# How to prove that $f(x) = \int_{0}^{\infty} f_{\lambda} (x) \, d\lambda$ where $f_{\lambda}$ is the eigenfunctions of $\Delta$

On Euclidean space $\mathbb R^n$, how to prove that $$f(x) = \int_{0}^{\infty} f_{\lambda} (x) \, d\lambda ,$$

where $\Delta f_{\lambda} (x) = -\lambda^2 f_{\lambda}(x)$, whith $\Delta$ is the Laplacian on $\mathbb R^n$, and $f_{\lambda}$ given by $$f_{\lambda} (x) = (2\pi)^{-n} \lambda^{n-1} \int_{\mathbb S^{n-1}} \widehat{f}(\lambda u) \, e^{i \, \lambda \, x .u} \, du.$$ Thanks you in advance

• Hint: integration with Polar coordinates – Milen Ivanov Jan 18 '16 at 23:04
• @ M. Ivanov, yes, by writing the Fourier inversion formula on Euclidean space in polar coordinates, we obtain $f(x) = \int_{0}^{\infty} f_{\lambda} (x) \, d\lambda ,$ but, why $f_{\lambda}$ is a eigenfunction of $\Delta$. – Z. Alfata Jan 19 '16 at 10:40

As discussed in the comments, we are only left to show that $\Delta f_{\lambda}(x) = - \lambda^2 f_{\lambda}(x)$ The following proof will work if $f \in C^2(\mathbb{R}^n)$ and it, together with all of its partial derivatives up to second order are in $L^1$, so that differentiation under the sign is justified. Also, we need $\widehat{f} \in L^1$, so that all the operations are justified.
Under these coditions we can interchange the Laplace operator and the integral $$\Delta f _\lambda (x) = (2 \pi)^{-n} \lambda^{n-1}\int_{\mathbb{S}^{n-1}} \Delta (\widehat{f}(\lambda u) e^{\lambda x\cdot u}) d \sigma(u)$$ Because the differentiation is with respect to $x_i, 1\leq i \leq n$ we get
$$\Delta f _\lambda (x) = (2 \pi)^{-n} \lambda^{n-1}\int_{\mathbb{S}^{n-1}} \widehat{f}(\lambda u) e^{i\lambda x\cdot u} (i \lambda)^2 (u_1^2 + \cdots + u_n ^2) d \sigma(u)$$ We are on the sphere so $u_1^2 + \cdots + u_n = 1$ and the result follows. The rest is just an integration in Polar coordinates and the Fourier inversion formula.