Implicit function theorem: what if for a given x_i=x^0_i, there are more y_j=y^k_j?
In the case of m+n variables we have n equations which must be satisfied for for the theorem to be applicable, althought (I understand that) there might be more than one solutions. But if for a given there are more than one satisfying the n equations (and I do not find any statement in the theorem preventing this) how can the theorem ensure that there is one and only one set of solutions which are continuous, satisfy and for which ? I mean how can f be a function?...Of course I must be missing something...
thanks a lot
Re: Implicit function theorem: what if for a given x_i=x^0_i, there are more y_j=y^k_
let's look at a very simple example: when m = n = 1. define:
F(x,y) = x2+y2 - 1.
consider now (a,b) such that F(a,b) = 0. suppose that a = √2/2, for example. well, we have two choices for b, both of which work: b = √2/2, and b = -√2/2.
if we use (a,b) = (√2/2,√2/2) we get the function (defined only in a neighborhood of √2/2 in R) f(x) = √(1 - x2).
if we use (a,b) = (√2/2,-√2/2), we get the function f(x) = -√(1 - x2).
if we try to define f in a neighborhood of a = 1, we have a problem: neither half of the circle works. why is this so? let's look at the Jacobian matrix DF:
at the point (1,0), the square (1x1) matrix 2b is not invertible, we have b = 0, and the inverse of the 1x1 matrix [2b] does not exist, since det([2b]) = 2b = 2(0) = 0.
this is the important part you're missing about the implicit function theorem, there is a condition on how F is changing with respect to the y-variables, the nxn matrix (∂Fm+i/∂yj) MUST be invertible.
the second key point is "which solution" of F(x1,....,xm,y1,....,yn) = 0 we choose (and there could be several), depends on which point:
(a1,...,am,b1,....,bn) we're at, the implicitly defined function f is only defined LOCALLY (it may be impossible to find a function f that works everywhere).
here is a more "grown-up example":
let F(x,y,z,w) = (x2+y2-z2-w2,x2+y2+z+w) this is a function from R4 to R2.
clearly F(1,1,-1,-1) = (0,0). let's check the following matrix determinant:
this tells us that (1,1,-1,-1) is a "bad point" to try to implicitly define a function f with F(x,y,f(x,y)) = (0,0).
on the other hand, we also have F(√2/2,√2/2,-1,0) = (0,0) and for this point the matrix above is:
, which has non-zero determinant, so this is a "good point".