1

What precisely is the set of necessary and sufficient conditions for the Fourier transform of a function $f : \mathbb{R} \mapsto \mathbb{C}$ to exist?

  • Obviously there’s the sort of tautological answer that the criterion is: $\int f(x) e^{-i\omega x}dx$ converges. But I’m hoping for something more “reductive” like “$f$ is absolutely integrable and has countably many discontinuities” (not literally that, necessarily, but something of that character)
  • I suppose the answer might depend on what kind of integral we’re assuming in the definition of Fourier transform (Riemann, Lebesgue, etc.)

Note: I’ve seen some other similar existing questions on Math SE which I don’t regard as duplicates because the answers don’t fully address what I’m asking. For example, the answers may give sufficient conditions which aren’t strictly necessary conditions.

NikS
  • 1,980
  • 2
    So far I know any measurable function that belongs to any Lebesgue space $L_p$ has its corresponding Fourier transform in $L_q$ space, where $1/p+1/q=1$, at least for $1\leq p<\infty$. – Dr Potato Apr 18 '25 at 03:18
  • 1
    I do not aware that there exist some suffcient and necessary condition to make the Fourier thansform of some function to be well-defined. But in Fouier analysis, we may regard Fourier transform as a linear operator and hence, we can extend the definition of Fourier transform on some good class, like Schwarz function, to some general class, like temperature distribution. For example, for $f\in L^2$, the Fourier transform of $f$ may be not converge pointwise, but we can find the Fourier transform of $f$ in $L^2$ class. – Mr.xue Apr 18 '25 at 03:20
  • The largest space on which the Fourier transform can be extended to is the space of tempered distributions. You are therefore asking: which functions are tempered distributions? Which is the content of this old question of mine. – Giuseppe Negro Apr 18 '25 at 04:21
  • Well, I think this also depends on what you require that the Fourier transform of $f$ be. Are you just requiring it to also be a function with domain on $\mathbb{R}$? – Diana Pestana Apr 18 '25 at 06:58

0 Answers0