0
$\begingroup$

Let $f\in L^1(\mathbb{R})$ such that $$\int_0^\infty e^{-yt}e^{ixt}\widehat{f}(t)dt=0,……(i)$$ $$\int_{-\infty}^0 e^{yt}e^{ixt}\widehat{f}(t)dt=0, ……(ii)$$ for all $x\in \mathbb{R}, y>0.$

Questions:

  1. Is it true that $f=0?$
  2. If $f$ satisfies only condition $(i)$, then can we conclude that $\widehat{f}$ vanishes identically on $[0,\infty)?$

Note: I can prove 2.(hence 1.) assuming the extra condition that $\widehat{f}\in L^1(\mathbb{R})$.

Improve this question
$\endgroup$
4
  • 1
    $\begingroup$ If $x=0$ in (i), then the Laplace transform of $\hat f$ is 0 for $y>0$ and then $\hat f=0$ for $t>0$. $\endgroup$
    – Giorgio Metafune
    Commented Sep 8, 2022 at 11:34
  • $\begingroup$ @GiorgioMetafune To apply your argument we need to assume that $\widehat{f}$ is also integrable on $[0,\infty)$. But is the result true if $\widehat{f}$ is not integrable on $[0,\infty)$? $\endgroup$
    – user483450
    Commented Sep 8, 2022 at 11:48
  • $\begingroup$ Usually for the Laplace transform one assumes that $g$ is exponentially bounded, that is $|g(t)| \leq Me^{Ct}$ so that the transform is defined for $Re\, y >C$. In this case $g=\hat f$ is even bounded. If you want to keep integrability assumptions, you may apply the argument to $e^{-t} \hat f(t)$. $\endgroup$
    – Giorgio Metafune
    Commented Sep 8, 2022 at 13:16
  • 1
    $\begingroup$ Alternatively, $\lim_{y\to 0+}\int e^{-y|t|}e^{ixt}\widehat{f}(t) = f(x)$ (up to a factor perhaps), for example because multiplying $\widehat{f}$ with $e^{-y|t|}$ is the same as convolving $f$ with the Poisson kernel. $\endgroup$
    – Christian Remling
    Commented Sep 8, 2022 at 13:47

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .