On distribution functions of the derivatives of weakly differentiable mappings (Q1904206)

Given any measurable mapping \(w:\Omega\to {\mathbf R}^m\), with \(\Omega\subset{\mathbf R}^n\), the distribution \(\mu^w\) associated to \(w\) is defined as the Borel measure \(\mu^w(B)= {\mathcal L}^n(w^{-1}(B))\) for every \(B\in{\mathcal B}({\mathbf R}^m)\), where \({\mathcal B}({\mathbf R}^m)\) denotes the \(\sigma\)-algebra of all Borel subsets of \({\mathbf R}^m)\) and \({\mathcal L}^n\) is the \(n\)-dimensional Lebesgue measure. In the paper it is shown that, given a function \(u\) in the Sobolev space \(W^{2,1}(0,1)\) the distribution \(\mu^{(u,u')}\) (called by the author joint distribution) determines the distribution \(\mu^{u''}\). In particular, if \(u,v\in W^{2,1}(0,1)\) are such that \(\mu^{(u,u')}=\mu^{(v,v')}\) then necessarily we have \(\mu^{u''}=\mu^{v''}\). Moreover, \(\mu^{(u,u')}\) also determines \(\mu^{(u,u',u'')}\) and hence \(\mu^{(u',u'')}\). Therefore, if \(u\in W^{3,1}(0,1)\), the distribution \(\mu^{(u,u')}\) determines \(\mu^{u'''}\) and so on. As the author points out in a final remark, it would be interesting to investigate on the multidimensional analogue of the result above, whether or not the distribution \(\mu^{(u,Du)}\) determines \(\mu^{D^2u}\), \(Du\) being the gradient of \(u\) and \(D^2u\) the Hessian matrix of all second derivatives of \(u\).