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A fairly general form of the Inverse Function Theorem is:

Suppose $X, Y$ are Banach spaces, $U \subset X$ is open and $f:U \to Y$ is continuously differentiable. If for some $x \in U$ the derivative $Df(x)$ is invertible, then there exists a neighborhood $V \subset f(U)$ such that $f(x) \in V$ and a continuously differentiable function $g: V \to U$ such that $f(g(x)) = x$ for all $x \in V$.

A question I have had is whether there are any sufficient conditions such that the converse holds, i.e., if $Df(x)$ is not invertible then $f$ is locally not invertible.?

Thanks in advance.

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$f(x)=x^3$ is not a counterexample, because $f$ is not a diffeomorphism around $0$. Any diffeomorphism has nonvanishing derivative, by the inverse function theorem. Your "in other words" sentence has a different meaning from the preceding one. –  user31373 Jul 17 '12 at 0:42
    
@LeonidKovalev Thanks I see, because $x^{1/3}$ is not continuously differentiable. I will edit this "in other words" out of the question. –  user12014 Jul 17 '12 at 1:48

2 Answers 2

up vote 2 down vote accepted

I prefer to phrase the question in the positive way: Suppose that $f:U\to V$ is invertible and $C^1$ smooth. Under what additional assumptions can we conclude that $f$ is a diffeomorphism, equivalently, that $Df$ is always invertible?

Results of this kind do exist, but only in finite dimensions as far as I know, and in low dimension at that. Way back in 1936 H. Lewy proved that in two dimensions every harmonic homeomorphism is a diffeomorphism (harmonicity of $f=(f_1,f_2)$ means that $\Delta f_1=\Delta f_2=0$, i.e., the components are harmonic functions). There are generalizations to other elliptic PDE in two dimensions, but unfortunately everything seems to fall apart in higher dimensions. This talk by G. Alessandrini gives an overview of this subject.

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It suffices to require that the inverse function $g$ be surjective. Suppose there exists a neighborhood $V\subset f(U)$ such that $f(x)\in V$ and a surjective continuously differentiable function $g:V\to U$ such that $f(g(x))=x$ for all $x\in V$. Then for any $x\in U$ we have some $y\in V$ such that $g(y)=x$, so by Chain Rule $$\begin{align} I &= D_y\:\mathrm{Id}\\ &=D_{y}(f\circ g)\\ &=D_{x}f\circ D_yg \end{align}$$ thus $D_xf$ is invertible.

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