$$\Delta_{\mathbf{x}} :=\sum_{i=1}^n\frac{\partial^2}{\partial x_i^2}, \qquad \mathbf{x}=(x_1, \dots, x_n)\in\mathbb{R}^n$$ Suppose I want to make a coordinate shift, namely $\mathbf{x}\longmapsto\mathbf{y}:=\mathbf{x}-\mathbf{x}_0$. The claim is $$\Delta_\mathbf{x}= \Delta_{\mathbf y}$$ Any hint on how to proceed?
Is it enough to say that every linear operator is invariant under shift?
$\vec{x}= \vec{y}+\vec{x_o}$
Where $\vec{x_o}$ is a constant vector, hence applying the chain rule we get:
$ \dfrac{\partial}{\partial y_i}$ =$\sum_{j} \dfrac{\partial x_j}{\partial y_i}\dfrac{\partial }{\partial x_j} = \sum_{j}\delta_{ij}\dfrac{\partial }{\partial x_j}= \dfrac{\partial }{\partial x_i}$
So actually any differential operator is invariant under shift, and hence in particular the Laplacian.