Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. 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

I'm looking for a complete answer to this problem.

Let $U,V\subset\mathbb{R}^d$ be open sets and $\Phi:U\rightarrow V$ be a homeomorphism. Suppose $\Phi$ is differentiable in $x_0$ and that $\det D\Phi(x_0)=0$. Let $\{C_n\}$ be a sequence of open(or closed) cubes in $U$ such that $x_0$ is inside the cubes and with its sides going to $0$ when $n\rightarrow\infty$. Denoting the $d$-dimension volume of a set by $\operatorname{Vol}(.)$, show that $$\lim_{n\rightarrow\infty}\frac{\operatorname{Vol}(\Phi(C_n))}{\operatorname{Vol}(C_n)}=0$$

I know that $\Phi$ cant be a diffeomorphism in $x_0$, but a have know idea how to use this, or how to do anything different. Thanks for helping.

share|cite|improve this question
Use the fact that $D\Phi(x_0)$ approximates $\Phi$ near $x_0$ and that $\text{Vol}(D\Phi(x_0)(C_n))=\det(D\Phi(x_0))C_n=0$ – Alex Youcis Sep 15 '12 at 3:46
Do you mean $x_0\in U$, that $\Phi$ is differentiable in $U$ or some neighborhood of $x_0$, and $x_0\in C_n$? – Alex Becker Sep 15 '12 at 7:31
I think i made a mistake, i mean the cubes are open(or closed) in $U$ and $x_0$ is inside the cubes. – Integral Sep 15 '12 at 20:20
up vote 1 down vote accepted

Assume $x_0=\Phi(x_0)=0$, and put $d\Phi(0)=:A$. By assumption the matrix $A$ (or $A'$) has rank $\leq d-1$; therefore we can choose an orthonormal basis of ${\mathbb R}^d$ such that the first row of $A$ is $=0$.

With respect to this basis $\Phi$ assumes the form $$\Phi:\quad x=(x_1,\ldots, x_d)\mapsto(y_1,\ldots, y_d)\ ,$$ and we know that $$y_i(x)=a_i\cdot x+ o\bigl(|x|\bigr)\qquad(x\to 0)\ .$$ Here the $a_i$ are the row vectors of $A$, whence $a_1=0$.

Let an $\epsilon>0$ be given. Then there is a $\delta>0$ with $$\bigl|y_1(x)\bigr|\leq \epsilon|x|\qquad\bigl(|x|\leq\delta\bigr)\ .$$ Furthermore there is a constant $C$ (not depending on $\epsilon$) such that $$\bigl|y(x)\bigr|\leq C|x|\qquad\bigl(|x|\leq\delta\bigr)\ .$$ Consider now a cube $Q$ of side length $r>0$ containing the origin. Its volume is $r^d$. When $r\sqrt{d}\leq\delta$ all points $x\in Q$ satisfy $|x|\leq\delta$. Therefore the image body $Q':=\Phi(Q)$ is contained in a box with center $0$, having side length $2\epsilon r\sqrt{d}$ in $y_1$-direction and side length $2C\sqrt{d}\>r$ in the $d-1$ other directions. It follows that $${{\rm vol}_d(Q')\over{\rm vol}_d(Q)}\leq 2^d\ d^{d/2}\> C^{d-1}\ \epsilon\ .$$ From this the claim easily follows by some juggling of $\epsilon$'s.

share|cite|improve this answer
The problem is: $\Phi$ is not necessarily $C^1$. – Integral Oct 22 '14 at 18:59
@Integral: See my edit. It's even simpler now. – Christian Blatter Oct 22 '14 at 19:20

You should try using the change of variables formula for integration on Euclidean space: $$ \mathrm{Vol}(\Phi(C_n)) = \int_{\Phi(C_n)} dx = \int_{C_n} |\det(D\Phi)(x)| dx $$ However, it would seem that you should need some additional condition, in particular that $\Phi$ is continuously differentiable about $x_0$.

share|cite|improve this answer

Let $r_n$ be the sidelength of $C_n$. The definition of derivative gives you an upper bound on $|\Phi(x)-\Phi(x_0)|$ which says that $\Phi(C_n)$ is contained in a ball of radius about $r_n$ (same order of magnitude). Of course, this is not enough to show that $\mathrm{vol}(\Phi(C_n))/\mathrm{vol}(C_n)$ is small. The key additional fact is that the range of $D\Phi(x_0)$ is a proper subspace of $\mathbb R^d$. Let $V$ be this subspace. Upon a closer inspection, the definition of derivative will tell you that for every $x\in C_n$ the distance of $\Phi(x)$ to $V$ is much less than $r_n$, i.e., its ratio to $r_n$ tends to zero. This means that $\Phi(C_n)$ is contained in some cylinder-like shape, the volume of which you can estimate geometrically, "base times height".

share|cite|improve this answer

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.