# jV0kdr-10 - 24 SPRING 2012 10 Proof of the Implicit...

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24 SPRING 2012 10. Proof of the Implicit Function Theorem. Proof. The formula for g ( x ) will follow from the rest of the theorem, as in Exercise 9.1, by defining Φ : R n R n × R m by Φ ( x ) = ( x , g ( x )) , so that ( f Φ )( x ) = f ( x , g ( x )) = 0 for all x I , and then using the Chain Rule. Define F : U R n × R m by F ( x , y ) = ( x , f ( x , y )). Then F is C 1 on U , with (10.1) F ( x , y ) = Id 0 D I f ( x , y ) D II f ( x , y ) ; this also shows that F ( a , b ) is invertible, since D II f ( a , b ) is. Thus the Inverse Function Theorem provides open sets V and W such that ( a , b ) V U , F ( a , b ) W R n × R m , and F : V W is one-to-one and onto with C 1 inverse. By shrinking V and W if necessary we can assume that V = K × J for some open sets K R n and J R m . Let u and v denote the C 1 component functions of F 1 , so F 1 ( s , t ) = ( u ( s , t ) , v ( s , t )) for all ( s , t ) W . Then ( s , t ) = F ( F 1 ( s , t )) = F ( u ( s , t ) , v ( s , t )) = ( u ( s , t ) , f ( u ( s , t ) , v ( s , t ))) , which shows that (10.2) u ( s , t ) = s and f ( s , v ( s , t )) = t for all ( s , t ) W . Let I = { x R n | ( x , 0 ) W } . Then a I since ( a , 0 ) = F ( a , b ) W . Moreover, I is open, since for every ( x , 0 ) W there are open sets A R n and B R m such that ( x , 0 ) A × B W ; in particular, A × { 0 } W , so x A I .

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