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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?


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What about $f(t)=\chi_{[-1, 1]}(t)$? This function is certainly integrable but its Fourier transform is $\mathrm{sinc}(\omega)=2\frac{\sin(\omega)}{\omega}$ (up to constants! They depend on your favourite definition of F-transform), which is not $L^1$. I think that in engineering literature the first function is known as rect. This is an explicit example in the vein of Sam's suggestion below. EDIT Indeed, Sam explictly mentioned this function. I didn't read carefully enough! :-) – Giuseppe Negro Sep 20 '11 at 21:43
up vote 10 down vote accepted

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.

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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 $\hat{f} \not\in L^1$.

Original example:

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 $\hat{f}(\omega) \sim \vert \omega \vert^{-\frac{1}{2}}$.

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This seems awfully complicated! =) But +1 for effort. – Sam Sep 20 '11 at 21:20
@Sam I realized $\mathrm{e}^{-\vert x\vert}/\sqrt{\vert x\vert}$ is a much simpler example. – Sasha Sep 20 '11 at 21:21
@ByronSchmuland Thank you Byron! I have edited the post now – Sasha Sep 27 '11 at 14:41

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