Show the laplacian is rotationally invariant:
$\Delta(f\circ R)=(\Delta f)\circ R, \forall R\in SO_d(\mathbb{R})$.
Suggestion: You can use that the Fourier Transform (FT) of $f(Ax)$ is $det(A)^{-1}\hat{f}((A^{-1})^t\xi)$ (lineal composition) and that FT of $\Delta f(x)$ is $-4\pi^2|\xi|^2\hat{f}(\xi)$.
I'm completely lost. Pelase help
The Fourier transform of $f(Ax)$ is $$ \mathcal{F}\{f\circ A\}(\xi)= \frac{1}{(2\pi)^{n/2}}\int_{\mathbb{R}^n} f(Ax)e^{-ix\cdot \xi}dx. $$ Let $y=Ax$. Then $x=A^t y$ because $A^tA=AA^t=I$ by the definition of a symmetric orthogonal matrix. The Jacobian of this transformation is $|A^t|=1$, which gives \begin{align} \mathcal{F}\{f\circ A\}(\xi) & = \frac{1}{(2\pi)^{n/2}}\int_{\mathbb{R}^n}f(y)e^{-i(A^t y)\cdot \xi}dy \\ & = \frac{1}{(2\pi)^{n/2}}\int_{\mathbb{R}^n}f(y)e^{-iy\cdot(A\xi)}dy \\ & = (\mathcal{F}\{ f\}\circ A)(\xi)\end{align} In other words, $\mathcal{F}\{f\circ A\}=\mathcal{F}\{f\}\circ A$. Therefore, \begin{align} -\Delta (f\circ A) &= \mathcal{F}^{-1}\{|\xi|^2\mathcal{F}\{f\circ A\}\} \\ &= \mathcal{F}^{-1}\{ |A\xi|^2\mathcal{F}\{f\}\circ A\} \\ &= \mathcal{F}^{-1}\{ |\xi|^2\mathcal{F}\{f\}\}\circ A \\ &= (-\Delta f)\circ A. \end{align}