Given a set of the k first eigenvalues $ (\lambda\_i)\_ 1 ^k $ of some operator , and a set of the first k orthomormal eigenfunctions for these eigenvalues : $ ( \phi\_i ) $ .
Define: $ \Phi(x,y) = \sum\_{i=1}^k \phi\_i(x) \phi\_i(y) $ and then define the fourier transform of this function:
$ \hat \Phi (z,y)= (2 \pi)^{-n/2} \int\_{x \in \mathbb{R} ^ n } \Phi(x,y)e^{ix \cdot z} dx$.


Can someone explain me the second equality in the following:
$ z\_j \hat{\Phi} (z,y) = (2 \pi)^{-n/2} \int\_{\mathbb{R}^n } \Phi(x,y)z\_j e^{ixz} dx = 
(2 \pi)^{-n/2} \int\_{\mathbb{R}^n } \Phi(x,y)(-i) \frac{\partial}{\partial x\_j } e^{ix \cdot z } dx $ 


BTW- What does the notation $ z\_j $ means in this context?


Hope someone will be able to help me


Thanks in advance 



Some Functional Analysis Questions (Laplace Operator And Fourier Transform) 
I cannot comment, so i'll just answer it. I assume you are working on $L^2(R^{2n})$ with Lebesgue measure, right?


$\hat{z}\_j$ is the operator that multiplies with the $z\_j$-coordinate. Any $z \in \mathbf{R}^n$ is written as $(z\_1,\ldots,z\_j,\ldots,z\_n)$, and then we get something like $\big( \, \hat{z}\_jf \,\big)(z) = z\_j f(z)$. Working with $\Phi$, you can then "put it inside" the integral and derive the equality you mentioned.


We didn't get to use anything of the mentioned eigenvalues of this mysterious unnamed operator, though ;)


