Re: help needed Implicit function theorem

so is it even possible to get some estimate of difference between x_0
and x_1 assuming suitable conditions when sup_{x}|f_1(x)-f_0(x)|<eps or
is it not possible?

Robert Israel wrote:
In article <1162777237.775064.152190@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
Kaushik <standshik@xxxxxxxxx> wrote:
thank you prof israel...... i was trying to generalize this result a
little bit assuming that the functions in the function class are
lipschitz continuous with lipschitz constant say M. further when we
replace f_0 by f_1, neighborhood is defined as
sup_{x} |f_0(x)-f_1(x)|<eps ( epsilon)

so now if (x_0,f_0) is solutionfor the equation F(f,x)=f(x)=0 then we
have f_0(x_0)=0.

now we are trying to solve the equation f_1(x)=0.

now for all x, f_1(x)-eps<f_0(x)<f_1(x)+eps, putting x= x_0, we have,

f_1(x_0)-eps<0<f_(x_0)+eps

now if we assume that there exist y,z such that

f_1(z)=f_1(x_0)-eps<0<f_1(x_0)+eps=f_1(y)


then for the extreme right equality we have
eps=f_1(y)-f_1(x_0)<=M|y-x_0|

or (eps/M)<=|y-x_0|


similarly for teh extreme left equality we have (eps/M)<=|z-x_0|

clearly, solution x_1 such that f_1(x_1)=0 will lie in beween the
points y and z, i.e. between [y,z] or [z,y] depending on the
increasing/decreasing behavior of the function f_1.

can we then say that the new solution will x_1 will lie in between,
x_0-(eps/M)<=x_1<=x_0+(eps/M) ???

No. Consider the case f_0(x) = 0 for all x. Having
|f_1(x)| < eps everywhere doesn't tell you anything about where
f_1(x) = 0.

Robert Israel israel@xxxxxxxxxxx
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada

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