1
$\begingroup$

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.

$\endgroup$

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.