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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:

Given a Schwartz function $f \in S$ that is the density of some prob. measure, one can perfectly write for the Heaviside function $\theta$.

$\int \theta \operatorname{d\mathbb{P}}(x) = <\theta,f> = <\theta, F^{-1}Ff> = <F^{-1} \theta, Ff>$

where $F$ and $F^{-1}$ denote the (inverse) Fourier transform.

Now $F^{-1}\theta$ is well known:

$\frac{1}{2\pi i}[P.V.(\frac{1}{x}) + i\pi \delta(x)]$

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 $L^p$, $p<\infty$). $\alpha$ 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.

I want to weaken the restrictions of test functions $f$, but I need to justify

$<\theta,f> = < F^{-1} \theta, Ff>$

it holds if f is a:

  • Schwartz function
  • $L^1$ function, $Ff$ also $L^1$ and in $C_\alpha$

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 $Ff \in C_\alpha$.

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 $|f(x)-\psi_n(x)| < b_n(x)$ 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 $b_n < B$; $B \in L^1$. However, I think this is again a too strong assumption - it rules out many densities (the fat tailed I am interested in).

I am not sure how to "save" the equation $<\theta,f> = < F^{-1} \theta, Ff>$.

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2  
This site will render tex code, if you put it between $'s. –  David Hansen May 27 '11 at 21:05
    
added the fourier-analysis tag. –  Peter Humphries Jul 3 '11 at 11:04

1 Answer 1

OK, the proof doesn't seem so hard using density arguments and mollifiers! Needs a last check, but I think I got it.

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This would belong better as an update to your main post, since answers shouldn't be used for comments. (Of course if you have a solution then by all means write it up here, as an answer, and you can then accept your own answer.) –  Yemon Choi Jun 19 '11 at 10:56
    
I'll do that soon. –  Pierre Jul 7 '11 at 21:42

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