3

Let $A\subset \mathbb{R}^n$ a Borel set of finite Lebesgue measure. They define $A^*$ to be the ball centered at 0 with the same measure that $A$.

The symmetric-decreasing rearrangement of a measurable function $f:\mathbb{R}^n \to \mathbb{R}$ is then defined by

$$f^*(x):=\int_0^{\infty} \chi_{\{|f|>t\}^*}(x)dt,$$

by comparison to the "layercake" representation of $f$, namely $$f(x)=\int_0^{\infty} \chi_{\{f>t\}}(x)dt.$$

Note that one can equally define $f^*$ to be the radial symmetric, decreasing function such that the level sets of $f^*$ and $f$ have the same measure (or volume).

I don't know why, but it always occurs to me that the rearrangement process is a regularizing process.

Here's my question: If $f$ is $k$ times continuously differentiable, does it follow that $f^*$ possesses the same regularity?

Guy Fsone
  • 25,237
henryforever14
  • 2,319

1 Answers1

1

In general the answer is no, it does not. A counter-example is the following. Given parabola $f$ (which is differentiable infinitely many times) on the interval $(-1,1)$, it has a symmetric decreasing rearrangement $f^*$ which is not differentiable at $0$. So for functions from $C^k$ we can guarantee just $PC^k$ (partially continuous of degree $k$). But if given function $f$ is monotone then, I think, your suggestion would be true.

epimorphic
  • 3,270
KolyaByk
  • 11
  • I'm guessing that "partially continuous" means piecewise continuous? Also, the guess in your last sentence isn't correct. We can consider the parabola $x \mapsto x^2$, this time on $(0,1)$; its rearrangement has the same shape as the $f^*$ from your answer, with the cusp at $0$. – epimorphic Aug 15 '15 at 16:54
  • Yes, you are right, of course (at both). So, if we have monotone function then its derivatives must be equal to 0 at the point where f attains its maximum. – KolyaByk Aug 16 '15 at 11:43
  • It seems that the counter-example of the parabola on (-1,1) or on (0,1) works because in fact the function is taken zero outside the interval. In particular it is discontinuous. Is this correct? But what about the case of a mapping $f\in C^{k}_c(\mathbb{R}^n)$? – CSI Aug 18 '15 at 20:40