fWa4tB-08

fWa4tB-08 - 20 SPRING 2012 8. Proof of The Inverse Function...

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20 SPRING 2012 8. Proof of The Inverse Function Theorem Proof. By Exercises 7.3 and 7.4, we may assume that a = 0 , f ( 0 )= 0 ,and f ° ( 0 ) = Id. Now, since f ° is continuous on U ,( f Id) ° = f ° Id is, so there exists ° > 0suchthat ° ( f Id) ° ( x ) ( f Id) ° ( 0 ) ° = ° ( f Id) ° ( x ) ° < 1 2 whenever ° x 0 ° = ° x ° < ° . Making ° smaller if necessary, we can assume B ° ( 0 ) U , so that V = B ° ( 0 )i saconv exopen neighborhood of 0 on which f Id is C 1 . Thus for any x , z V , the Mean Value Inequality gives ° ( f Id)( x ) ( f Id)( z ) °≤ 1 2 ° x z ° , whence ° f ( x ) f ( z ) ° = ° ( x z )+( f ( x ) x ) ( f ( z ) z ) ° ≥° x z °−° ( f Id)( x ) ( f Id)( z ) ° ≥° x z °− 1 2 ° x z ° = 1 2 ° x z ° . It follows that f is one-to-one on V , since f ( x )= f ( z )forces ° x z ° =0 . Ita lsofo l lows that f 1 : f ( V ) V is continuous, since for any y = f ( x )and w = f ( z )in f ( V ), the above inequality gives ° f 1 ( y ) f 1 ( w ) °≤ 2
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fWa4tB-08 - 20 SPRING 2012 8. Proof of The Inverse Function...

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