Re: Uniqueness of implicit functions

From: Mikhail V. Sokolov (mmm_spam_at_rambler.ru)
Date: 12/16/04 

Date: 16 Dec 2004 08:15:01 -0500

On 15 Dec 2004 19:45:01 -0500, Robert Israel wrote:
>>May be we should start with easier case:
>>Consider a "regular" (say, continuous) function y=f(x) of real
>>variable x on some interval B. It is required to find all functions
>>g(x,y) such, that equation g(x,y)=0 has y=f(x) as the only solution,
>>when x is from B.
>
>g(x,y) = h(x,y-f(x)) where h is any function (continuous if you're
>requiring g to be continuous) such that, for x in B, h(x,y) = 0 if
and
>only if y = 0.

 Dear Robert, thank you for the reply.
Please, explain, how one can prove, that all such functions have the
form g(x,y) = h(x,y-f(x)).

Is there a way to generalized the result for the following case:
Suppose that a function f(x1,...,xn)=0 (of real variables x1,....,xn
on some connected set B^n) defines a "reqular" curve in R^(n-1). It is
required to find all functions g(x1,...xn) such, that equation
g(x1,...,xn)=0 has this curve as the only solution in B^n.

Thanks, Mikhail.

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