Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
2  
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
add comment

2 Answers

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.

share|improve this answer
add comment

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

share|improve this answer
    
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.