Fourier transformation diagonalize the Laplacian operator

Question

How the following point is illustrated?

The Fourier transformation diagonalizes the Laplacian operator.

Answer 1

Denote the dot product of vectors $x,y,s$ by $x\cdot s,\;y\cdot s$. Then $f\in L^2(\mathbb{R}^3)$ can be written in terms the Fourier transform and its inverse: $$ f = \frac{1}{(2\pi)^{3}}\int_{\mathbb{R}^3} \left(\int_{\mathbb{R}^3} f(y)e^{-is\cdot y}dy\right)e^{is\cdot x}ds $$ This is a "continuous" eigenfunction expansion in terms generalized eigenfunctions of the Laplace operator $\Delta$: $$ -\Delta_{x} e^{is\cdot x}=|s|^2 e^{is\cdot x}. $$ These "eigenfunctions" are not really eigenfunctions because they are not square-integrable. However, integral sums of such eigenfunctions are square-integrable.