NOTES WEEK 15 DAY 1 SCOT ADAMS

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1 NOTES WEEK 15 DAY 1 SCOT ADAMS We fix some notation for the entire class today: Let n P N, W : R n, : 2 P N pw q, W : LpW, W q, I : id W P W, z : 0 W 0 n. Note that W LpR n, R n q. Recall, for all T P LpR n, R n q, that rt s P R nˆn denotes the matrix of T. Recall, for all A P R nˆn, that L A P LpR n, R n q denotes g the linear fÿ transformation corresponding to A and that A 2 : e n nÿ is i 1 j 1 the L 2 -norm of A. Define } } P N pw q by }T } rt s 2. Define DET P P n pw, Rq and ADJ P P n 1 pw, W q by by DETpT q det rt s and ADJpT q L adj rt s. Let W ˆ : tt P W DETpT q 0u. Let R : ADJ DET : W W. Then domrrs W ˆ and, for all T P W, we have RpT q T 1. All of this notation is fixed for the entire class. Note, by HW#55, that, for all T P W, for all w P W, we have T pwq 2 ď rt s 2 w 2, i.e., we have T pwq ď }T } w. LEMMA 0.1. Let X, Y and Z be finite dimensional vector spaces, let f : X Y and g : Y Z. Let p P X and let k P N 0. Assume p P domrd k fs and fppq P domrd k gs. Then p P domrd k pg fqs. A 2 ij Proof. This is the Faà di Bruno result, proved in an earlier class. LEMMA 0.2. Let U, X, Y and Z be finite dimensional vector spaces, let f : U X and g : U Y. Let P BpX, Y, Zq. Let p P U and let k P N 0. Assume p P domrd k fs and p P domrd k gs. Then p P domrd k pf gqs. Proof. Let B : P BpX, Y, Zq. For all x P X, for all y P Y, we have Bpx, yq x y. The function B : X ˆ Y Ñ Z is a polynomial, and so domrd k Bs X ˆ Y. By k applications of HW#48, for all u P U, we have Dupf, k gq pduf, k Dugq. k So, since p P domrd k fs and since p P domrd k gs, we get p P domrd k pf, gqs. So, since f g B pf, gq, and Date: May 2, 2017 Printout date: May 4,

2 2 SCOT ADAMS since pf, gqppq P X ˆ Y domrd k Bs, by Faà di Bruno (Lemma 0.1), we see that p P domrd k pf gqs, as desired. FACT 0.3. For all k P N 0, we have domrd k Rs W ˆ. Proof. We have domrd k Rs Ď domrrs W ˆ, and so we wish to prove W ˆ Ď domrd k Rs. We wish to P W ˆ, T P domrd k Rs. Let T P W ˆ be given. We wish to show: T P domrd k Rs. Define P BpR, W, W q by t w tw. Let r : Rzt0u Ñ R be defined by rptq 1{t. Then we have R rr pdetqs ADJ. So, by Lemma 0.2, it suffices to show both T P domrd k pr pdetqqs and T P domrd k padjqs. Because DET : W Ñ R and ADJ : W Ñ W are both polynomials, we conclude that domrd k pdetqs W domrd k padjqs. Therefore we have T P W ˆ Ď W domrd k padjqs, and it remains to show that T P domrd k pr pdetqqs. By freshman calculus, domrd k rs Rzt0u. Since T P W ˆ, by definition of W ˆ, DETpT q 0. Then T P W ˆ Ď W domrd k pdetqs and DETpT q P Rzt0u domrd k rs, so, by Faà di Bruno (Lemma 0.1), T P domrd k pr pdetqqs, as desired. For any function γ : R W, for all t P R, recall that γ 1 ptq lim hñ0 rγpt ` hqs rγptqs. h LEMMA 0.4. Let a, b P R, let s ě 0 and let γ : R W. Assume a ă b. Assume that γ is continuous on ra, bs and differentiable on pa, bq. Assume, for all t P pa, bq, that γ 1 ptq ď s. Then rγpbqs rγpaqs ď pb aq s. Proof. Let v : rγpbqs rγpaqs. We wish to show that v ď pb aq s. Either v 0 or v 0. Case 1: v 0. Proof in Case 1: Since b a ą 0 and s ě 0, we get pb aq s ě 0. Then v 0 ě pb aqs. End of proof in Case 1. Case 2: v 0. Proof in Case 2: Let u : v{ v. Let P SBFpW, Rq be the dot product. Let f : ra, bs Ñ R be defined by fptq rγptqs u. By the Mean Value Theorem, choose c P pa, bq such that f 1 pcq rfpbqs rfpaqs. b a

3 NOTES WEEK 15 DAY 1 3 We have f 1 pcq rγ 1 pcqs u, so, by Cauchy-Schwarz, f 1 pcq ď γ 1 pcq u. By assumption, γ 1 pcq ď s. Also, u v { v 1. Then we have f 1 pcq ď f 1 pcq ď γ 1 pcq u ď s 1 s. Then f 1 pcq ď s. So, since b a ą 0, we get pb aq pf 1 pcqq ď pb aq s By definition of f, we have rfpbqs rfpaqs prγpbqs rγpaqsq u. So, since rγpbqs rγpaqs v, we conclude that rfpbqs rfpaqs v u. Since u v{ v, we see that v u pv vq{ v. Then rfpbqs rfpaqs v u pv vq{ v v 2 { v v. Then v rfpbqs rfpaqs pb aq pf 1 pcqq ď pb aq s, as desired. End of proof in Case 2. REMARK 0.5. Let β : R Ñ W, let η : W Ñ W and let t P R. Then pη βq 1 ptq pd βptq ηqpβ 1 ptqq. Proof. To be added. LEMMA 0.6. Let f : W W and let U be an open ball in W. Assume, for all x P U, that }pd x fq I} ď 1{2. Then both of the following are true: (1) f U : U Ñ W is a n.t. and P U, D x f P W ˆ. Proof. Proof of (1): Let φ : f U. We wish to show: φ : U Ñ W is a n.t. Let η : φ I. We wish to show: φ : U Ñ W is a n.c. That is, we wish to show, for all p, q P U, that rηppqs rηpqqs ď r1{2s p q. Let p, q P U be given. We wish to prove: rηppqs rηpqqs ď r1{2s p q. Let β : R Ñ W be defined by βptq p1 tqp ` tq. Then we have βp0q p and βp1q q. Let γ : η β. Then γp0q ηppq and γp1q ηpqq. So, by Lemma 0.4, it suffices to show, for all t P p0, 1q, that γ 1 ptq ď r1{2s p q. Let t P p0, 1q be given. We wish to show that γ 1 ptq ď r1{2s p q, i.e., that pη βq 1 ptq ď r1{2s q p. By assumption, for all x P U, we have }pd x fq I} ď 1{2, and, in particular, D x f. For all x P U, because φ agrees with f on U and because U is a nbd of x, it follows that D x φ D x f. Because I P LpW, W q, it follows, for all x P W, that D x I I. For all x P U, we have D x η pd x φq pd x Iq, so since pd x φq pd x Iq pd x fq I, we conclude that D x η pd x fq I; in particular, D x η. Then, for all x P U, we have }D x η} }pd x fq I} ď 1{2.

4 4 SCOT ADAMS Let ξ : βptq. Since U is an open ball in W, choose c P W and r ą 0 such that U Bpc, rq. Then p, q P U Bpc, rq, so p c ă r and q c ă r. We have ξ βptq p1 tqp ` tq and c p1 tqc ` tc. Subtracting these two equations, we get ξ c p1 tqpp cq ` tpq cq. Then ξ c ď p1 tq p c ` t q c ă p1 tqr ` tr r. Then ξ P Bpc, rq U. Recall that, for all x P U, we have D x η and }D x η} ď 1{2. Then D ξ η and }D ξ η} ď 1{2. By Remark 0.5, we conclude that pη βq 1 ptq pd βptq ηqpβ 1 ptqq. For all t P R, we have βptq p1 tqp ` tq. Then, for all t P R, we have β 1 ptq p ` q q p. So, since D βptq η D ξ η, it follows that pd βptq ηqpβ 1 ptqq pd ξ ηqpq pq. Then pη βq 1 ptq pd ξ ηqpq pq. Then pη βq 1 ptq ď }D ξ η} q p ď r1{2s q p. End of proof of (1). Proof of (2): Let x P U be given. We wish to show that D x f P W ˆ. By assumption, }pd x fq I} ď 1{2. By definition of } }, we have }pd x fq I} rpd x fq Is 2. Also, we have rpd x fq Is rd x fs ris. Then rd x fs ris 2 rpd x fq Is 2 }pd x fq I} ď 1{2. Then, by HW#56, we conclude that rd x fs is invertible. Equivalently, we have detrd x fs 0. By definition of DET, we have DETpD x fq detrd x fs. Since DETpD x fq detrd x fs 0, by definition of W ˆ, it follows that D x f P W ˆ, as desired. End of proof of (2). LEMMA 0.7. Let U and V be open subsets of W, φ : U Ñ V a homeomorphism. Assume φpzq z, D z φ I. Then D z pφ 1 q I. Proof. Because D z φ I, we get φ T z I P OW W p1q. We wish to show that pφ 1 q T z I P OW W p1q, or, equivalently, that I pφ 1 q T z P OW W p1q. Since φpzq z, we get z φ 1 pzq and φ T z φ. Since φ 1 pzq z, we get pφ 1 q T z φ 1. We wish to show that I pφ 1 q P OW W p1q. Claim: φ 1 P O W W p1q. Proof of claim: Let η : φ I. Then we have η φ I pφ T z q I P OW W p1q, and we conclude that η { P Oˆ W R. We wish to prove that φ 1 { P Oˆ W W. Choose a pnbd U 0 of z in W such that U 0 Ď U and such that, for all x P U 0, ηpxq { x ď 1{2. Let V 0 : φ pu 0 q. Then pφ 1 q pv 0 q U 0. Recall that U and V are open subsets of W, that φ : U Ñ V is a homeomorphism and that φpzq z. So, since U 0 is a pnbd of z in W and since U 0 Ď U, it follows that V 0 is a pnbd of z in W and V 0 Ď V. It therefore suffices to show, for all y P V 0, that φ 1 pyq { y ď 2. Let y P V 0 be given. We wish to show that φ 1 pyq { y ď 2. Let x : φ 1 pyq. We wish to show that x { y ď 2, i.e., that x ď 2 y.

5 NOTES WEEK 15 DAY 1 5 We have x φ 1 pyq P pφ 1 q pv 0 q U 0, so ηpxq { x ď 1{2, i.e., ηpxq ď x {2. We have φpxq φpφ 1 pyqq y. Then ηpxq pφ Iqpxq rφpxqs x y x. Let ˇˇ ˇˇ P N prq denote absolute value. Since norms are distance semidecreasing, we get ˇˇ y x ˇˇ ď y x. Then ˇ y x ˇˇ ď y x ηpxq ď x { 2. Then x p x {2q ď y ď x ` p x {2q, i.e., x {2 ď y ď 3 x {2. Since x {2 ď y, we get x ď 2 y, as desired. End of proof of claim. We have φ I P OW W p1q. By the claim, φ 1 P O W W p1q. it follows that pφ Iq pφ 1 q P row W p1qs ro W W p1qs Ď OW W p1q. We have pφ Iq pφ 1 q pφ pφ 1 qq pi pφ 1 qq id V pφ 1 q I pφ 1 q. Then I pφ 1 q pφ Iq pφ 1 q P OW W p1q, as desired. LEMMA 0.8. Let U and V be open subsets of W, φ : U Ñ V a homeomorphism. Assume φpzq z, D z φ P W ˆ. Then D z pφ 1 q pd z φq 1. Proof. Let L : pd φq 1 z. Then L P W ˆ, so L : W Ñ W is a homeomorphism and Lpzq z. Let V 1 : L pv q and let φ 1 : L φ. Then φ 1 : U Ñ V 1 is a homeomorphism. Also, φ 1 pzq Lpφpzqq Lpzq z. By the Chain Rule, we have D z φ 1 pd z Lq pd z φq. As L is linear, we have D z L L. By definition of L, we have D z φ L 1. Then pd z Lq pd z φq L L 1 I. Then D z φ 1 I. Then, by Lemma 0.7 (with φ replaced by φ 1 ), we see that D pφ 1 z 1 q I. Since φ 1 L φ, we get L 1 φ 1 φ, and so φ 1 pl 1 φ q 1 1 φ 1 1 L. So, by the Chain Rule, we have D pφ 1 z q rd pφ 1 z 1 qs rd z Ls. Also, rd pφ 1 z 1 qs rd z Ls I L L pdφq 1. We conclude that D pφ 1 z q pdφq 1, as desired. We leave it as an unassigned homework problem to show, for any φ : W W, for any x P W, that D z pφ T x q D x φ. LEMMA 0.9. Let U and V be open subsets of W, φ : U Ñ V a homeomorphism. Assume domrdφs U, imrdφs Ď W ˆ. Then Dpφ 1 q R pdφq pφ 1 q. Proof. Let y P V be given. We wish to show: pdpφ 1 qq pyq pr pdφq pφ 1 qq pyq.

6 6 SCOT ADAMS Let x : φ 1 pyq. We wish to show: pdpφ 1 qqpyq RppDφqpxqq. That is, we wish to show: D pφ 1 y q RpD x φq. By definition of R, we wish to show: D pφ 1 y q pd φq 1 x. Let U 0 : U x and V 0 : V y. Define α : U 0 Ñ U and β : V 0 Ñ V by αpuq u ` x and βpvq v ` y. Let φ 0 : φ T x. Then, by definition of φ T x, we have φ 0 pβ 1 q φ α. So, since α : U 0 Ñ U, φ : U Ñ V and β 1 : V Ñ V 0 are all homeomorphisms, we conclude that φ 0 : U 0 Ñ V 0 is a homeomorphism. We have φ 0 pzq rφpx ` zqs rφpxqs rφpxqs rφpxqs z. Then, by Lemma 0.8, we get D pφ 1 z 0 q pd z φ q 1 0. We leave it as an exercise for the reader to show that pφ T x q 1 pφ 1 q T y. By the unassigned homework problem mentioned just before the statement of this lemma, we have both D z pφ T x q D x φ and D ppφ 1 z q T y q D pφ 1 y q. We have φ 0 φ T x and φ 1 0 pφ T x q 1 pφ 1 q T y. Then D z φ 0 D z pφ T x q D x φ. Also, D pφ 1 z 0 q D ppφ 1 z q T y q D pφ 1 y q. Then as desired. D y pφ 1 q D z pφ 1 0 q pd z φ 0 q 1 pd x φq 1, The equation Dpφ 1 q R pdφq pφ 1 q in Lemma 0.9 will be called the Inverse Function Formula. Recall that R : W ˆ Ñ W and that, for all T P W ˆ, we have RpT q T 1. Consequently, Lemma 0.9, because of the assumptions domrdφs U and imrdφs Ď W ˆ, the Inverse Function Formula implies that domrdpφ 1 qs V. DEFINITION Let X and Y be finite dimensional vector spaces. Let f : X Y. Let S Ď X. By f is infinitely differentiable on S, we P N 0, S Ď domrd k fs. DEFINITION Let X and Y be finite dimensional vector spaces. Let f : X Y. By f is a diffeomorphism, we mean that all of the following hold: f is injective, domrfs is open in X, imrfs is open in Y, f is infinitely differentiable on domrf s and f 1 is infinitely differentiable on imrfs.

7 NOTES WEEK 15 DAY 1 7 DEFINITION Let X and Y be finite dimensional vector spaces. Let f : X Y. Let p P X. By f is a diffeomorphism near p, we mean that there exists a nbd U of p in W such that U Ď domrfs and such that f U is a diffeomorphism. For example p q 2 : R Ñ R is not injective, and is therefore not a diffeomorphism. However, p q 2 : R Ñ R is a diffeomorphism near 2, because p q 2 p1, 3q is a diffeomorphism. In fact, p q 2 : R Ñ R is a diffeomorphism near any nonzero real number. We can now state the Inverse Function Theorem: THEOREM Let f : W W and let p P W. Assume that f is infinitely differentiable on a nbd of p. Assume that D p f P W ˆ. Then f is a diffeomorphism near p. Proof. This will be proved in the next class.