I am looking for an example of a function $f: \mathbb R \rightarrow \mathbb R$ such that $f \in L^1$ in the sense that $\int_{\mathbb R} |f| < \infty$ but its Fourier transform $\hat f$ is not in $L^1$. Does anyone have one?
Thanks.
I am looking for an example of a function $f: \mathbb R \rightarrow \mathbb R$ such that $f \in L^1$ in the sense that $\int_{\mathbb R} |f| < \infty$ but its Fourier transform $\hat f$ is not in $L^1$. Does anyone have one?
Thanks.
Note that any function whose fourier transform is in $L^1$ must be equal to a continuous function almost everywhere, since $\mathcal F(\mathcal F(f)) = f$ a.e. in this case. This follows from the inversion formula and because the Fourier transform of a function is continuous.
This gives us many examples of functions you are looking for. For example $f(x) = \chi_{[-1,1]}(x)$ must necessarily be such a function.
Added: The function $f(x) = \vert x \vert^{-1/2} \mathrm{e}^{-\vert x \vert}$ is a simple example (much simpler than the original example I proposed). Its Fourier transform is:
$ \hat{f}(\omega) = \sqrt{\frac{1}{\sqrt{\omega ^2+1}}+\frac{1}{\omega ^2+1}} $ and has asymptote $\hat{f}(\omega) \sim \vert \omega \vert^{-1/2}$ for large $\vert \omega \vert$, thus $\hat{f} \not\in L^1$.
An example would be $ f(x) = \left\{ \begin{array}{cc} x^{-1/4} \mathrm{e}^{-x} & x > 0 \\ \vert x \vert^{-1/2} \mathrm{e}^{x} & x < 0 \end{array} \right. $ It is clear that $\int_\mathbb{R} \vert f(x) \vert \mathrm{d} x < \infty$. The Fourier transform $ \hat{f}(\omega) = \frac{\sqrt{1-i \omega }-\sqrt{1+i \omega }}{ \sqrt{8 (1+\omega ^2)}}+\frac{1}{2} \sqrt{\frac{1}{\sqrt{\omega ^2+1}}+\frac{1}{\omega ^2+1}}+\frac{\Gamma \left(\frac{3}{4}\right)}{\sqrt{2 \pi } \, (1-i \omega )^{3/4}} $ The integral $\int_\mathbb{R} \vert \hat{f}(\omega) \vert \mathrm{d} \omega $ diverges because $\hat{f}(\omega) \sim \vert \omega \vert^{-\frac{1}{2}}$.