Decay of the Fourier Transform of non-integrable functions.

Question

An integrable function generally means an $L^1$ function. If we stick to functions from $\mathbb{R}\to\mathbb{C}$, then $f$ is integrable means that $$ \|f\|_1 = \int\limits_\mathbb{R} |f(t)|dt < \infty. $$ In this case the Fourier transform of $f$ which we denote by $F$ is well defined as an integral and we have the standard inequality (for $\xi\in\mathbb{R}$), $$ |F(\xi)| \leq \|f\|_1. $$

If $f$ takes on positive values, then I expect we can modify the proof of the Rieman-Lebesgue lemma to show that $$ |F(\xi)| \leq \frac{\|f\|_1}{|\xi|} $$ for $|\xi| \gg 0$.

My Question is whether analogous bounds exist if we assume just that $$ \int\limits_\mathbb{R} f(t) dt < \infty $$ and not necessarily that $f\in L^1(\mathbb{R})$.

Thank you.

Answer 1

The result you want to extend to the case $f\not\ge0$ is already false for $f>0$:

There exists $f\in L^1(\Bbb R)$ with $f>0$ such that $\hat f(\xi)\ne O\left(\frac1{|\xi|}\right).$

I "knew" immediately that $f>0$ could not give that bound on $\hat f$, but I thought the counterexample might be tricky. No, we can do this with no work at all:

Say $f>0$, $\int f<\infty$, $\int f^2=\infty$. Since $\hat f$ is continuous, if we had $\hat f(\xi)=O\left(\frac1{|\xi|}\right)$ it would follow that $\hat f\in L^2$, hence $f\in L^2$, contradiction.