Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

The Inverse Function Theorem provides sufficient conditions to determine when a function is defined implicitly by a relation. I would like to know some ways to determine when no such function is defined.

Below is a link to a specific example and conjecture.

http://math.stackexchange.com/questions/46750/how-to-prove-the-implicit-function-theorem-fails

share|improve this question

4 Answers 4

In the same spirit as Michael's answer: Over the complex variables, there is a theorem due to W.F. Osgood (published in his "Lehrbuch der Funktionentheorie", 2. Aufl., Bd. 2, Parte 1, Leipzig 1929) about solvability of $w=f(z)$ for a system of holomorphic functions on a neighborhood of a point $a \in \mathbb{C}^n$, which is an isolated point of the set $\{z:f(z)=b:=f(a)\}$. This is nicely discussed in B.V. Shabat's book "Complex analysis" (part II, section 14, item 44-although I am not sure if this is included in the English translation of the book)-in terms of resultants and Weierstrass' Preparation Theorem. Shabat also refers to: M. Herve, "Several Complex Variables. Local Theory", Oxford 1963. And there is some information (on the topic of failure of the implicit function theorem) in the book by Aizenberg and Yuzhakov on residue theorems.

share|improve this answer

A natural approach would be to classify such singular points by the deficiency of the rank of the Jacobian. If the deficiency is one, you can solve for all but one of the variables and reduce the problem to a scalar equation. The rest is then quite straightforward: The equation f(x)=y, with f(0)=0 is solvable for x in a neighborhood of 0 if the leading term in the Taylor expansion of f is odd; it is not always solvable if the leading term is even. If the deficiency in the rank of the Jacobian is two, you end up with a system of two equations, generally quadratic at leading order. Discussing the solvability of such a system is still a manageable task.

share|improve this answer

In this context, a remarkable phenomenon to consider is also lack of uniqueness, that may be considered an instance of bifurcation.

share|improve this answer
    
@Pietro: I'm sorry, I don't quite follow. In the context, what is non-unique? Just to see if I understand correctly, are you referring to the case where the implicit relation is multi-valued (for example the relation from $x$ to $y$ near $(1,0)$ for the function $f(x,y) = x^2 + y^2 = 1$)? Thanks. –  Willie Wong Jun 25 '11 at 11:58
    
Yes: e.g. for $f:\Lambda \times X\to Y$, I mean non-uniqueness, for a given $\lambda\in\Lambda $, of a point $x$ such that $f(\lambda,x)=0$. If this happens in any nbd of some $\lambda_0$, we say $\lambda_0$ is bifurcation point for the equation $f(\lambda,x)=0$. I meant to recall that a relevant case (thus not a pathology but an important phenomenon) where the picture of the IFT does not hold is this, bifurcation. –  Pietro Majer Jun 28 '11 at 16:36

Just a comment about the link you posted: G=0 implies $4uv=2y^2-2x^2$ so when you substitute in $F=0$ you simply get $x^2+y^2+u^2+v^2=0$ which can be only satisfied by $x=y=u=v=0$ (if you work in $\mathbb{R}^4$...)

share|improve this answer
    
Yeah that result was mentioned I guess it was kind of assumed the variables are Real. I elaborated a little to say that this means you can't have x or y defined as functions of u and v in an open n-hood of any point (u,v) because the only point that's a candidate is (0,0) and even at this point any variation in u or v will no longer satisfy the system. What I'm curious about is if there are other ways to make conclusions like that. –  LaLone Jun 25 '11 at 2:06

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.