Let $O(d)$ be the space of orthogonal $d \times d$ matrices over $\mathbb{R}$. These matrices act on $C_c^2(\mathbb{R}^d)$, the set of twice continuously differentiable functions $f : \mathbb{R}^d \to \mathbb{R}$ with compact support, via $(f \circ A)(x) := f(Ax)$, where $A \in O(d)$. I am trying to show that for all $f \in C_c^2(\mathbb{R}^d) $ and for all $A \in O(d)$, we have $\Delta(f \circ A) = (\Delta f)\circ A$. Here, $\Delta$ denotes the Laplace operator.
I am aware of the hint in the answer given to this related question here, however I am unable to complete the proof. Further hints on how exactly to apply the chain rule, as well as a full proof would be very much appreciated. Thanks!
Edit: Until now, I have the following: Let $A := (a_{ij})$, $x := (x_1,...,x_d)$. Since $AA^\top = I$, we have $$ \sum_{j=1}^d a_{ij}a_{kj} = \delta_{ik}, $$ where $\delta_{ik}$ is the Kronecker Delta. We thus have $$(f \circ A)(x) = f(Ax) = f \left( \sum_{i=1}^da_{1i}x_i,..., \sum_{i=1}^da_{di}x_i\right).$$ We denote $z_i = g_i(x_1,...,x_d) = \sum_{k=1}^da_{ik}x_k$. With the chain rule $$ \frac{\partial f}{\partial x_j} = \sum_{i=1}^d \frac{\partial f}{\partial z_i} \frac{\partial z_i}{\partial x_j},$$ we get that \begin{align} \frac{\partial}{\partial x_j}(f \circ A)(x) &= \frac{\partial}{\partial x_j}f(z_1,...,z_d)\\ &= \sum_{i=1}^d \frac{\partial f(z_1,...,z_d)}{\partial z_i} \frac{\partial g_i(x_1,...,x_d)}{\partial x_j}\\ &= \sum_{i=1}^d a_{ij} \frac{\partial f(z_1,...,z_d)}{\partial z_i}. \end{align} In the next step we get \begin{align} \frac{\partial^2}{\partial x_j^2}(f \circ A)(x) = \frac{\partial}{\partial x_j} \left( \sum_{i=1}^d a_{ij} \frac{\partial f(z_1,...,z_d)}{\partial z_i} \right) = \sum_{i=1}^d a_{ij} \frac{\partial^2 f(z_1,...,z_d)}{\partial x_j \partial z_i}, \end{align} but I'm not sure if this last step is valid. Now, if we compute $(\Delta f) \circ A$, we get \begin{align} ((\Delta f)\circ A)(x) &= (\Delta f)(Ax)\\ &= \left( \sum_{j=1}^d \frac{\partial^2f}{\partial x_j^2} \right) (z_1,....z_d)\\ &= \sum_{j=1}^d \frac{\partial^2f(z_1,...,z_d)}{\partial x_j^2}, \end{align} and since we already computed the first and second partial derivative of $f$, we get that $\Delta(f \circ A) = (\Delta f) \circ A$.
I have two problems:
- I haven't used the fact that $A$ is orthogonal.
- I'm unsure whether or not I'm using the definition of the chain rule correctly.
Any help is appreciated!
I think part of your confusion stems from the use of $\partial /\partial_{z_j}$. It might be easier to see it, when you use the partial derivative $\partial_j$ (meaning you take the partial derivative with respect to the $j$th slot of the function). You have correctly computed
\begin{align} \frac{d}{dx_j} (f\circ A)(x) &= \frac{d}{dx_j} f( \sum_{i=1}^d a_{1i} x_i, \dots, \sum_{i=1}^d a_{di} x_i ) \\ &= \sum_{k=1}^d (\partial_k f)( \sum_{i=1}^d a_{1i} x_i, \dots, \sum_{i=1}^d a_{di} x_i ) \cdot \frac{d}{dx_j} \left( \sum_{\ell=1}^d a_{k\ell} x_\ell\right) \\ &= \sum_{k=1}^d a_{kj} \cdot (\partial_k f)( \sum_{i=1}^d a_{1i} x_i, \dots, \sum_{i=1}^d a_{di} x_i ) \end{align}
Repeating the same argument with $\partial_k f$ instead of $f$ leads to
\begin{align} \frac{d^2}{dx_j^2} (f\circ A)(x) &= \sum_{k=1}^d a_{kj} \cdot \frac{d}{dx_j} (\partial_k f)( \sum_{i=1}^d a_{1i} x_i, \dots, \sum_{i=1}^d a_{di} x_i ) \\ &= \sum_{k=1}^d a_{kj} \sum_{\ell=1}^d a_{\ell j} (\partial_\ell \partial_k f)( \sum_{i=1}^d a_{1i} x_i, \dots, \sum_{i=1}^d a_{di} x_i ) \end{align}
Thus, using your equation for the coefficients we get
\begin{align} \Delta (f\circ A)(x) &= \sum_{j=1}^d \frac{d^2}{dx_j^2} (f\circ A)(x) \\ &= \sum_{j=1}^d \sum_{k=1}^d a_{kj} \sum_{\ell=1}^d a_{\ell j} (\partial_\ell \partial_k f)( \sum_{i=1}^d a_{1i} x_i, \dots, \sum_{i=1}^d a_{di} x_i ) \\ &= \sum_{k,\ell=1}^d \left( \sum_{j=1}^d a_{kj} a_{\ell j} \right) (\partial_\ell \partial_k f)( \sum_{i=1}^d a_{1i} x_i, \dots, \sum_{i=1}^d a_{di} x_i ) \\ &= \sum_{k,\ell=1}^d \delta_{k,\ell} (\partial_\ell \partial_k f)( \sum_{i=1}^d a_{1i} x_i, \dots, \sum_{i=1}^d a_{di} x_i ) \\ &= \sum_{k=1}^d (\partial_k^2 f)( \sum_{i=1}^d a_{1i} x_i, \dots, \sum_{i=1}^d a_{di} x_i ) \\ &= (\Delta f) ( \sum_{i=1}^d a_{1i} x_i, \dots, \sum_{i=1}^d a_{di} x_i ) \\ &= (\Delta f)(Ax) \\ &= ((\Delta f)\circ A)(x) \end{align}