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I am trying to find the Fourier transform of $|x|$ in the sense of distributions in its simplest form. Here is what I have done so far:

Let $$f(x)=|x|=\lim_{a\rightarrow 0}\frac{1-e^{-a|x|}}{a},$$ then the Fourier transform is given by $$\begin{aligned} \hat{f}(\xi)&=\int_{-\infty}^\infty f(x)e^{-2\pi i x \xi}dx \\ &=\lim_{a\rightarrow 0}\frac{1}{a}\left(\delta(\xi)-\frac{2a}{a^2+4\pi^2\xi^2}\right). \end{aligned}$$ Using the identity (see here), $$\delta(\xi)=\lim_{a\rightarrow 0}\frac{1}{\pi}\frac{a}{a^2+\xi^2},$$ we know that $$2\pi\delta(2\pi\xi)=\lim_{a\rightarrow0}\frac{2a}{a^2+4\pi^2\xi^2}.$$ Hence, using the identity, $$\delta(b x)=\frac{1}{|b|}\delta(x),$$ we know that $$\hat{f}(\xi)\stackrel{?}{=}\lim_{a\rightarrow0}\frac{1}{a}[\delta(\xi) - \delta(\xi)].$$ This doesn't seem right... Can you see where I have gone wrong and do you know how to calculate $\hat{f}(\xi)$ in its simplest form?

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    $\begingroup$ The Fourier transform should satisfy $\langle f,\check \varphi \rangle = \langle \hat f, \varphi\rangle$ for a test function $\varphi$. To see what $\hat f$ should be, you need to compute $\langle f, \check \varphi\rangle$. $\endgroup$
    – tomasz
    Commented Jul 22, 2020 at 17:41
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    $\begingroup$ It turns out that, for any test function $\varphi$ in the Schwartz space, the pairing $\langle \hat{f}, \varphi \rangle$ satisfies $$ \langle \hat{f}, \varphi \rangle = -\frac{1}{2\pi^2} \int_{0}^{\infty} \frac{\varphi(x)-2\varphi(0)+\varphi(-x)}{x^2} \, \mathrm{d}x.$$ For this reason, we may formally write $$\hat{f}(\xi)=-\frac{1}{2\pi^2} \operatorname{p.v.}\frac{1}{\xi^2}. $$ $\endgroup$ Commented Jul 22, 2020 at 18:05
  • $\begingroup$ @SangchulLee I obtained the same result $\lim_{\varepsilon\to 0^+}\int_{|x|\ge \varepsilon}\frac{\phi(x)-\phi(0)}{x^2}\,dt$ by applying the result I derived in THIS ANSWER using a regularization approach. I've posted my solution herein and look forward to your feedback. ;-) Stay safe and healthy my friend! $\endgroup$
    – Mark Viola
    Commented Jul 22, 2020 at 19:46

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So, a way to compute it is to write $|x| = x\mathop{\mathrm{sign}}(x)$. By definition, we have $$ \langle \mathcal{F}(|x|),\varphi\rangle = \langle |x|,\mathcal{F}(\varphi)\rangle = \langle x\mathop{\mathrm{sign}}(x),\mathcal{F}(\varphi)\rangle $$ Since $x∈ C^\infty$, we can then write $$ \langle x\mathop{\mathrm{sign}}(x),\mathcal{F}(\varphi)\rangle = \langle \mathop{\mathrm{sign}}(x),x\,\mathcal{F}(\varphi)\rangle = \frac{1}{2i\pi}\langle \mathop{\mathrm{sign}}(x),\mathcal{F}(\varphi')\rangle $$ where I used the formula for the Fourier transform of a derivative. Now, by definition again, and then using the fact that $\mathcal{F}(\mathop{\mathrm{sign}}(x)) = 1/{i\pi} \,\mathrm{P}(\tfrac{1}{x})$ (the principal value of $1/x$) we get $$ \frac{1}{2i\pi}\langle \mathop{\mathrm{sign}}(x),\mathcal{F}(\varphi')\rangle = \frac{1}{2i\pi}\langle \mathcal{F}(\mathop{\mathrm{sign}}(x)),\varphi'\rangle \\ = \frac{-1}{2\pi^2}\langle \mathrm{P}(\tfrac{1}{x}),\varphi'\rangle = \frac{1}{2\pi^2}\langle \mathrm{P}(\tfrac{1}{x})',\varphi\rangle $$ so that $$ \mathcal{F}(|x|) = \frac{1}{2\pi^2} \mathrm{P}(\tfrac{1}{x})' = \frac{-1}{2\pi^2} \mathrm{P}(\tfrac{1}{x^2}) $$ where $\mathrm{P}(\tfrac{1}{x^2})$ is the Hadamard finite part of $\tfrac{1}{x^2}$. Away from $0$, we can thus say that $$ \mathcal{F}(|x|) = \frac{-1}{2\pi^2x^2} $$ (if I did not make mistakes in the constants and signs ...)

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    $\begingroup$ (+1) I computed it in a different way but obtained the same answer. So I guess we are both right or wrong... $\endgroup$ Commented Jul 22, 2020 at 18:17
  • $\begingroup$ Thank you. It looks like the Fourier transform of $|x|$ is the same as Fourier transform of $$\lim_{a\rightarrow0}-\frac{e^{-a|x|}}{a}.$$ Is that right? What happens to the contribution from the $1/a$ term? $\endgroup$
    – Peanutlex
    Commented Jul 22, 2020 at 18:27
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    $\begingroup$ @Peanutlex That limit doesn't exist. $\endgroup$
    – aschepler
    Commented Jul 22, 2020 at 18:50
  • $\begingroup$ @ll3.14 Well done! I've posted a solution that relies on the results from an answer I posted HERE. In that answer, I used a regularization approach to evaluate the Fourier Transform of $xH(x)$. Since $\text{sgn}(x)=2H(x)-1$, application of the former result leads to effectively the same result as yours, modulo a constant die to our different definitions of the FT. ;-) $\endgroup$
    – Mark Viola
    Commented Jul 22, 2020 at 19:41
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EDIT: I reread the post and wanted to present an edit to the original posted solution that addresses directly the concern about the OP's analysis. To that end we proceed with the addendum.

You were on the right track! In fact, if one begins with the regularization $f(x)=|x|=\lim_{a\to 0}\frac{1-e^{-a|x|}}{a}$, then one finds that in distribution

$$\mathscr{F}\{f\}(\omega)=\lim_{a\to 0^+}\frac1a\left(2\pi\delta(\omega)-\frac{2a}{a^2+\omega^2}\right)$$

To evaluate this distributional limit, we begin with a test function $\phi(\omega)$ and find

$$\begin{align} \langle \mathscr{F}\{f\},\phi\rangle&=\lim_{a\to0^+}\frac1a\left(2\pi\phi(0)-\int_{-\infty}^\infty \frac{2a\phi(\omega)}{a^2+\omega^2}\,d\omega\right)\\\\ &=\lim_{a\to 0^+}\int_{-\infty}^\infty \left(-\frac{2(\phi(\omega)-\phi(0))}{a^2+\omega^2}\right)\,d\omega\\\\ &=-\lim_{\varepsilon\to 0^+}\int_{|\omega|\ge \varepsilon}\frac{2(\phi(\omega)-\phi(0))}{\omega^2}\tag{1E} \end{align}$$

So, we find that

$$\mathscr{F}\{f\}(\omega)=-\frac2{\omega^2}\tag{2E}$$

where we interpret the distribution in $(2E)$ in the sense of $(1E)$

Note that we have used the convention $\mathscr{F}\{f\}(\omega)=\int_{-\infty}^\infty f(x)e^{i\omega x}\,dx$. Had we used instead the convention $\mathscr{F}\{f\}(\omega)=\int_{-\infty}^\infty f(x)e^{i2\pi \xi x}\,dx$, then $(2E)$ would be replaced with $-\frac1{2\pi^2 \xi^2}$



In This Answer, I showed that the Fourier Transform of $f(t)=tH(t)$, where $H(t)$ denotes the Heaviside function, is given by

$$\mathscr{F}\{f\}(\omega)=-\frac1{\omega^2}+i\pi \delta'(\omega)\tag1$$

where the distribution $d(\omega)=\displaystyle -\frac1{\omega^2}$ in $(1)$ is interpreted to mean

$$\langle d, \phi\rangle=-\lim_{\varepsilon\to0^+}\int_{|\omega|\ge\varepsilon}\frac{\phi(\omega)-\phi(0)}{\omega^2}\,d\omega\tag2$$

where $\phi(\omega)$ is a Schwartz function.


Using $g(t)=t\text{sgn}(t)=2tH(t)-t$ along with $\mathscr{F}\{t\}(\omega)=i2\pi \delta'(\omega)$ and $(1)$, we find that

$$\begin{align} \mathscr{F}\{g\}&=-\frac2{\omega^2}\tag3 \end{align}$$

where again $(3)$ is defined analogously to $(2)$.

And we are done!

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  • $\begingroup$ @md2perpe I've edited to conform to the convention you mentioned. $\endgroup$
    – Mark Viola
    Commented Jul 22, 2020 at 19:55
  • $\begingroup$ Thank you very much this is really helpful. $\endgroup$
    – Peanutlex
    Commented Jul 22, 2020 at 20:02
  • $\begingroup$ @peanutlex Good catch. I've edited accordingly. $\endgroup$
    – Mark Viola
    Commented Jul 22, 2020 at 20:38
  • $\begingroup$ And you're welcome. My pleasure. If you have a read if the linked answer, it uses a regularization similar to the one you had tried. Have a look and see if it addresses directly your question. $\endgroup$
    – Mark Viola
    Commented Jul 22, 2020 at 20:40
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    $\begingroup$ @Peanutlex I added a section ("EDIT") to address directly your analysis. You were on the right track and I've presented a way forward. Again, please let me know how I can improve my answer. $\endgroup$
    – Mark Viola
    Commented Aug 21, 2020 at 15:58
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From $\operatorname{sign}'=2\delta$ we get $i\xi\,\widehat{\operatorname{sign}}(\xi)=2,$ from which we can conclude $$ \widehat{\operatorname{sign}}(\xi) = -2i\operatorname{pv}\frac{1}{\xi}+C\delta(\xi). $$ Since $\operatorname{sign}$ is odd, so must be $\widehat{\operatorname{sign}},$ which forces $C=0.$

Now, $f(x) = x \operatorname{sign}(x),$ so $$ \hat{f}(\xi) = i\frac{d}{d\xi}\widehat{\operatorname{sign}}(\xi) = i\frac{d}{d\xi}\left(-2i \operatorname{pv}\frac{1}{\xi}\right) = -2 \operatorname{fp}\frac{1}{\xi^2}. $$

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