0
$\begingroup$

As I read in this post the Fourier transform of $\psi(\lambda) = \log{|\lambda|}$ must be interpreted in distributional sense and it is given by: $$\mathscr{F}\{\psi\}(x)=-2\pi \gamma \delta(x)-\pi \text{PV}\left(\frac1{|x|}\right)$$

where we interpret $\text{PV}\left(\frac1{|x|}\right)$ to mean that for any $\phi\in \mathbb{S}$ (Schwartz space), for any $\nu >0$:

$$\int_{-\infty}^\infty \phi(x) \text{PV}\left(\frac1{|x|}\right)\,dx=\int_{|x|\le \nu}\frac{\phi(x)-\phi(0)}{|x|}\,dx+ \int_{|x|\ge \nu}\frac{\phi(x)}{|x|}\,dx$$

It follows that the convolution of $\mathscr{F}\{\psi\} * f$ is well defined as soon as $f$ decays fast enough around zero and at infinity.

Now my question is: if you consider a measure $\mu$ that is compactly supported on $[a,b]$ with $a,b >0$ (therefore it is zero before $a$ and after $b$) is the convolution $\mathscr{F}\{\psi\} * \mu$ still well defined?

Ps: I posted before this question on MSE but I realised it was not the suitable website.

Improve this question
$\endgroup$
6
  • 1
    $\begingroup$ See Theorem 4.3 here: mat.univie.ac.at/~stein/lehre/SoSem09/distrvo.pdf $\endgroup$
    – Christian Remling
    Commented Mar 28 at 14:11
  • 1
    $\begingroup$ A bit more informally, you can also just observe that the product $\widehat{\mu}(t)\psi (t)$ is still tempered since $\widehat{\mu}$ is bounded. $\endgroup$
    – Christian Remling
    Commented Mar 28 at 14:12
  • $\begingroup$ @ChristianRemling Thank you very much for your answer! One more question: if I truncate the domain of $\psi(t)$ near zero and consider $\hat{\mu}(t)\psi(t) 1_{|t|>a}$ for some $a>0$, is the resulting distribution in $L^2$? $\endgroup$
    – Grandes Jorasses
    Commented Mar 28 at 16:03
  • 1
    $\begingroup$ No, not in general because $\psi$ gets large for large $|t|$ and $\widehat{\mu}$ is not guaranteed to have any decay, so we cannot automatically conclude that $\widehat{\mu}\psi\in L^2(|t|>a)$. $\endgroup$
    – Christian Remling
    Commented Mar 28 at 16:13
  • 1
    $\begingroup$ Yes, that would work if you restrict to $|t|<a$. $\endgroup$
    – Christian Remling
    Commented Mar 28 at 16:58

0

Reset to default

You must log in to answer this question.