Some Functional Analysis Questions (Laplace Operator And Fourier Transform) - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T13:31:46Z http://mathoverflow.net/feeds/question/100956 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/100956/some-functional-analysis-questions-laplace-operator-and-fourier-transform Some Functional Analysis Questions (Laplace Operator And Fourier Transform) jason mfash 2012-06-29T16:39:00Z 2012-06-29T16:58:24Z <p>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: $\hat \Phi (z,y)= (2 \pi)^{-n/2} \int_{x \in \mathbb{R} ^ n } \Phi(x,y)e^{ix \cdot z} dx$.</p> <p>Can someone explain me the second equality in the following: $z_j \hat{\Phi} (z,y) = (2 \pi)^{-n/2} \int_{\mathbb{R}^n } \Phi(x,y)z_j e^{ixz} dx = (2 \pi)^{-n/2} \int_{\mathbb{R}^n } \Phi(x,y)(-i) \frac{\partial}{\partial x_j } e^{ix \cdot z } dx$ </p> <p>BTW- What does the notation $z_j$ means in this context?</p> <p>Hope someone will be able to help me</p> <p>Thanks in advance </p> http://mathoverflow.net/questions/100956/some-functional-analysis-questions-laplace-operator-and-fourier-transform/100959#100959 Answer by Yul Otani for Some Functional Analysis Questions (Laplace Operator And Fourier Transform) Yul Otani 2012-06-29T16:58:24Z 2012-06-29T16:58:24Z <p>I cannot comment, so i'll just answer it. I assume you are working on $L^2(R^{2n})$ with Lebesgue measure, right?</p> <p>$\hat{z}_j$ 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 <code>$\big( \, \hat{z}_jf \,\big)(z) = z_j f(z)$</code>. Working with $\Phi$, you can then "put it inside" the integral and derive the equality you mentioned.</p> <p>We didn't get to use anything of the mentioned eigenvalues of this mysterious unnamed operator, though ;)</p>