0
$\begingroup$

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

$\endgroup$
  • $\begingroup$ Hint: integration with Polar coordinates $\endgroup$ – Milen Ivanov Jan 18 '16 at 23:04
  • $\begingroup$ @ 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 $\Delta$. $\endgroup$ – Z. Alfata Jan 19 '16 at 10:40
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ @ M. Ivanov, thank you very much for your help $\endgroup$ – Z. Alfata Jan 20 '16 at 9:50
  • $\begingroup$ You're welcome :) $\endgroup$ – Milen Ivanov Jan 20 '16 at 19:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.