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Can the derivative of a BV function $f:\mathbb{R}^n\to\mathbb{R}^n$ be controlled by the symmetric part of the derivative $\frac{1}{2}(Df+(Df)^T)$?

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This is not a full answer, but some comments that might put you on the right track.

If a function $f=(f_1,\ldots,f_n):\mathbb{R}^n\to\mathbb{R}^n$ is sufficiently smooth and has compact support, then $$ f_k=\frac{2}{n\omega_n}\sum_{1\leq i\leq j\leq n}\left(\epsilon_{jk}*\frac{\partial K_{ij}}{\partial x_i}-\epsilon_{ij}*\frac{\partial K_{ij}}{\partial x_k}+\epsilon_{ki}*\frac{\partial K_{ij}}{\partial x_j}\right), $$ where $$ \epsilon_{ij}=\frac{1}{2}\left(\frac{\partial f_i}{\partial x_j}+\frac{\partial f_j}{\partial x_i}\right) \quad \text{are components of} \quad \epsilon(f)=\frac{1}{2}\left(Df+(Df)^T\right) $$ and $K_{ij}(x)=x_ix_j/|x|^n$ see page 63 in [1]. Taking the derivative of the above formula we obtain the representation of the derivative of $f_k$ as a singular integral applied to $\epsilon(f)$ (the derivative of $\partial K_{ij}/\partial x_i$ has nonintegrable singularity and the convolution needs to be interpreted as a singular integral, see [1])). Thus for sufficiently smooth functions we can represent $Df$ as a singular integral of $\epsilon(f)$. In this way one can prove the so called Korn inequality: $\Vert Df\Vert_p\leq C\Vert \epsilon(f)\Vert_p$ for $1<p<\infty$.

The problem is that in the case of $BV$ functions the derivative is a Radon measure and singular integrals do not act well on measures. However, from the fact that $\epsilon(f)$ is a measure (functions such that $\epsilon(f)$ is a Radon measure are called functions of bounded deformation) one can conclude that $f$ is approximately differentiable a.e. so some estimates are possible (see Corollary 1 page 66 in [1]).

See also [2] for a more recent development and related references.

[1] P. Hajłasz, On approximate differentiability of functions with bounded deformation.. Manuscripta Math. 91 (1996), 61–72.

[2] J. Verdera, Capacitary differentiability of potentials of finite Radon measures.. arXiv:1812.11419.

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  • $\begingroup$ Thank you. This is very helpful. $\endgroup$ – Riku Apr 10 '19 at 17:14

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