Inverse Function Theorem and its Applications Sachchidanand Prasad Outline Statement of the Theorem ....................................... 2 Mean Value Inequality......................................... 2 Contraction Mapping Theorem.................................... 2 Proof of the Theorem 1......................................... 3 Holomorphic Inverse Function Theorem in one Complex Variable ............... 5 Implicit Function Theorem....................................... 6 An approximation of inverse function in one dimension...................... 7 Application of Inverse Function Theorem in Manifold....................... 8 References ................................................ 9 1 Statement of the Theorem n n 1 Theorem 1 (Inverse Function Theorem). Suppose U ⊆ R is open, x0 2 U; f : U ! R is C n and Df(x0) is invertible. Then there is a neighborhood V ⊆ U; W ⊆ R of x0 and f(x0) = y0 respectively and a C 1 function g : W ! V (see Figure 1) such that f(g(y)) = y and g(f(x)) = x; 8 x 2 V and 8 y 2 W: Moreover, Dg(f(x)) = (Df(x))−1 f W V f(x0) = y0 x0 g U Figure 1 Before seeing the proof of the above theorem we will recall some of the theorems that we are going to use in our proof. Mean Value Inequality n n Theorem 2. Let f : U ⊆ R ! R be differentiable on the open set U. Let x; y 2 U such that [x; y]1 ⊆ U. Then kf(x) − f(y)k ≤ max kDf(c)k · kx − yk: (1) c2[x;y] Proof. See [3, p. 248]. Contraction Mapping Theorem Definition 1 (Contraction Map). Let (X; d) be a metric space and φ : X ! X be any function. If there exists a number c 2 (0; 1) such that for all x; y 2 X d(φ(x); φ(y)) ≤ c d(x; y); (2) then φ is said to be a contraction map of X. 1line joining x and y 2 Theorem 3. If X is a complete metric space, and if φ is a contraction map of X, then there exists a unique x 2 X such that φ(x) = x, i.e. φ has a unique fixed point. Proof. See [2, p. 220]. Proof of the Theorem 1 Without loss of generality we can assume the following: (For an explicit definition see [1, p. 185]) 1. x0 = 0 (By setting xnew = x − x0). 2 2. f(0) = 0 (by setting fnew(x) = f(xnew + x0) − f(x0)): −1 3 3. Df(0) = I By setting fnew(x) = (Df(0)) f(x) . Let (x) = x − f(x). Clearly 2 C 1. Observe that, D (0) = O. Since 2 C 1 and D (0) = O, implies there exists an r > 0 such that, 1 kxk ≤ r =) kD (x)k ≤ : (3) 2 x r r 2 0 y 8 x 2 B(0; r); kD (x)k < 1=2 Figure 2 Using Theorem 2, r kxk ≤ r =) k (x)k ≤ max kD (c)k kxk ≤ (4) c2[0;x] 2 n r Fix y 2 R with kyk < 2 and define φy(x) = x − f(x) +y (5) | {z } (x) 2 Observe that xnew + x0 2 U: 3Here f is the new function that we have defined in step 2. 3 Using Equation (4), kxk ≤ r =) kφy(x)k = kx − f(x) + yk ≤ kx − f(x)k + kyk < r: So, due to continuity of φy, we can say that φy : B[0; r] ! B[0; r]. For any x; y 2 B[0; r]; 1 kφ (x) − φ (y)k = k (x) − (y)k ≤ kx − yk (follows from Theorem 2) y y 2 Now using Theorem 3, φy is a contraction map. So, φy has a unique fixed point, say xy 2 B[0; r]: So, there exists a unique xy 2 B[0; r] such that, f(xy) = y: r Since kyk < 2 =) xy 2 B(0; r): Now define, r W = 0; ;V = f −1(W ) \ (0; r) and g : W ! V as g(y) = x : (See Figure 3) B 2 B y f B(0; r) r W = B(0; r=2) V x0 y0 g U Figure 3 Clearly, (f ◦ g)(y) = y;(g ◦ f)(x) = x; 8x 2 V and 8y 2 W: Now we will prove that g 2 C 1: g is continuous : Let y; z 2 W: Let g(y) = u and g(z) = v. Consider, 1 k (u) − (v)k ≤ ku − vk 2 1 =) k(f(u) − f(v)) − (u − v)k ≤ ku − vk 2 =) ku − vk ≤ 2kf(u) − f(v)k =) kg(y) − g(z)k ≤ 2ky − zk < " g is differentiable : Let y 2 W and g(y) = x. We will prove that Dg(y) = (Df(x))−1. Let Df(x) = A. Since W is open, 9k such that y+k 2 W: Set g(y+k) = x+h =) h = g(y+k)−g(y). 4 We want g(y + k) − g(y) − A−1k ! 0 as k ! 0: kkk A(g(y + k) − g(y)) − k () ! 0 as k ! 0: kkk Ah − k Ah − (f(x + h) − f(x)) () = ! 0 as k ! 0: kkk kkk (f(x + h) − f(x)) − Ah khk () − ! 0 as k ! 0: (6) khk kkk Since, khk ≤ 2kkk; h ! 0 if k ! 0: So, the limit in (6) is 0 as h ! 0: Hence, g is differentiable. And, Dg(y) = (Df(g(y)))−1 (7) −1 Now, since Dg is a composition of the function y Df(g(y)) and A A on the space of invertible matrices. Since g is continuous and f 2 C 1;Df is continuous. As Dg = i ◦ Df ◦ g, where i denotes the inverse of a matrix and the composition of continuous functions is continuous hence y Dg(y) is continuous. Remark: In the above theorem, if f 2 C k then g 2 C k: n n 1 n Corollary 1. If f : R ! R is C and has the property that for every a 2 R ;Df(a) is non-singular. Then f is ana open map. n Proof. Let U ⊂ R be any open set. We will show that f(U) is open. Let y 2 f(U) =) 9 a 2 U such that f(a) = y: According to the hypothesis, Df(a) is non-singular, so using Theorem 1, there exist neighborhoods V ⊂ U; W of a and f(a) respectively and a C 1 function g : W ! V such that f ◦ g = idW : So, W = f(g(W )) which is open and x 2 W =) g(x) 2 V =) x = f(g(x)) 2 U =) W ⊂ f(U): Hence, f is an open map. Holomorphic Inverse Function Theorem in one Complex Variable Theorem 4. Let f : C ! C be a holomorphic map such that f(0) = 0 and Df(0) is non-singular. Then there exists open sets U; V around the origin and g : V ! U such that f ◦ g = idV ; g ◦ f = idU and g is holomorphic. 2 2 2 Proof. Let f(z) = u(x; y) + iv(x; y) where u; v : R ! R. So we can write the function f : R ! R given by f(x; y) = (u(x; y); v(x; y)). Given that, • f(0; 0) = (0; 0) • f is holomorphic implies f 2 C 1: ux(0; 0) −vx(0; 0) 2 2 • Df(0; 0) = and det Df(0; 0) = ux(0; 0) + vx(0; 0) 6= 0: vx(0; 0) ux(0; 0) 5 Hence from the real inverse function theorem there exists neighborhoods U; V around the origin 1 and g : V ! U such that f ◦ g = idV and g ◦ f = idU : From inverse function theorem g will be C . Now we will show that g is holomorphic function. Let f 0(0) = α. Consider −1 −1 −1 ux(0; 0) −vx(0; 0) 1 Re(α) Im(α) Re(α ) −Im(α ) Dg(0; 0) = = 2 = −1 −1 : vx(0; 0) ux(0; 0) jαj −Im(α) Re(α) Im(α ) Re(α ) As, Dg(0; 0) is C−linear, g is holomorphic Implicit Function Theorem x Theorem 5. Suppose U ⊂ n+m is open and F : U ! m is C 1. Writing a vector in n+m as , R R R y n m x0 @F x0 with x 2 R and y 2 R , suppose that F = 0 and the m × m matrix @y is non-singular. y0 y0 1 Then there are neighborhoods V of x0 and W of y0 and a C function φ : V ! W so that x F = 0; x 2 V and y 2 W () y = φ(x): y Moreover, @F x −1 @F x Dφ(x) = − : @y φ(x) @x φ(x) f y0 W Z g x0 V Figure 4 n+m n+m Proof. Define f : R ! R by 2 x 3 x f = x : y 4F 5 y Since the linear map 2 3 I O x0 Df = 4 @F x0 @F x0 5 y0 @x @y y0 y0 6 n m is invertible. Using Theorem 1 there are neighborhoods V ⊂ R of x0, W ⊂ R of y0, and n+m 1 Z ⊂ R of 0 and a C function g : Z ! V × W so that g is the inverse of f on V × W (See Figure 4). Now we define φ : V ! W by x x = g ; φ(x) 0 Since g is C 1 so is φ(x): Now, 2 x 3 x x x x x = f = f g = =) F = 0: 4F 5 φ(x) 0 0 φ(x) φ(x) 02 x 31 x x x On the other hand, if F = 0; x 2 V and y 2 W; then = g f = g x = y y y @4F 5A y x g =) y = φ(x): 0 x Now we will calculate the derivative of φ.