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

  • 1
    $\begingroup$ $\dfrac{\partial}{\partial x_j} e^{ix \cdot z} = i z_j e^{i x \cdot z}$ $\endgroup$ Jun 29, 2012 at 17:22

1 Answer 1


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 ;)

  • $\begingroup$ Thanks for your quick reply ! ! The problem is that I can't figure out how to "derive" the equality I mentioned... What are they doing in order to get it? Thanks again !!! $\endgroup$ Jun 29, 2012 at 17:06
  • $\begingroup$ I'm not sure of what you mean... "derive" in the sense of derivative, in the sense motivating the appearance of the formula, or in the sense of getting from the RHS to the LHS? If so, on which step? Can you understand it by going backwards? For instance, $\frac{\partial}{\partial x_j} e^{i x z} = \frac{\partial}{\partial x_j} e^{i x_1 z_1}e^{i x_2 z_2}\ldots e^{i x_n z_n} = iz_j e^{i x z}$. Just a matter of understanding that $x z$ means the usual inner product in $\mathbf{R}^n$. $\endgroup$
    – Yul Otani
    Jun 29, 2012 at 17:36
  • $\begingroup$ Great ! That's excatly what I was missing. Thanks a lot ! $\endgroup$ Jun 29, 2012 at 18:10

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .