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4 votes
1 answer
597 views

Meaning of Alberti rank-one theorem

Heuristically what does Alberti's rank-one theorem imply about the structure of a $\mathrm{BV}$ vector field $\boldsymbol{b}$? Is it rigorously fair to say that the level lines of $\boldsymbol{b}$ ...
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4 votes
0 answers
151 views

Estimating the size of $\Omega_r=\{x\in\Omega: \text{dist}(x,\partial\Omega)<r \}$

Let $\Omega$ be a bounded domain in $\Bbb R^n$. Define $$ \Omega_r=\{x\in\Omega: \text{dist}(x,\partial\Omega)<r \}, $$ i.e. it the ring of thickness $r$ at the boundary of $\Omega$. Intuitively, ...
BigbearZzz's user avatar
  • 1,245
4 votes
1 answer
266 views

Prove $\int_{\mathbb R^N \setminus \Omega} |x - z|^{-N-s} dz \approx dist(x,\partial \Omega)^{-s}$, $s \in (0,2)$

Let $\Omega \subset \mathbb R^N$ and $s \in (0,2)$. Under what assumptions on $\partial \Omega$ do we have $$\int_{\mathbb R^N \setminus \Omega} |x - z|^{-N-s} dz \approx \mathrm{dist}(x,\partial \...
Jun's user avatar
  • 303
3 votes
1 answer
274 views

Function square-integrable

Let $f$ be an arbitrary function in $L^2(0,\infty)$ and consider the function $$(g_f)(y) = \frac{1}{y-x_0} \int_{0}^{\infty} f(x) \left(\frac{xy}{(x^2+y^2+1)}\right)^2 \ dx$$ where $x_0$ is an ...
Andrea Tauber's user avatar
1 vote
0 answers
92 views

Alberti rank-one theorem and reduction of the study of BV function to the two-dimensional case

By Alberti rank-one theorem, could it be possible to reduce the study of a function $u \in BV(\mathbb{R}^N, \mathbb{R}^N)$ to the study of a function $\tilde{u} \in BV(\mathbb{R}^2, \mathbb{R}^2)$? At ...
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