MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $B$ be an infinite-dimensional Banach space, and let $M\subset\mathbb{R}^n$ be a neighborhood of the origin in $\mathbb{R}^n$.

Suppose that $I:M\to B$ is a real-analytic function with $I(0)=0$ and such that the derivative of $I$ at $0$ has maximal rank.

Is it true that there exist neighborhoods $U,V\subset B$ of $0$ and a real-analytic diffeomorphism $\phi:U\to V$ such that $\phi\circ I$ is the restriction of a linear map $\mathbb{R}^n\to B$? If so, what is a good reference?

EDIT: I asked this question in the real-analytic setting, but might as well have done so in the complex-analytic case.

share|cite|improve this question
up vote 2 down vote accepted

I cannot give you a reference, but the answer ought to be yes.

To simplify notation, identify $\mathbb{R}^n$ with its image in $B$ under the derivative $I'(0)$. That image, being finite-dimensional, is the range $pB$ of a finite rank, bounded projection $p$. Replacing $M$ by a smaller neighbourhood if necessary, we can assume that $p\circ I$ is a diffeomorphism of $M$ onto a neighbourhood $N$ of $0$ in $pB$. Let $h\colon N\to M$ be its inverse, and define $\phi\colon p^{-1}(N)\to p^{-1}(M)$ by $$\phi(w)=x+w-I(x),\qquad x=h(pw).$$ If $w\in pB$ then $h(p(I(w)))=w$, so $\phi(I(w))=w+I(w)-I(w)=w$ – i.e., $\phi\circ I$ is the identity on $pB$.

To find the inverse of $\phi$, note that if $y=\phi(w)$ and $x=h(pw)\in pB$ then $py=x+pw-p(I(h(pw))=x$, so $w=y-x+I(x)=y-p(y)+I(py)$, i.e., $$\phi^{-1}(y)=y-py+I(py).$$

Edit: On second thought, it would have been more natural to define $\phi^{-1}$ first, with the requirement that its restriction to $pB$ be $I$. Letting its restriction to $(1-p)B$ be the identity is the simplest way to make it a diffeomorphism.

share|cite|improve this answer
Yes, this looks good. I was sort of hoping for a reference to the literature (as I'm essentially asking for a version of the constant rank theorem), so I will leave the question open for a while in case someone has a pointer! – Lasse Rempe-Gillen May 1 '10 at 21:53

Your Answer


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.