# Control the derivative of a BV function by its symmetric part

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

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.

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]).