Consider the differential operator $D:=\sum_{i=1}^nx_i\frac{\partial}{\partial x_i}$. Then I claim that $D(u\circ A)(x)=D(u)(Ax)$. The proof is as follows: First note that $Du(x)=\langle x, \nabla_x u(x)\rangle$ with the gradient $\nabla_x$. By the chain rule we have that $\nabla_x (u\circ A)(x)=A^T\nabla_yu(y)|_{y=Ax}$. Thus $$ \langle x,\nabla_x(u\circ A)(x)\rangle=\langle x,A^T\nabla_yu(y)|_{y=Ax}\rangle=\langle Ax,\nabla_yu(y)|_{y=Ax}\rangle=\langle y,\nabla_yu(y)\rangle|_{y=Ax} $$ proving that $D(u\circ A)(x)=D(u)(Ax)$. Is there a mistake in the proof? I would rather expect only invariance under orthogonal transformations. Thanks in advance.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.