Suppose that $T:\mathbb{R}^k\rightarrow\mathbb{R}^k$ is a transformation which is differentiable at a point $x\in\mathbb{R}^k$. Let $\lambda$ be the Lebesgue measure on $\mathbb{R}^k$. Denote by $B(x,r)$ the open ball centred at $x$ with radius $r$. Now let $\mu$ be another measure on $\mathbb{R}^k$ such that $$\lim\limits_{r\rightarrow 0}\frac{\mu(B(Tx,r))}{\lambda(B(Tx,r))}=a<\infty$$ Define $$Q(r)=\begin{cases}\frac{\mu(T(B(x,r)))}{\lambda(T(B(x,r)))}&\text{if }\lambda(T(B(x,r)))\neq0\\ a &\text{otherwise}\end{cases}$$ Can we say anything about $\lim\limits_{r\rightarrow0}Q(r)$? What if we just assume that for $r>0$, $\lambda(T(B(x,r)))\neq0$?

  • $\begingroup$ Are you supposing that $T$ is continuous in some neighbourhood of $x$? This would be a very useful hypothesis. $\endgroup$ – Crostul Jul 7 '15 at 9:19
  • $\begingroup$ I was trying not to assume that. I can prove it if we assume that T is one-to-one on some neighbourhood of $x$ but i can't see how continuity would help. $\endgroup$ – user61496 Jul 7 '15 at 9:42

Well, "can we say anything" is fuzzy enough that one can't give a definitive no. But it seems to me that we can't say anything that's anything like the sort of thing you want to say, unless you strengthen the hypotheses.

EDIT: No, two counterexamples, illustrating it seems to me different sorts of things that can go wrong.

EDIT: No, three examples. The first two seem to me to indicate you need to assume some smoothness for $T$. The third shows that even if you assume $T$ is infinitely differentiable on $\mathbb R^n$ you may also want to assume that the derivative of $T$ at $a$ is non-singular (I suspect this is the case you have in mind).

EDIT: Well, for 100 points the OP should get his or her money's worth. An example where $T$ is linear (and non-singular), showing that we also need to strengthen the hypotheses on $\mu$; the other examples were all about bad $T$.

First example:

Take $\mu=\lambda$, so $a=1$. There exists $T$ satisfying all your hypotheses such that $\lambda(T(B(0,r)))=0$ for every $r>0$ (making the second limit $0$, but only because of the escape clause you inserted in the definition; the second limit does not exist in any interesting sense).

Write $\mathbb R\setminus\{0\}$ as the disjoint union of measurable sets $E_n$ which tend to $0$ in the sense that $\sup_{x\in E_k}|x|\to0$. And also so that $dist(0,E_k)>0$, and in fact $$\frac{diam(E_k)}{dist(0,E_k)}\to0.$$Choose $x_k\in E_k$, and let $$T=\sum_kx_k\chi_{E_k}.$$Then $T(0)=0$, $T$ is differentiable at the origin, in fact $|x-T(x)|=o(x)$, but $\lambda(T(\mathbb R^n))=0$.

That's pretty cheap. Could be modified to give more interesting badness I suspect, but it seems to me this version is enough to show that you really need more hypotheses to get an interesting positive result.

Second example: Again we're going to have $a=1$. This time we're going to get $\lambda(T(B(0,r)))\sim r$ but $\mu(T(B(0,r)))=0$. Take $n=2$, and regard $\mathbb R^2$ as the same as $\mathbb C$ just for the sake of notation.

Let $$S=\{re^{it}\,:\,0\le r\le 1,r^2\le t\le 2\pi-r^2\}.$$There exists a map $T:\mathbb C\to\mathbb C$ such that $T(0)=0$, $T(\mathbb C)\subset S$, and $|z-T(z)|=o(z)$, so $T$ is differentiable at the origin. For example map $re^{it}$ to $re^{i\phi_f(t)}$ for suitable $\phi_r$. (Or rather do that for $0\le r\le 1$ and do whatever you want for $r>1$.) If you define $T$ that way then $$\frac{\lambda(T(B(0,r)))}{\lambda(B(0,r))}\to1.$$

Now $\mu$ is going to be the measure supported on the positive real axis $[0,\infty)$, such that $$\mu((\alpha,\beta))=\pi(\beta^2-\alpha^2).$$Then $\mu(B(0,r))=\lambda(B(0,r))$ for all $r$, while $\mu(T(B(0,r)))=0$.

Third example. $n=1$, $a=0$, $T(x)=x^2$, $$\mu(E)=\lambda(E\cap(0,\infty)).$$Now the first limit is $1/2$ while the second is $1$.

Example the fourth: As in the second example consider $\mathbb R^2=\mathbb C$, and let $\mu$ be supported on the positive real axis with $\mu((\alpha,\beta))=\pi(\beta^2-\alpha^2)$. Let $T(x+iy)=2x+iy$. First limit is $1$, second is $2$.

  • $\begingroup$ Are there any counterexamples in which $T$ is differentiable on a neighbourhood of $x$ but $T'(x)\neq 0$? $\endgroup$ – user61496 Jul 14 '15 at 16:28
  • $\begingroup$ For $n>1$ seems like the relevant condition is more "$DT(0)$ is non-singular". But anyway, seems to me yes, at least for $n>1$. Realized this the other day, decided not to add a fourth example. Haven't worked out the details, but I bet this is right: Take $n=2$. Define $T(x,y)=(2x,y)$. Now your hypothesis has to do with $\mu$ of disks while the conclusion talks about $\mu$ of ellipses, so we just need to rig up a $\mu$ that can see the difference, shouldn't be hard. This one shows you also need stronger hypotheses on $\mu$ (unless you want to assume that $DT(0)$ is the identity or something) $\endgroup$ – David C. Ullrich Jul 14 '15 at 16:39
  • $\begingroup$ Otoh since you mention "$T'$", maybe you were asking about $n=1$. Seems clear to me that if $n=1$ and $T'(0)\ne0$ then the result is finally true; fairly easy in fact. $\endgroup$ – David C. Ullrich Jul 14 '15 at 16:41
  • $\begingroup$ @user61496 See edit to answer... $\endgroup$ – David C. Ullrich Jul 14 '15 at 18:04
  • $\begingroup$ @user929304 Erm, thanks. Did you notice who posted that post? $\endgroup$ – David C. Ullrich Jun 13 '16 at 14:20

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