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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.

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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.

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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 $x_0$.

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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".

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