Re: help needed Implicit function theorem

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) ???







Robert Israel wrote:
In article <1162749676.597967.325910@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
Kaushik <standshik@xxxxxxxxx> wrote:
hi,
sorry i have already posted this in another group but i desperately
need an answer. can anybody help me with this problem? suppose x is in
R, f,g belongs
to a differntiably continuous function space G, where any element s in
G is a function from x to R, i.e. s:x -->R

You seem to have garbled this description. The elements of G
should be functions from R (or some subset of R) to R, not just
from a single point x to R. And I suspect you mean a space of
continuously differentiable functions, not "differntiably
continuous function space".

Now suppose i have a relation
F(f,x)=f(x)-g(x)=0 where g is a fixed function from G, and f is a
variable function from G. suppose, the solution of this equation is
(f_0,x_0).


my question is, can i use some variation of implicit function like
theorem to figure out that if i change f_0 in its close
neighborhood to

f_1 ( that is select another function in close neighborhood), how does
that affect x_0 to x_1, where (f_1,x_1) is the new solution of the
relation for this newly chosen function f_1. that is locally, can we
use an implicit function like result?

For convenience, we may take g = 0 (i.e. replace f by f - g;
presumably your function space is a linear space) and x_0 = 0.
Thus f_0(0) = 0, and you want to solve f_1(x) = 0.
Suppose e.g. there are a and b with a < 0 < b and
f_0(a) < 0 < f_0(b); in particular this will be true if
(f_0)'(0) > 0. Then if f_1 is close enough to f_0 that
f_1(a) < 0 < f_1(b), you can use the Intermediate Value Theorem
to get f_1(c) = 0 for some c between a and b.

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

.

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