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Unformatted text preview: Lecture 7 We continue our proof of the Inverse Function Theorem. As before, we let U be an open set in R n , and we assume that 0 U . We let f : U R n be a C 1 map, and we assume f (0) = 0 and that Df (0) = I . We summarize what we have proved so far in the following theorem. Theorem 2.18. There exists a neighborhood U 0 of 0 in U and a neighborhood V of 0 in R n such that 1. f maps U 0 bijectively onto V 2. f 1 : V U 0 is continuous, 3. f 1 is differentiable at . Now, we let U be an open set in R n , and we let f : U R n be a C 2 map, as before, but we return to our original assumptions that a U , b = f ( a ), and Df ( a ) : R n R n is bijective. We prove the following theorem. Theorem 2.19. There exists a neighborhood U 0 of a in U and a neighborhood V of b in R n such that 1. f maps U 0 bijectively onto V 2. f 1 : V U 0 is continuous, 3. f 1 is differentiable at b . Proof. The map f : U R n maps a to b . Define U = U a = { x a : x U } . Also define f 1 : U R n by f 1 ( x ) = f ( x + a ) b , so that f 1 (0) = 0 and Df 1 (0) = Df ( a ) (using the Chain Rule)....
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 Fall '04
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