Which conditions must fulfill $f(t)$ to have an absolute-integrable Fourier Transform $F(w)$: $\int\limits_{-\infty}^\infty |F(w)| dw < \infty$? Which conditions must fulfill $f(t)$ to have an absolute-integrable Fourier Transform $F(w)$: $\int\limits_{-\infty}^\infty |F(w)| dw < \infty$?

*

*At first, thinking in $f(t)$ as an arbitrary one variable function which has a Fourier Transform

*Secondly, for a specific $f(t)$ that is time-limited and continuous, so its also compact-supported and bounded ($\sup_t |f(t)| < \infty$), and has unlimited bandwidth.

*Third, adding to the conditions of point (2) that it has a Fourier Transform that fulfill the Riemann-Lebesgue Lemma, and also has finite-energy $\int_{t_0}^{t_F} |f(t)|^2\,dt < \infty$, is absolute-integrable $\int_{t_0}^{t_F} |f(t)|\,dt < \infty$, and has zero values at the edges of its domain $f(t_0)=f(t_F)=0$.

*Same as third, but requiring now that $f(t_0)\neq 0$ and/or $f(t_F) \neq 0$.

The actual order of requirement following how restrictive they are is $(3)\rightarrow (4)\rightarrow (2)\rightarrow(1)$.
 A: *

*

*

*Notice first that if $f$ is such that its Fourier transform is positive, then
$$
\int_{-\infty}^\infty |F| = \int_{-\infty}^\infty F =  f(0) \leq \|f\|_{L^\infty}
$$
so in this case it is sufficient to be bounded. But this is a rather special case.




*

*In general, at least $1/2$ derivative has to be in $L^2$. More precisely, by the Cauchy-Schwarz inequality, for any $s>1/2$
$$
\int_{-\infty}^\infty |F| \leq \left(\int_{-\infty}^\infty (1+|2\pi w|^2)^s\,|F(w)|^2\,\mathrm d w\right)^{1/2} \left(\int_{-\infty}^\infty (1+|2\pi w|^2)^{-s}\,\mathrm d w\right)^{1/2}.
$$
Since $s>1/2$, the last integral $C = \int_{-\infty}^\infty (1+|w|^2)^{-s}\,\mathrm d w$ is finite, so as $(1+|w|^2)^{s/2}\,F(w)$ is the Fourier transform of $(1-\Delta)^s\,f$ and the Fourier transform preserves the $l^2$ norm
$$
\int_{-\infty}^\infty |F| \leq C\, \|(1-\Delta)^{s/2}f\|_{L^2} = C\, \|f\|_{H^s}.
$$
This space $H^s$ is the Sobolev space of order $s$. Since every function with integrable function transform is bounded, the order $s>1/2$ is optimal in the sense that one can find unbounded functions in $H^{1/2}$. Of course it is sufficient to take for example $s=1$, in which case the above inequality becomes
$$
\int_{-\infty}^\infty |F| \leq C\, \left(\int_{-\infty}^\infty |f|^2 + |f'|^2\right)^{1/2}.
$$


*For the reader that knows Besov spaces, one can get slightly more precise by replacing $H^s$ with $s>1/2$ by $B^{1/2}_{2,1}$ which is an intermediate space between $H^s$ and $H^{1/2}$. But this is a subject with a huge literature, you can find papers easily on the subject by typing "Integrability of Fourier transform" in Google for instance. Also, the space of functions with integrable Fourier transforms is sometimes called a Wiener algebra. You can find a survey on these questions called "The Wiener algebra of absolutely convergent Fourier integrals: an overview".


*Being continuous and converging to $0$ at infinity is also not sufficient (i.e. the converse of Riemann-Lebesgue Lemma is false), see A Fourier transform of a continuous $L^1$ function.


*The fact that the function is compactly supported will not change anything as what is needed is regularity.


*Similarly, to have a bounded integral or to have a bounded energy will not help.


*Worse than 3, so still not sufficient.
