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July 15th, 2018, 07:40 PM   #1
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Prove the eigenvalues $\lambda$ of $\lambda \phi_j(x)= \int_G{ K(x-y)\phi_j(y)dy}$ is

Prove the eigenvalues $\lambda$ of $\lambda \phi_j(x)= \int_G{ K(x-y)\phi_j(y)dy}$ 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 2R-perodic.

When I try to devide the convolution by $\phi_j(x)$, I have
$\lambda=\int_G{K(x-y)\phi_j(y) / \phi_j(x)dy}=(2R)^{n/2}\int_G{K(x-y)\phi_j(y-x)dy}$.

Let $t=x-y$ 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?
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July 16th, 2018, 01:52 PM   #2
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You'll have better luck with mathematics stack exchange

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