I need some help in how to find Fourier transform for a distribution in $R^2$ for the function $1/(y-x)$ ? As far as I know it coincides with Fourier transform for the function, but I`m confused how to find it? Any help please.

Thank you

  • $\begingroup$ There is a small but significant issue, that your function is not locally integrable, so it's unclear what the distribution is of which you want the Fourier transform. Do you intend some sort of principal value integral? $\endgroup$ Mar 28 '17 at 21:09
  • $\begingroup$ I think yes. I`m kind of confused how to start solving it. $\endgroup$
    – Danny
    Mar 28 '17 at 21:40

It's not clear what the groundrules are for you, but the way that I would genuinely try to understand this is the following.

First, since Fourier transform respects rotations, it suffices to find the Fourier transform of what at first appears to be $u(x,y)=1/x$. That is, this function/distribution is constant in $y$.

But $1/x$ is not locally integrable, so it is not at all clear what we mean by declaring this a distribution, insofar as it is understood (often tacitly) that a function-as-distribution is really "integrate-against-that-function". So a not locally $L^1$ function is ambiguous at best.

For this example, a standard (and appropriate/useful) re-interpretation is via a principal-value integral: the distribution (in one variable) attached to $1/x$ is not integrate against $1/x$, but is $u(f)=\lim_{\varepsilon\to 0^+} \int_{|x|\ge \varepsilon} {f(x)\over x}\;dx$. Note that this is not a literal integral, but is some sort of extension of integral. It only makes literal sense for $f$ with some sort of good behavior at $0$, etc.

There is another exercise which is to show that the Fourier transform of this integral is (a constant multiple of) the sign function $\sigma$. I can't tell how much of the question this sub-question is meant to be.

Thus, the Fourier transform of $u\otimes 1$ is $\widehat{u}\otimes \widehat{1}=c\cdot \sigma\otimes \delta$ (because the Fourier transform of $1$ is $\delta$), and rotating back to the original, this is $c\cdot \sigma(x-y)\otimes \delta(x+y)$. (Modulo details! :)

  • 1
    $\begingroup$ I think I understand what you mean, but still I need to think about it. This is actually a one part of a bigger question that I have. I already know about p.v(1/x) , thanks a lot of you and I`ll come back to it after thinking. $\endgroup$
    – Danny
    Mar 28 '17 at 21:56
  • $\begingroup$ how that will change if we make the function 1/y(y-x) ? $\endgroup$
    – Danny
    Mar 29 '17 at 20:19
  • $\begingroup$ @Danny, then I'd still want to be able to separate variables by making a linear change of variables to $1/xy$. Then the corresponding distribution has to be a sort of doubled-up p.v. integral $u\otimes u$, with two epsilons, one for $x$, one for $y$. Then $\widehat(u\otimes u)=c\cdot \sigma\otimes\sigma$. The undo the linear change of variables. $\endgroup$ Mar 29 '17 at 20:29
  • $\begingroup$ I think what makes me confused is the cross sign that you are using. what do you mean actually by it? why don`t we just use a translation and rotation applied immediately to p.v.(1/x)? $\endgroup$
    – Danny
    Mar 31 '17 at 3:12
  • $\begingroup$ Ah, sorry, $f\otimes g$ means $f(x)\,g(y)$, so $(f\otimes g)(x,y)=f(x)\,g(y)$, but we don't have to use dummy arguments. $\endgroup$ Mar 31 '17 at 11:30

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