Inverse Function Theorem writeup for MATH 4604 (Advanced Calculus II) Spring 2015

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1 Inverse Function Theorem writeup for MATH 4604 (Advanced Calculus II) Spring 2015 Definition: N := {1, 2, 3,...}. Definition: For all n N, let i n : R n R n denote the identity map, defined by i n (x) = x. For all n N, let I n R n n denote the n n identity matrix. Definition: For all n N, 0 n := (0,..., 0) R n. For all m, n N, by 0 n m R n m we mean the zero matrix. Definition: For all n N, for all r > 0, for all x R n, we define B r (x) := { y R n y x < r }, and B r (x) := { y R n y x r }. Definition: Let m, n N, let M R n m. Let l [1, n] and k [1, m] be integers. Then the (l, k) entry of M is denoted M lk. Definition: Let m, n N and let M R n m. Let v 1,..., v n R m be the rows of the matrix M, so, for all integers l [1, n], v l = (M l1,..., M lm ). Then L M : R m R n will denote the homogeneous linear function defined by L M (w) = (v 1 w,..., v n w). Definition: Let f and g be functions. Then f g [[0, x 0 ]] means: [ f and g are both defined at x 0 ] and [ f(x 0 ) = g(x 0 ) ]. Definition: Let l, m, n N, let D, E R m, let f : D R n, let g : E R n and let x 0 R m. Then f g [[l, x 0 ]] means: both f g [[0, x 0 ]] and, for all integers k [1, l], for all integers j 1,..., j k [1, m], we have j1...j k f j1...j k g [[0, x 0 ]]. Also, f g [[ l, x 0 ]] means: both f g [[0, x 0 ]] and, for all integers k [1, l], for all integers j 1,..., j k [1, m], we have j 1 j k j1...j k f j1...j k g [[0, x 0 ]]. Also, f g [[ l, x 0 ]] means: both f(x 0 ) = g(x 0 ) and, for all integers k [1, l], for all integers j 1,..., j k [1, m], we have j 1 j k j1...j k f j1...j k g [[0, x 0 ]]. Also, f g [[l, x 0 ]] means: x 0 (Int D) (Int E) and f(x 0 ) = g(x 0 ) and [f(x)] [g(x)] lim x x 0 x x 0 l = 0 n. 1

2 Definition: Let l, m, n N, let D R m and let f : D R n. Let U R m be an open set. We say that f is C l on U if: both U D and j 1,..., j l {1,..., m}, j1...j l f is continuous on U. Definition: Let m, n N, let D R m and let f : D R n. Let U R m be an open set. We say that f is C 0 on U if: both U D and f is continuous on U. We say that f is C on U if: for all l N, f is C l on U. Definition: Let m, n N, let D R m, let ν : D R n be a function and let x 0 R m. We say that ν vanishes continuously at x 0 if all three of the following hold: x 0 Int D, ν(x 0 ) = 0 n and ν is continuous at x 0. Remark (apprvan): Let m, n N, let D, E R m, let f : D R n, let g : E R n, let x 0 R m and let l N. Then f g [[l, x 0 ]] iff there exists an open neighborhood U of x 0 and there exists ν : U R vanishing continuously at x 0 such that U D E and such that, for all x U, we have: [f(x)] [g(x)] = [ν(x)][ x x 0 l ]. Proof: This is Remark (apprvan) from the Differentiability writeup. QED Definition: Let m, n N, let D R m and let f : D R n. For all x R m, f is differentiable at x means: x Int D and there exists a polynomial P : R m R n such that f P [[1, x]]. Note: The polynomial P from the preceding definition can be chosen to have degree 1, as follows: Assuming the degree of P is > 1, let P 0 : R m R n be defined by P 0 (x) = P (x + x 0 ), then let Q 0 be obtained from P 0 by dropping all terms of degree > 1, then let Q : R m R n be defined by Q(x) = Q 0 (x x 0 ), then replace P by Q. Definition: Let m, n N, let D R m and let f : D R n. Let M R n m and let x 0 R m. Let D be the set of x R m such that f is differentiable at x. Then f : D R n m is defined by: f (x 0 ) = M means both f is differentiable at x 0 and l {1,..., n}, j {1,..., m}, ( j f l )(x) = M lj. Remark (CL): Let m, n N, let D R m, let f : D R n and let x 0 R m. Assume f is differentiable at x 0. Define C, L : R m R n by C(x) = f(x 0 ) and L(x) = L f (x 0 )(x x 0 ). Then f C + L [[1, x 0 ]] and f C + L [[1, x 0 ]]. Proof: This is Remark (CL) from the Differentiability writeup. QED Definition: Let n N, let D R n, let h : D R n and let S D. Then h is a half-contraction on S means: for all v, w S, we have [h(w)] [h(v)] [1/2][ w v ]. Remark (hcunifc): Let n N, let D R n, let h : D R n and let S D. Assume that h is a half-contraction on S. Then h is uniformly continuous on S. 2

3 Proof: Given ε > 0. We wish to show that there exists δ > 0 such that, for all v, w S, we have: [ w v < δ ] [ [h(w)] [h(v)] < ε ]. Define δ := 2ε. Given v, w S. Assume that w v < δ. We wish to prove that [h(w)] [h(v)] < ε. We have [h(w)] [h(v)] [1/2][ w v ] < [1/2] δ = ε, as desired. QED Definition: Let n N. For all v, w R n, we define [v, w] := {(1 t)v + tw t [0, 1]}. For all S R n, S is convex means: for all v, w S, we have [v, w] S. Definition: Let m, n N and let M R n m. Then we define { } M := max L M (v) v Rm, v = 1, n m n m M 1 := M jk, and M := Mjk 2. j=1 k=1 j=1 k=1 Exercise (opnormprop): Let m, n N and let M R n m. Prove: (a) For all v R m, L M (v) M v. (b) For all v R m, for all x R n, x [L M (v)] x M v. (c) We have: M M 1 m n [ M ]. Hint for (c): For all integers j [1, n], k [1, m], let E jk denote the n m matrix whose (j, k) entry is 1, and whose other entries are all 0; show that E jk = 1. From M = M jk E jk, by using the triangle inequality, show that M M 1. Then, using jk Cauchy-Schwarz, show that M 1 m n [ M ]. Fact (1by1prod): Let m, n N and let M R n m. Let a 1,..., a m R and b 1,..., b n R. Let A := [ a 1 a 2 a m ] R 1 m and B := [ b 1 b 2 b n ] R 1 n denote the row vectors with entries a 1,..., a m and b 1,..., b m, respectively. Let v := (a 1, a 2,..., a m ) R m and w := (b 1, b 2,..., b n ) R n be the corresponding vectors. Then AMB t R 1 1 is the 1 1 matrix whose entry is v (L M (w)). Proposotion (1by1CR): Let m, n N and let D R m. Let α : [0, 1] D, h : D R n and β : R n R be functions. Let f := β h α : [0, 1] R. Let t (0, 1) and assume: α is differentiable at t, h is differentiable at α(t) and β is differentiable at h(α(t)). By ( β)(h(α(t))) R n, we will denote the vector corresponding to the row vector β (h(α(t))) R 1 n. By α(t) R m, we will denote the vector corresponding to the column vector α (t) R m 1. Then f is differentiable at t and f (t) R is given by the formula: f (t) = [ ( β) (h(α(t) ] [ L h (α(t)) ( α(t))) ]. Proof: By the Chain Rule, [β (h(α(t)))][h (α(t))][α (t)] R 1 1 is the 1 1 matrix whose entry is f (t). The result therefore follows from Fact(1by1prod). QED 3

4 Exercise (linnonlinlin): Let m, n N, let D R m and let h : D R n. Let v, w R m and let x R n. Assume that [v, w] D and assume that h is differentiable on [v, w]. Define f : [0, 1] R by f(t) = x (h((1 t)v + tw) ). Show, for all t (0, 1), that f (t) = x [ L h ((1 t)v+tw) ( v + w)) ]. Hint: Define α : [0, 1] D and β : R n R by α(t) = (1 t)v + tw and β(y) = x y. Then f = β h α. Apply Proposition (1by1CR). Remark (implieshc): Let n N, let D R n, let h : D R n and let S D. Assume that S is convex, that h is differentiable on S and that h 1/2 on S. Then h is a half-contraction on S. Proof: Given v, w S. Let u := w v and let x := [h(w)] [h(v)]. We wish to prove that x [1/2] u. Define f : [0, 1] R by f(t) = x (h((1 t)v + tw) ). By Exercise(linNonlinLin), for all t (0, 1), f is differentiable at t and we have f (t) = x [L h ((1 t)v+tw)(u))]. So, by (b) of Exercise (opnormprop), we have: f (t) x h ((1 t)v + tw) u. So, since h 1/2 on S, we conclude, for all t (0, 1), that f (t) x [1/2] u. By the Mean Value Theorem, choose t (0, 1) such that [f(1)] [f(0)] 1 0 = f (t). Then f (t) = [f(1)] [f(0)] = [x (h(w))] [x (h(v))] = x [(h(w)) (h(v))] = x x = x 2. Then x 2 = f (t) x [1/2] u. Then x [1/2] u, as desired. QED Theorem (HCFixPt): Let n N, let D R n, let h : D R n and let S D. Assume that h is a half-contraction on S. Assume that S is nonempty and closed. Assume that h(s) S. Then there exists x S such that h(x) = x. Proof: Choose x 0 S. For all k N, let x k := h k (x 0 ) S. Let a := x 0 x 1. For all k N, we have x k x k+1 [1/2][ x k 1 x k ] [1/2] 2 [ x k 2 x k 1 ] [1/2] k [ x 0 x 1 ] = Then, for all k, l N, we have: k l implies a 2 k. x k x l x k x k x l 1 x l a a 2 k + + [ 2 l 1 1 < a 2 k + 1 ] 2 k+1 + a = 2 k 1. 4

5 Then, for all s, t N, we have x s x t a 2 [min{s,t}] 1. It follows that x 1, x 2,... is Cauchy, and therefore convergent. Let x := lim j x j. Since S is closed and since x 1, x 2, x 3... S, it follows that x S. We wish to prove that h(x) = x. By Remark (hcunifc), we see that h is continuous on S. Then as desired. QED h(x) = lim j [h(x j)] = lim j x j+1 = x, Definition: Let n N, let D R n, let g : D R n and let S D. approximate identity on S means: i n g is a half-contraction on S. Then g is an Remark (aiunifc): Let n N, let D R n, let g : D R n and let S D. Assume that g is an approximate identity on S. Then g is uniformly continuous on S. Proof: By Remark (hcunifc), i n g is uniformly continuous on S. So, since i n is also uniformly continuous on S, we conclude that i n (i n g) is uniformly continuous on S. That is, g is uniformly continuous on S. QED Remark (impliesai): Let n N, let D R n, let g : D R n and let S D. Assume that S is convex, that g is differentiable on S and that I n g 1/2 on S. Then g is an approximate identity on S. Proof: Let h := i n g. We wish to show that h is a half-contraction on S. We have h = I n (g ). Then h 1/2 on S, so, by Remark (implieshc), we see that h is a half-contraction on S. QED Remark (aiinj): Let n N, let D R n, let g : D R n and let S D. Assume that g is an approximate identity on S. Then g is injective on S. Proof: Given v, w S. Assume that g(v) = g(w). We wish to prove that v = w. Let h := i n g. Then h is a half-contraction on S. We have [ h(w) ] [ h(v) ] = [ w (g(w)) ] [ v (g(v)) ] = w v. Then w v = [h(w)] [h(v)] [1/2][ w v ]. Subtracting [1/2][ w v ] from both sides gives: [1/2][ w v ] 0. Multiplying both sides by 2 gives: w v 0. So, as w v 0, we get w v = 0. Then w v = 0 n, so w = v, as desired. QED Corollary (ainonsing): Let n N and let A R n n. Assume that I n A 1/2. Then A is nonsingular. Proof: Let g := L A : R n R n. Then g = A on R n. Then I n (g ) = I n A 1/2 on R n. It follows, from Remark (impliesai), that g : R n R n is an approximate identity 5

6 Then, by Remark (aiinj), we see that g : R n R n is injective. That is, L A : R n R n is injective. It follows that A is nonsingular. QED Exercise (translateai): Let n N, let D R n, let g : D R n, let S D and let x 0, y 0 R n. Assume that g is an approximate identity on S. Define γ : D x 0 R n by γ(x) = [g(x + x 0 )] y 0. Show that γ is an approximate identity on S x 0. Remark (openstart): Let n N, let δ > 0, let D R n, let x 0 D and let g : D R n. Assume that g is an approximate identity on B 2δ (x 0 ). Then g( B 2δ (x 0 )) B δ (g(x 0 )). Proof: Let y 0 := g(x 0 ). We wish to show that g( B 2δ (x 0 )) B δ (y 0 ). Equivalently, we wish to show that [g( B 2δ (x 0 ))] y 0 B δ (0 n ). Define γ : D x 0 R n by γ(x) = [g(x+x 0 )] y 0. By Exercise (translateai), we see that γ is an approximate identity on B 2δ (0 n ). We have γ( B 2δ (0 n )) = [g( B 2δ (x 0 ))] y 0. We therefore wish to show that γ( B 2δ (0 n )) B δ (0 n ). Let R := B δ (0 n ) and S := B 2δ (0 n ). Then R and S are closed subsets of R n and, by the triangle inequality, R + R S. We wish to show that R γ(s). Given y R. We wish to prove: y γ(s). That is, we wish to prove: There exists x S such that y = γ(x). Let f := i n γ. Then f is a half-contraction on S, i.e., on B 2δ (0 n ). Moreover, f(0 n ) = (i n γ)(0 n ) = 0 n [γ(0 n )] = 0 n 0 n = 0 n. Since f is a half-contraction on B 2δ (0 n ) and since f(0 n ) = 0 n, it follows that f(b 2δ (0 n )) B δ (0 n ). That is, f(s) R. Define h : S R n by h(z) = y + [f(z)]. As f is a half-contraction on S and y is a constant, we see that h is a half-contraction on S. Moreover, h(s) = y+[f(s)] R+R S. By Theorem (HCFixPt), choose x S such that h(x) = x. We wish to show that y = γ(x). Because x = h(x) = y + [f(x)] = y + [(i n γ)(x)] = y + x [γ(x)], it follows that γ(x) = y, as desired. QED Corollary (openimage): Let n N, let D R n, let g : D R n and let U D. Assume that g is an approximate identity on U and that U is an open subset of R n. Then g(u) is also an open subset of R n. Proof: Given y 0 g(u). We wish to show that there exists δ > 0 such that B δ (y 0 ) g(u). Choose x 0 U such that y 0 = g(x 0 ). Since U is open, choose ε > 0 such that B ε (x 0 ) U. Let δ := ε/3. We wish to show that B δ (y 0 ) g(u). We have B 2δ (x 0 ) B 3δ (x 0 ) = B ε (x 0 ) U. Then g is an approximate identity on B 2δ (x 0 ). So, by Remark (openstart), we have g( B 2δ (x 0 )) B δ (g(x 0 )). Then B δ (y 0 ) B δ (y 0 ) = B δ (g(x 0 )) g( B 2δ (x 0 )) g(u), as desired. QED Theorem (aihomeom): Let n N, Let D R n, let g : D R n and let W D. Assume that W is open. Assume that g is an approximate identity on W. Then g(w ) is open. Moreover, g W : W g(w ) is a homeomorphism. Proof: Let X := g(w ). Let f := g W : W X. By Corollary (openimage), X is open. We wish to show that f : W X is a homeomorphism. By Remark (aiinj), f : W X is injective. As X = g(w ) = f(w ), we see that f : W X is surjective. By Remark (aiunifc), f : W X is uniformly continuous, and therefore continuous. It remains to show that f 1 : X W is continuous. Let V W and assume that V is open. We wish to show that (f 1 ) 1 (V ) is open. That is, we wish 6

7 to show that f(v ) is open. Since g is an approximate identity on W, it follows that g is an approximate identity on V, so, by Corollary (openimage), we see that g(v ) is open. So, since f(v ) = g(v ), we conclude that f(v ) is open, as desired. QED Lemma (invid): Let n N, let U, W R n both be open and let g : U W be a homeomorphism. Assume that g i n [[1, 0 n ]]. Then g 1 i n [[1, 0 n ]]. Proof: As g i n [[1, 0 n ]], we get g(0 n ) = i n (0 n ) = 0 n. Then 0 n = g 1 (0 n ), and it remains [g 1 (y)] y [g 1 (y)] y to prove that lim = 0 n, or, equivalently, that lim = 0. y 0 n y y 0 n y Let R := i n g. Since g i n [[1, 0 n ]], by Remark (apprvan), choose R : U R n vanishing continuously at 0 n such that, for all x U, we have R(x) = [ R(x)][ x ]. For all x U, we have g(x) = (i n R)(x) = x [ R(x)][ x ]. Replacing x by g 1 (y), we see, for all y W, that y = [g 1 (y)] [ R(g 1 (y))][ g 1 (y) ]. Define ε : W R n by ε(y) = R(g 1 (y)). Then, for all y W, we have y = [g 1 (y)] [ε(y)][ g 1 (y) ], so [ g 1 (y) ] y = [ g 1 (y) ][ε(y)]. [ g 1 (y) ] [ ε(y) ] We therefore wish to prove that lim = 0. y 0 n y Since g : U W is a homeomorphism, it follows that g 1 : W U is continuous. In particular, g 1 is continuous at 0 n. So, since g 1 (0 n ) = 0 n, we conclude that g 1 vanishes continuously at 0 n. Since R and g 1 both vanish continously at 0 n, we conclude that ε vanishes continously at 0 n, as well. Choose δ > 0 such that, for all y B δ (0 n ), we have ε(y) < 1/2. By the Squeeze Theorem, it suffices to prove, for all y (B δ (0 n ))\{0 n }, that 0 [ g 1 (y) ] [ ε(y) ] y < 2 [ ε(y) ]. Given y (B δ (0 n ))\{0 n }. It suffices to prove that g 1 (y) < 2 [ y ]. As g 1 (y) = y + [ε(y)][ g 1 (y) ], we get g 1 (y) y + [ ε(y) ][ g 1 (y) ]. Because y B δ (0 n ), we get ε(y) < 1/2. Then g 1 (y) < y + [1/2][ g 1 (y) ]. Then, subtracting [1/2][ g 1 (y) ] from both sides, we get: [1/2] [ g 1 (y) ] < y. Multiplying both sides by 2 then yields: g 1 (y) < 2 [ y ], as desired. QED Exercise (linderiv): Let m, n N, let M R n m and let x R m. Show that L M : R m R n is differentiable at x and that (L M ) (x) = M. Lemma (00Diff): Let n N, let U, V R n both be open and let f : U V be a homeomorphism. Assume that 0 n U V and that f(0 n ) = 0 n. Assume that f : U V is differentiable at 0 n. Then f 1 : V U is differentiable at 0 n. Proof: Let A := f (0 n ), let B := A 1, let W := L B (V ) and let λ := L B V : V W. Then L A W : W V is a continuous inverse to λ, it follows that λ : V W is a homeomorphism. We have λ(0 n ) = L B (0 n ) = 0 n. Also, by Exercise (linderiv), we see 7

8 that L B is differentiable at 0 n and that (L B ) (0 n ) = B. Because λ agrees with L B on V, because V is an open neighborhood of 0 n and because L B is differentiable at 0 n, it follows that λ is differentiable at 0 n and that λ (0 n ) = (L B ) (0 n ). Then λ (0 n ) = (L B ) (0 n ) = B. Let g := λ f : U W. Then g(0 n ) = λ(f(0 n )) = λ(0 n ) = 0 n. Also, g : U W is a composition of homemomorphisms, and so is a homeomorphism. Also, by the Chain Rule, g is differentiable at 0 n and g (0 n ) = [λ (f(0 n ))][f (0 n )]. We have λ (f(0 n )) = λ (0 n ) = B and f (0 n ) = A. Then g (0 n ) = BA = A 1 A = I n. By Remark (CL), we see that g 0 + L In [[1, 0 n ]]. So, as L In = i n, we get g i n [[1, 0 n ]]. Then by Lemma (invid), g 1 i n [[1, 0 n ]]. So, since i n : R n R n is a polynomial of degree 1, we conclude that g 1 is differentiable at 0 n. We have λ f = g, so f = λ 1 g, so f 1 = g 1 λ. We know that λ is differentiable at 0 n, that λ(0 n ) = 0 n and that g 1 is differentiable at 0 n. By the Chain Rule, f 1 is differentiable at 0 n. QED Proposition (GenDiff): Let n N, let U, V R n both be open, let f : U V be a homeomorphism and let x 0 U. Assume that f : U V is differentiable at x 0 and that f (x 0 ) is nonsingular. Then f 1 : V U is differentiable at f(x 0 ). Proof: Let y 0 := f(x 0 ). We wish to show that f 1 : V U is differentiable at y 0. Let U 0 := U x 0 and V 0 := V y 0. Define α : U U 0 by α(x) = x x 0. Define β : V V 0 by β(y) = y y 0. Also, α : U U 0 is a homeomorphism, so α 1 : U 0 U is a homeomorphism. Also, for all x U 0, we have α 1 (x) = x + x 0. Also, β is a homeomorphism, so β 1 : V 0 V is a homeomorphism. Also, for all y V 0, we have β 1 (y) = y + y 0. Because α is a restriction, to the open set U, of a polynomial, it follows that α is differentiable on U. Similarly, β is differentiable on V. Similarly, α 1 is differentiable on U 0. Similarly, β 1 is differentiable on V 0. In partciular, α is differentiable at x 0, β is differentiable at y 0, α 1 is differentiable at 0 n and β 1 is differentiable at 0 n. Let f 0 := β f α 1 : U 0 V 0. Then f 0 : U 0 V 0 is a composition of homeomorphisms, so is a homeomorphism. Also, f 0 (0 n ) = β(f(α 1 (0 n ))) = β(f(x 0 )) = β(y 0 ) = 0 n. By the Chain Rule, we conclude that f 0 is differentiable at 0 n. Then, by Lemma (00Diff), it follows that f0 1 is differentiable at 0 n. Since f0 1 = α f 1 β 1, we see that α 1 f0 1 β = f 1. Then, by the Chain Rule, f 1 : V U is differentiable at y 0, as desired. QED Fact (invsmooth): Let n N and define N := {M R n n det M 0}. mapping M M 1 : N R n n is C. Then the Proof: The transposed-cofactor map trcof : R n n R n n and the determinant map det : R n n R are both polynomials and are therefore C. Then, by the Quotient Rule, trcof/ det : N R n n is also C. For all M R n n, we have M 1 = (trcof/ det)(m). The result follows. QED Fact (opnormcontin): Let n N. Then M M : R n n R is uniformly continuous with respect to the L 2 norm on R n n. Proof: This is Problem 50 from the Homework. Fact (opnormcontin): Let n N, let U, V R n be both be open, let f : U V and let 8

9 l N { }. Then f : U V is a C l -diffeomorphism means: f : U V is bijective, f : U V is C l and f 1 : V U is C l. Note: A C 1 homeomorphism need not be a C 1 -diffeomorphism. For example, the function x x 3 : R R is a C 1 homemorphism, but its inverse x 3 x : R R is not differentiable at 0. This example shows that, without the assumption that f (x 0 ) is nonsingular, Proposition (GenDiff) fails. Lemma (bootstrap): Let l, n N, let U, V R n both be open and let f : U V be a C l homeomorphism. Assume, for all x U, that f (x) is nonsingular. Assume that f 1 : V U is C l 1. Then f 1 : V U is C l. Proof: Let g := f 1 : V U. We wish to show that g is C l. By Proposition (GenDiff), g : V U is differentiable, and so it suffices to prove that g : V R n n is C l 1. Let N := {M R n n det M 0}. Then f (U) N. Define ι : N R n n by ι(m) = M 1. By Fact (invsmooth), we know that ι : N R n n is C. By assumption, g : V U is C l 1. Because f : U V is C l, it follows that f : U N is C l 1. We have i n V = f g, so, by the Chain Rule, for all y V, we have (i n V ) (y) = [f (g(y))][g (y)]. As V is open, for all y V, we have (i n V ) (y) = i n(x). Because i n = L In, it follows, for all y R n, that i n(y) = I n. Then, for all v Y, we have I n = i n(y) = (i n V ) (y) = [f (g(y))][g (y)], and so, since f (g(y)) is nonsingular, we get [f (g(y))] 1 = g (y), i.e., we get ι(f (g(y))) = g (y). Then ι f g = g : V R n n. So, since g : V U is C l 1, and since f : U N is C l 1, and since ι : N R n n is C, it follows that g : V R n n is C l 1, as desired. QED Theorem (reginv): Let n N, let l N { }, let U, V R n both be open and let f : U V be a C l homeomorphism. Assume that f is nonsingular on U. Then f : U V is a C l -diffeomorphism. Proof: Case 1: l N. Proof in Case 1: Because f : U V is a homeomorphism, it follows that f 1 : V U is C 0. Then, by Lemma (bootstrap), f 1 : V U is C 1. Then, by Lemma (bootstrap), f 1 : V U is C 2. Then, by Lemma (bootstrap), f 1 : V U is C 3. Continuing, we eventually see that f 1 : V U is C l. Then f : U V is a C l -diffeomorphism, as desired. End of proof in Case 1. Case 2: l =. Proof in Case 2: Since f : U V is C 1 and C 2 and C 3 and, it follows, from Case 1, that f : U V is a C 1 -diffeomorphism and a C 2 -diffeomorphism and a C 3 -diffeomorphism and. Then f 1 : V U is C 1 and C 2 and C 3 and.... Then f 1 : V U is C. Then f : U V is a C -diffeomorphism. End of proof in Case 2. QED Theorem (InvFnThm): Let n N, let l N { }, let D R n be open, let x 0 D and let f : D R n be C l. Assume that f (x 0 ) is nonsingular. Then there exist open neighborhhoods U of x 0 and V of f(x 0 ) such that U D, f(u) = V and f U : U V is a C l -diffeomorphism. 9

10 Proof: Let A := f (x 0 ) R n n. Then A is nonsingular. Let B := A 1 R n n. Let g := L B f : D R n. Then g is C l on D, and g (x 0 ) = I n. We have [g(x 0 )] I n = 0. Also, recall that D is open. So, by continuity of x [g(x)] I n : U R, choose δ > 0 such that B δ (x 0 ) D and such that, for all x B δ (x 0 ), we have [g(x)] I n < 1/2. Let U := B δ (x 0 ). Then U D. Also, for all x U, we have [g(x)] I n 1/2. Then, by Remark (impliesai), g is an approximate identity on U. Also, by Corollary (ainonsing), g is nonsingular on U. Let W := g(u) and let g 0 := g U : U W. By Theorem (aihomeom), W is open and g 0 : U W is a homeomorphism. We have g 0 = g on U. So, since g is C l on D and since U D, it follows that g 0 is C l on U. Since g 0 = g on U, we conclude that g 0 = g on U. So, since g is nonsingular on U, it follows that g 0 is nonsingular on U. Then, by Theorem (reginv), g 0 : U W is a C l -diffeomorphism. We have L A g = L A L B f = f. Let V := f(u) = L A (g(u)) = L A (W ). Because L A : R n R n is a homeomorphism and because W is open, it follows that L A (W ) is open, i.e., that V is open. So since f(x 0 ) f(u) = V, we see that V is an open neighborhood of f(x 0 ). Let f 0 := f U : U V. It remains to show that f 0 : U V is a C l -diffeomorphism. Let λ := L A W : W V. Then λ : W V is a C -diffeomorphism. So, because g 0 : U W is a C l -diffeomorphism, it follows that λ g 0 : U V is a C l -diffeomorphism. Because L A g = f, it follows that (L A g) U = f U, i.e., that λ g 0 = f 0. Then f 0 : U V is a C l -diffeomorphism, as desired. QED Definition: Let m, n N, let W R m be open and let f : W R n. Then f is open means: for all U W, [ U is open ] [ f(u) is open ]. Corollary (nonsingopen): Let n N, let W R n be opne and let g : W R n be C 1. Assume that g is nonsingular on W. Then g : W R n is open. Proof: Let D W be open. We wish to show that g(d) is open. Let f := g D : D R n. Then f : D R n is C l and f(d) = g(d). We wish to show that f(d) is open. Given y 0 f(d). We wish to show that there exists a neighborhood V of y 0 such that V f(d). Choose x 0 D such that f(x 0 ) = y 0. Since f = g on D, since D is an open subset of W, it follows that f = g on D. Since x 0 D W, we see that g (x 0 ) is nonsingular. Then f (x 0 ) is nonsingular. By Theorem (InvFnThm), choose open neighborhoods U of x 0 and V of f(x 0 ) such that U D and such that f(u) = V. As y = f(x 0 ), we see that V is an open neighborhood of y 0, and it remains to show that V f(d). We have U D and f(u) = V, so V = f(u) f(d), as desired. QED In Corollary (nonsingopen), we can replace C 1 by differentiable and the result is still true. However, the proof requires techniques from Algebraic Topology that are outside the scope of this course. (The basic result that is needed is: Let n N, let r > 0 and let x R n. Let B := B r (x 0 ) and let B denote the boundary in R n of B. Let δ := r/100. Let f : B R n be a continuous function such that, for all x B, we have: [f(x)] x < δ. Then f( B ) B δ (x 0 ). Proof: Say, for a contradiction, that p B δ (x 0 ) and p / f( B ). Let A : B B denote the antipodal map, defined by A(x) = x 2(x x 0 ). Then, 10

11 for all x B, we have [A(x)] x = πr. Fix a continuous function r : R n \{p} B such that, for all y [B r+δ (x 0 )]\[ B r δ (x 0 ) ], we have [r(y)] y < 10δ. Then, for all x B, we have [r(f(x))] x [r(f(x))] [f(x)] + [f(x)] x < 10δ + δ = 11δ; so, since 11δ = (11/100)r < πr, we see that r(f(x)) A(x). Then (r f) ( B) : B B is homotopic to the identity map B B. On the other hand, because B is contractible and because r f : B B is continuous, it follows that (r f) B : B B is homotopic to a constant map. Thus the identity map B B is homotopic to a constant map. Thus the identity map H n 1 ( B) H n 1 ( B) is equal to zero, contradiction.) In Corollary (nonsingopen), we can replace C 1 by continuous, provided we also replace g is nonsingular on W by g : W R n is injective. However, the proof requires techniques from Algebraic Topology that are outside the scope of this course. (The result that is needed is called Invariance of Domain.) It is a consequence of the Mean Value Theorem that, for any function f : R R, if, for all t R, we have f (t) 0, then f is injective. By the following Example (noninj), this fact from one-variable calculus does not directly generalize to the multivariable setting; however, in Proposition (diffapproxid) below, we expose a kind of generalization. Example (noninj): For all integers n 2, there exists a C function g : R n R n such that g is nonsingular on R n, but g is not injective. Proof: Define g : R n R n by g(r, s, t 1,..., t n 2 ) = (e r [cos s], e r [sin s], t 1,..., t n 2 ). Then g is C. Also, for all r, s, t 1,..., t n 2 R, we have det(g (r, s, t 1..., t n 2 )) = e 2r 0. Finally, g(0, 0, 0,..., 0) = g(0, 2π, 0,..., 0), so g is not injective. QED Theorem (injdiffeo): Let n N, let l N { }, let U R n be open and let f : U R n be C l and injective. Assume that f is nonsingular on U. Then f(u) is open and f : U f(u) is a C l -diffeomorphism. Proof: By Corollary (nonsingopen), f : U R n is open. Let V := f(u). Then V is open, and it remains to show that f : U V is a C l -diffeomorphism. Since f : U V is injective, surjective, continuous and open, we conclude that f : U V is a homeomorphism. So, as f is nonsingular on U, by Theorem (reginv), we see that f : U V is a C l -diffeomorphism, as desired. QED By Example (noninj), if we drop, from Theorem (injdiffeo), the assumption that f is injective, then the result is no longer true. This is a difference between one-variable calculus and multi-variable calculus. In a one-variable calculus course, it is not unusual to ask students to show, say, that, with f : R R defined by f(x) = x + x 3, then f : R R is bijective. There is something analogous in multi-variable calculus: Proposition (diffapproxid): Let n N, let l N { } and let f : R n R n be C l. Assume that I n f 1/2 on R n. Then f : R n R n is a C l -diffeomorphism. Proof: By Corollary (ainonsing), f is nonsingular on R n. So, by Theorem (reginv), it suffices to show that f : R n R n is a homeomorphism. By Remark (impliesai), f is an approximate identity on R n. Define τ : R n R n by τ(x) = x [f(0 n )]. Let g := τ f : R n R n. Since f = τ 1 g and since τ : R n R n is 11

12 a homeomorphism. it suffices to show that g : R n R n is a homeormophism. Since f is an approximate identity on R n and since τ : R n R n is distance-preserving, we see that g is an approximate identity on R n. Then, by Theorem (aihomeom), it follows that g : R n g(r n ) is a homeomorphism. It therefore suffices to show that g(r n ) = R n. We have g(0 n ) = τ(f(0 n )) = [f(0 n )] [f(0 n )] = 0 n. So, since g is an approximate identity on R n, it follows, for all δ > 0, that g is an approximate identity on B 2δ (0 n ); then, by Remark (openstart), we get g( B 2δ (0 n ) ) B δ (0 n ). Taking the union over all δ > 0, we get g(r n ) R n. So, as g(r n ) R n, we get g(r n ) = R n, as desired. QED For a specific application of Proposition (diffapproxid), define f : R 2 R 2 by, say, f(s, t) = ( s + e s2 t , t + e s4 t We leave it as an exercise to show, for all s, t R, that I 2 [f (s, t)] 1 1/2. Then, by (c) of Exercise (opnormprop), for all s, t R, we have I 2 [f (s, t)] 1/2. Thus, I 2 f 1/2 on R 2. So, by Proposition (diffapproxid), since f : R 2 R 2 is C, it follows that f : R 2 R 2 is a C -diffeomorphism. ). 12