# Some Functional Analysis Questions (Laplace Operator And Fourier Transform)

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

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

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.
• 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 $\mathbf{R}^n$. Jun 29, 2012 at 17:36