HW 3 part 2 Let H_{1}(x) = F (x + a) − F (a). Then DH_{1}(0) = Let H(x) = [DF (a)]^{−1}H_{1}(x) H(0) = and by the chain rule, DH(0) = By 8, 9, and note, H is a diffeomorphism iff F is a diffeomorphism. Thm 6.4 (Inverse Function Theorem): Suppose F : W^{open} ⊂ R^{n} → R^{n} ∈ C^{r}. Suppose for a ∈ W, det(DF (a)) ≠ 0. Then there exists U such that a ∈ U^{open}, V = F (U) is open, and F : U → V is a C^{r}-diffeomorphism. Moreover, for x ∈ U and y = F (x), DF_{y}^{−1} = (DF_{x})^{−1} Proof: Without loss of generality, assume F (0) = 0 and DF (0) = I Claim: There exists a nbhd U_{1} of 0 such that F |_{U_{1}} is 1:1. Method 1: Suppose for all n ∈ N, there exists an 0 < h_{n} < \frac{1}{n} such that F (h_{n}) = F (0) Then \frac{F (0+h_{n})−F (0)−Ih_{n}}{||h_{n}||} = But lim_{h→0} \frac{F (0+h)−F (0)−Ih}{||h||} = Hence

Claim: DF_{y}^{−1} = (DF_{x})^{−1} for y in some nbhd V_{1} of 0. Let c ∈ F^{−1}(V_{1}) and let T = DF (c). Since F is differentiable, there exists r(x, c) such that

Let y = F (x) and d = F (c). Then y − d = T [F^{−1}(y) − F^{−1}(d)] + ||F^{−1}(y) − F^{−1}(d)||r(F^{−1}(y), F^{−1}(d)) Thus F^{−1}(y) − F^{−1}(d) = T^{−1}(y − d) − ||F^{−1}(y) − F^{−1}(d)||T^{−1}r(F^{−1}(y), F^{−1}(d)) Let R(y, d) = \frac{F^{−1}(y)−F^{−1}(d)−T^{−1}(y−d)}{||y−d||} = −\frac{||F^{−1}(y)−F^{−1}(d)||}{||y−d||}T^{−1}r(F^{−1}(y), F^{−1}(d)) IF F^{−1} is continuous, then lim_{y→d}r(F^{−1}(y), F^{−1}(d)) = IF \frac{||F^{−1}(y)−F^{−1}(d)||}{||y−d||} is bounded, then lim_{y→d}R(y, d) = Hence DF^{−1}(d) = Thus we need to show F^{−1} is continuous and \frac{||F^{−1}(y)−F^{−1}(d)||}{||y−d||} is bounded.