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July 15th, 2018, 06:40 PM  #1 
Newbie Joined: Jun 2018 From: Viet Nam Posts: 3 Thanks: 0  Prove the eigenvalues $\lambda$ of $\lambda \phi_j(x)= \int_G{ K(xy)\phi_j(y)dy}$ is
Prove the eigenvalues $\lambda$ of $\lambda \phi_j(x)= \int_G{ K(xy)\phi_j(y)dy}$ is $\int_G{K(x)\phi_{j}(x)dx}$, with $\phi_j(x)=(2R)^{n/2}exp(i\pi j. \frac{x}{R}), j \in \mathbb{Z}^n, x, y \in \mathbb{R}^n, G=\{x \in \mathbb{R}^n: x_i\leq R,i=1,...,n\} $ and $K(x)$ is 2Rperodic. When I try to devide the convolution by $\phi_j(x)$, I have $\lambda=\int_G{K(xy)\phi_j(y) / \phi_j(x)dy}=(2R)^{n/2}\int_G{K(xy)\phi_j(yx)dy}$. Let $t=xy$ assume that $t\in G$, so $dt=dy$ and $ \lambda=  (2R)^{n/2}\int_{G}{K(t)\phi_{j}(t)dt}$. What's wrong with me? 
July 16th, 2018, 12:52 PM  #2 
Global Moderator Joined: May 2007 Posts: 6,555 Thanks: 600  

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$lambda, $lambda$, eigenvalues, intg, kxyphijydy$, phijx, prove 
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