follows tht L P W ˆ, so L : W Ñ W is diffeomorphism, Lpzq z nd D z L L. By the Chin Rule, D z pl f 0 q pd z Lq pd z f 0 q. Since D z L L nd D z f 0 L 1, we see tht pd z Lq pd z f 0 q L L 1 I. Then D z pl f 0 q I. Also, by Fà di Bruno, L f 0 is infinitely differentible on nbd of z. Let f 1 : L f 0. Then D z f 1 I nd f 1 is infinitely differentible on nbd of z. Also, we hve f 1 pzq Lpf 0 pzqq Lpzq z. Becuse pl 1 q f 1 f 0, by Fà di Bruno, it suffices to show tht f 1 is diffeomorphism ner z. Tht is, we wish to show tht there exists n open neighborhood U of z such tht U Ď domrf 1 s nd such tht f 1 U is diffeomorphism. Let F : Df 1 : W W. We hve F pzq pdf 1 qpzq D z f 1 I. As f 1 is infinitely differentible on nbd of z, we get z P domrd 2 f 1 s. We hve domrd 2 f 1 s domrddf 1 s domrdf s. Then z P domrdf s. Then domrf s is nbd of z nd F is continuous t z. Since domrf s is nbd of z, choose δ 1 ą 0 such tht Bpz, δ 1 q Ď domrf s. Since F is continuous t z, choose δ 2 ą 0 such tht, for ll x P domrf s, we hve r x z ă δ 2 s ñ r }rf pxqs rf pzqs} ď 1{2 s. Since f 1 is infinitely differentible on nbd of z, choose δ 3 ą 0 such tht f 1 is infinitely differentible on Bpz, δ 3 q. Then, for ll k P N 0, we hve Bpz, δ 3 q Ď domrd k f 1 s. In prticulr, we see tht Bpz, δ 3 q Ď domrf 1 s. Let δ : mintδ 1, δ 2, δ 3 u, U : Bpz, δq. Then U Ď Bpz, δ 3 q Ď domrf 1 s, nd we wish to show tht f 1 U is diffeomorphism. Let φ : f 1 U. We wish to show tht φ is diffeomorphism. Since U Bpz, δq, we know tht U is n open bll in W. Also, for ll x P U, we hve x P Bpz, δ 1 q Ď domrf s nd x z ă δ 2, nd so, by choice of δ 2, we hve }rf pxqs rf pzqs} ď 1{2. Then, by Lemm 0.2 (with f replced by f 1 ), both of the following re true: (1) φ : U Ñ W is n.t. nd (2) @x P U, D x f 1 P W ˆ. Let V : imrφs. By (1) bove, nd by Theorem 0.3, we conclude both tht V is open in W nd tht φ : U Ñ V is homeomorphism. Then φ is 1-1. Also, domrφs U, so domrφs is open in W. Also, imrφs V, so imrφs is open in W. Since U Bpz, δq Ď Bpz, δ 3 q, we conclude tht

4 SCOT ADAMS f 1 is infinitely differentible on U. As U is open in W, for ll k P N 0, we hve D k φ pd k f 1 q U. Then φ is infinitely differentible on U. Tht is, φ is infinitely differentible on domrφs. It remins only to show tht φ 1 is infinitely differentible on imrφs. Tht is, we wish to show tht φ 1 is infinitely differentible on V. We hve domrdφs domrpdf 1 q Us, so, s f 1 is infintely differentible on U, we get domrdφs U. We hve imrdφs imrpdf 1 q Us, so, by (2) bove, we get imrdφs Ď W ˆ. Then, by Lemm 0.1, we see tht Dpφ 1 q R pdφq pφ 1 q. It suffices to show, for ll k P N, tht domrd k pφ 1 qs V. We rgue by induction on k. We hve domrdpφ 1 qs domrr pdφq pφ 1 qs V. It therefore suffices to show, for ll k P N, tht p domrd k pφ 1 qs V q ñ p domrd k`1 pφ 1 qs V q. Let k P N be given. We wish to prove: p domrd k pφ 1 qs V q ñ p domrd k`1 pφ 1 qs V q. Assume domrd k pφ 1 qs V We wish to prove domrd k`1 pφ 1 qs V. Since domrd k`1 pφ 1 qs Ď domrφ 1 s V, we need only show tht domrd k`1 pφ 1 qs Ě V. Tht is, we wish to prove, for ll y P V, tht y P domrd k`1 pφ 1 qs. Let y P V be given. We wish to prove: y P domrd k`1 pφ 1 qs. Since domrd k`1 pφ 1 qs domrd k Dpφ 1 qs, we seek to show: y P domrd k Dpφ 1 qs. By the induction ssumption, V domrd k pφ 1 qs. Since φ : U Ñ U is infinitely differentible on U, we see tht U domrd k`1 φs. Then U domrd k pdφqs. Since R : W ˆ Ñ W ˆ is infinitely differentible on W k, we get W ˆ domrd k Rs. We hve y P V domrd k pφ 1 qs. Also, φ 1 pyq P U domrd k pdφqs. Also, we hve pdφqpφ 1 pyqq P imrdφs Ď W ˆ domrd k Rs. Then, by Fà di Bruno (twice), we get y P domrd k pr pdφq pφ 1 qqs. So, since Dpφ 1 q R pdφq pφ 1 q, we get y P domrd kdpφ 1 qs. This ends our exposition of the Inverse Function Theorem. We now begin mesure nd integrtion. For the mteril tht follows, you will be expected to know definitions nd to be ble to nswer some bsic true/flse questions, but you will not be sked to provide ny proofs.

NOTES WEEK 15 DAY 2 5 DEFINITION 0.5. Let R be set of sets. By R is ring or ring of sets, we men: for ll A, B P R, A Y B, A X B, AzB P R. DEFINITION 0.6. Let S be set of sets. Then we define xsy fin > : tyf F Ď S is finite nd nonempty nd pirwise-disjointu. By convention, for ny set S of sets, we hve H P xsy fin >. DEFINITION 0.7. Let S be set of sets. By S is semiring or semiring of sets, we men: for ll A, B P S, A X B P S nd AzB P xsy fin >. For the rest of this lecture we define I : t r, bq, b P R, ă b u, nd we define l : I Ñ r0, 8q by lpr, bqq b. For the rest of this lecture, for ll d P N, we define S d : t I 1 ˆ ˆ I d I 1,..., I d P I u, nd we define λ d : S d Ñ r0, 8q by λ d pi 1 ˆ ˆ I d q rlpi 1 qs rlpi d qs. We discussed, with pictures, why, for ll d P N, S d is semiring. REMARK 0.8. Let S be semiring of sets. Then xsy fin > is ring. Let d P N. For the rest of this lecture, we define R d : xs d y fin >. Then, by Remrk 0.8, R d is ring. In clss, we described S d s set of lego blocks, nd R d s the set of ll lego ssemblies. DEFINITION 0.9. Let S be set of sets nd let µ : S Ñ r0, 8q. Then, by µ is finitely dditive or f.., we will men: for ny finite, pirwise-disjoint F Ď S, we hve: «ď ı ď F P S ñ µ F ÿ ff µpf q. F PF REMARK 0.10. For ll d P N, we hve: λ d : S d Ñ r0, 8q is f..

6 SCOT ADAMS A function whose domin is set of sets nd whose trget is r0, 8q will be referred to s set-function. If set A is in the domin nd the imge of A is v, then we often think of v s the volume of A, but sometimes cll it the mesure of A, especilly if the set A is not three-dimensionl. (In two dimensions, re is sometimes used, nd in one dimenion, length. The generic term is mesure.) THEOREM 0.11. Let S be semiring of sets. Let λ : S Ñ r0, 8q be f.. set function. Let R : xsy fin >. Then there exists unique f.. set function µ : R Ñ r0, 8q such tht µ S λ. Proof. Omitted, but we explined this by sying: if you know the size of ll your blocks, then you know the size of ll your ssemblies. For the rest of this lecture, for ll d P N, we let µ d : R d Ñ r0, 8q be the unique f.. set function such tht µ d S d λ d. Then µ d computes the mesure (clled volume, when d 3) of ny lego ssembly. Let d P N. There re mny sets in R d tht we cre bout tht re not in R d. Tht is, mny importnt sets re not lego ssemblies. For exmple, in two dimensions, disk is not in R 2. In three dimensions, few yers go, my son built very ominous-looking Deth Str out of legos, but it s relly only pproximtely bll. How cn we compute the res of disks nd volumes of blls? More generlly, how cn we compute the mesures of curvy sets. Here re the key: DEFINITION 0.12. Let d P N. For ny bounded B Ď R d, we define µ d pbq : sup t µ d paq A P R d, A Ď B u nd µ d pbq : inf t µ d pcq C P R d, C Ě B u. The disk of rdius r ą 0 bout x P R 2 is denoted Bpx, rq. REMARK 0.13. @x P R 2, @r ą 0, µ 2 pbpx, rqq πr 2 µ 2 pbpx, rqq. DEFINITION 0.14. Let d P N nd let B Ď R d. By B is Riemnn mesurble, we men: both p B is bounded q nd p µ d pbq µ d pbq q. Sometimes Jordn is used in plce of Riemnn. We refer to µ d s Riemnn outer mesure nd µ d s Riemnn inner mesure. Then set is Riemnn mesurble if it is bounded s hs the sme inner mesure s outer mesure. According to Remrk 0.13, ny disk in R 2 is Riemnn mesurble.

NOTES WEEK 15 DAY 2 7 DEFINITION 0.15. Let d P N. We define R d to be the set of ll Riemnn mesurble sets in R d. We define µ d : µ d R d µ d R d. So, for ny d P N, for ny Riemnn mesurble set B Ď R d, we hve µ d pbq µ pbq µ d dpbq. For ll d P N, the mpping µ d : R d Ñ r0, 8q is clled Riemnn mesure (or Jordn mesure ). Here re some of its bsic properties: LEMMA 0.16. Let d P N. Then: (1) R d (2) µ d is ring of sets, : R d Ñ r0, 8q is f.., (3) R d Ď R d nd µ d R d µ d. We now move from mesure to integrtion. DEFINITION 0.17. Let B be set nd let A Ď B. Then we define χ B A : B Ñ t0, 1u by # χ B 1, if x P A; Apxq 0, if x R A. We cll χ B A the chrcteristic function of A in B. DEFINITION 0.18. Let d P N. Let f : R d R. Then we define O f : t px, yq P R d ˆ R 0 ă y ă fpxq u. The set O f is clled the ordinte set of f. DEFINITION 0.19. Let d P N. Let f : R d R. Then we define ż f : rµ d`1po f qs rµ d`1po f qs. ż If f, then we sy tht f is Riemnn integrble. ż We cll f the Riemnn integrl of f. Note tht f is Riemnn integrble iff O f nd O f re both Riemnn mesurble in R d`1. So, if f is Riemnn mesurble, then O f nd O f must both be bounded in R d`1, nd it follows tht f is bounded nd compctly suported. The : r0, 1s Ñ t0, 1u is both bounded nd compctly supported, but is not Riemnn integrble. (Proof omitted.) Continuity implies integrbility: converse is not true, s χ r0,1s QXr0,1s

8 SCOT ADAMS LEMMA 0.20. Let, b P R. Assume ă b. Let f : r, bs R. Assume tht f is continuous. Then f is Riemnn integrble. DEFINITION 0.21. Let f : R R. Let ż, b P R. Assume ă b b ż nd ssume r, bs Ď domrfs. Then we define f : f χ R r,bs. We finish with the Fundmentl Theorem of Clculus: THEOREM 0.22. Let, b P R. Assume ă b. Let f : r, bs R. Assume f is continuous. Define Φ : r, bs Ñ R by Φpxq ż x f. Then (1) Φ is continuous on r, bs nd differentible on p, bq nd (2) @x P p, bq, Φ 1 pxq fpxq. COROLLARY 0.23. Let, b P R. Sy ă b. Let f, F : r, bs R. Assume f is continuous. Assume Then (1) F is continuous on r, bs nd differentible on p, bq nd (2) @x P p, bq, F 1 pxq fpxq. f rf pbqs rf pqs. Proof. Define Φ : r, bs Ñ R by Φpxq Φpbq f, so ż x f. Then Φpq 0 nd f rφpbqs rφpqs. Also, by Theorem 0.22, (1 ) Φ is continuous on r, bs nd differentible on p, bq nd (2 ) @x P p, bq, Φ 1 pxq fpxq. Let C : F Φ. By subtrcting (1 ) from (1) nd then subtrcting (2 ) from (2), we conclude tht (1 ) C is continuous on r, bs nd differentible on p, bq nd (2 ) @x P p, bq, C 1 pxq 0. Then, by the MVT, C is constnt. Then 0 rcpbqs rcpqs. Adding this to f rφpbqs rφpqs, we get f rf pbqs rf pqs. The bsic philosophy of Corollry 0.23 is tht to integrte f on n intervl r, bs, it s enough to move to n ntiderivtive F of f,

NOTES WEEK 15 DAY 2 9 compute F on the boundry t, bu of r, bs nd tke those boundry vlues F pq nd F pbq nd put them into some simple formul, nmely rf pbqs rf pqs. So problem on the one-dimensionl intervl r, bs gets reduced to problem on the zero-dimensionl boundry t, bu. We hd brief discussion of how to generlize Corollry 0.23 to higher dimensions. In two dimensions, the generliztion is clled Green s Theorem. In three nd higher dimensions, it s clled Stokes Theorem. I will not be testing this, nd won t even put forth ll the necessry definitions. I simply wish to show tht, with the right nottion, these theorems look lot like the Fundmentl Theorem of Clculus. The following is Green s Theorem on rectngle: THEOREM 0.24. Let, b, c, d P R. Sy ă b nd c ă d. Let R : r, bs ˆ rc, ds Ď R 2, R : p, bq ˆ pc, dq Ď R 2. Let f be 2-form on R nd let F be 1-form on R. Assume f is continuous. Assume (1) F is continuous on R nd differentible on R nd (2) on R, df f. ż ż Then f F. R BR To fully understnd Theorem 0.24, you need to know wht 2-form (like f) is, nd how to integrte it over two-dimensionl spce (like the rectngle R). You lso need to know wht the boundry of BR of the rectngle R is; it consists of four rrows, one from the point p, cq P R 2 to the point pb, cq P R 2, one from the point pb, cq P R 2 to the point pb, dq P R 2, one from the point pb, dq P R 2 to the point p, dq P R 2 nd one from the point p, dq P R 2 to the point p, cq P R 2. Note tht these directed rrows wind counterclockwise round the rectngle R. You lso need to know wht 1-form (like F ) is, nd how to integrte it over one-dimensionl spce (like BR). You lso need to understnd kind of differentition, denoted d, tht crries 1-forms to 2-forms. To sy tht F is n nti-derivtive of f is to sy df f. Once you hve full understnding of Theorem 0.24, you cn go on to other regions besides rectngle, nd then on to higher dimensions. I wish we hd hd time to do some of this.