Fourier transform of distributions with non-standard test functions - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T17:04:45Z http://mathoverflow.net/feeds/question/66214 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/66214/fourier-transform-of-distributions-with-non-standard-test-functions Fourier transform of distributions with non-standard test functions Pierre 2011-05-27T19:59:18Z 2011-07-17T11:22:12Z <p>This might be a quite simple question for function analysis standards, but it has some obstacles. I'll try to improve the readability a bit by not using the full tex code. A short motivation:</p> <p>Given a Schwartz function $f \in S$ that is the density of some prob. measure, one can perfectly write for the Heaviside function $\theta$.</p> <p>$\int \theta \operatorname{d\mathbb{P}}(x) = &lt;\theta,f> = &lt;\theta, F^{-1}Ff> =$</p> <p>where $F$ and $F^{-1}$ denote the (inverse) Fourier transform.</p> <p>Now $F^{-1}\theta$ is well known: </p> <p>$\frac{1}{2\pi i}[P.V.(\frac{1}{x}) + i\pi \delta(x)]$</p> <p>But: Smoothness is a way too restrictive property. The distribution $F^{-1}\theta$ is a well defined functional for test functions that have a bounded derivative in a neighbourhood $E$ around 0, and where $\sup_{x \in R} |x^\alpha (Ff)(x)|$ is finite (hence it is in $L^p$, $p&lt;\infty$). $\alpha$ and $E$ are fixed! One can define a norm on that space (let's call it $C_\alpha$) using infty norm and infty norm of the derivative in E. This is also the standard standard proof for $F^{-1}\theta$ being in $S'$, the dual of the Schwartz functions.</p> <p>I want to weaken the restrictions of test functions $f$, but I need to justify</p> <p>$&lt;\theta,f> = &lt; F^{-1} \theta, Ff>$</p> <p>it holds if f is a:</p> <ul> <li>Schwartz function</li> <li>$L^1$ function, $Ff$ also $L^1$ and in $C_\alpha$</li> </ul> <p>However it is not clear (to me), if it holds if the space of test functions is not closed under Fourier transform. I'd like it to hold for as much prob. densities as possible for which $Ff \in C_\alpha$.</p> <p>Of course Schwartz functions are dense in $L^1$, so I thought about using density arguments. $L^1$-convergence of the sequence (say $\psi_n$) of Schwartz functions implies only pointwise convergence under fourier transform. Too weak. I need some form of uniform convergence also under Fourier transform. I could achieve it if I assume that $|f(x)-\psi_n(x)| &lt; b_n(x)$ outside a compact set where b_n is some sequence of $L^1$ functions controlling the "tail behaviour" of the pointwise convergence. $b_n$ goes to zero pointwise for n to infty and $b_n &lt; B$; $B \in L^1$. However, I think this is again a too strong assumption - it rules out many densities (the fat tailed I am interested in). </p> <p>I am not sure how to "save" the equation $&lt;\theta,f> = &lt; F^{-1} \theta, Ff>$.</p> http://mathoverflow.net/questions/66214/fourier-transform-of-distributions-with-non-standard-test-functions/68204#68204 Answer by Pierre for Fourier transform of distributions with non-standard test functions Pierre 2011-06-19T08:46:08Z 2011-06-19T08:46:08Z <p>OK, the proof doesn't seem so hard using density arguments and mollifiers! Needs a last check, but I think I got it.</p>