Laplacian and Fourier transform 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}