1
$\begingroup$

Let $f\in L^1(\mathbb{R})$. We define $\phi_f(x)=\int_{\mathbb{R}} \hat{f}(\zeta)e^{2\pi i\zeta x}d\zeta$ if the improper Riemann integral is finite otherwise, $\phi_f(x)=\infty$.

  1. Does there exist any integrable function $f$ for which $\{x: \phi_f(x)=\infty\}$ enjoys positive measure?

  2. Suppose that $\{x: \phi_f(x)=\infty\}$ is a null set (its Lebesgue measure is zero). Could we conclude that $f(x)=\int_{\mathbb{R}} \hat{f}(\zeta)e^{2\pi i\zeta x}d\zeta$?

p.s. $\hat{f}$ is the Fourier transform of $f$.

Improve this question
$\endgroup$
8
  • $\begingroup$ We have $|e^{2\pi i\zeta x}\hat f(\zeta)|=|\hat f(\zeta)|$, so membership in $L^1(\mathbb R)$ doesn't seen to depend on $x$. So the set in your question is either empty or all of $\mathbb R$ for any $f$. $\endgroup$
    – Wojowu
    Commented Nov 22, 2020 at 16:39
  • $\begingroup$ For an integrable function with nonintegrable Fourier transform, see here $\endgroup$
    – Wojowu
    Commented Nov 22, 2020 at 16:44
  • $\begingroup$ Thanks, I revised the question. $\endgroup$
    – ABB
    Commented Nov 22, 2020 at 17:00
  • $\begingroup$ Could you be more precise about what you mean by "the integral does not exist"? Is your integral supposed to be the Lebesgue integral, in which case you need f-hat to be integrable; or are you taking some kind of improper Riemann integral? Note that since f is integrable, f-hat is continuous and bounded and vanishes at infinity $\endgroup$
    – Yemon Choi
    Commented Nov 22, 2020 at 17:17
  • 3
    $\begingroup$ Kolmogorov constructed an $L^1(T)$ function whose Fourier series diverges everywhere. That would strongly suggest (to me, at least) that there is a similar example in the continuous setting. $\endgroup$
    – Christian Remling
    Commented Nov 22, 2020 at 21:26

0

Reset to default

You must log in to answer this question.