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Let's say you are given $\omega_f \in \mathcal{S}'(\mathbb R)$ with \begin{align*} f \colon \mathbb{R} &\to \mathbb{R}\\ x &\mapsto x, \end{align*} and the definition of Fourier transform

\begin{align*} \hat{\phi}(\omega)=\int_{-\infty}^{+\infty}e^{-i\omega x}f(x)dx, \, \omega \in \mathbb{R}. \end{align*} Obviously, that function has no Fourier transform, but its corresponding tempered distribution $\omega_f$ does. Using \begin{align*} \widehat{ (\partial_j \phi)} (\omega) =& \,\, i\omega_j\hat{\phi}(\omega)\,\,\,\,\,\,\,\,\,\, (\star) \\ \widehat{1} =& \,\, 2\pi \delta(\omega) \end{align*} we find \begin{equation} \widehat{ (\partial_jf)}(\omega)= \widehat{1}\Rightarrow\hat{f}(\omega)=-2 \pi i \frac{\delta(\omega)}{\omega}. \end{equation} On the other hand, I know that \begin{align*} 2\pi\delta(\omega)=\int_{-\infty}^{\infty}e^{-i\omega x}dx \Rightarrow 2\pi \delta'(\omega)=-i\underbrace{\int_{-\infty}^{\infty}xe^{-i\omega x}dx}_{=:\,\hat{f}(\omega)} \end{align*} which gives \begin{equation} \hat{f}(\omega) = 2\pi i\delta'(\omega). \end{equation}

Why do this two results contradict eachother?

Is it because ($\star$) is not true for $\phi \in \mathcal{S}'(\mathbb R^n)$?

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2 Answers 2

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Writing $\frac{\delta(\omega)}{\omega}$ is not a good idea, because strictly speaking, it's not well defined in many cases. But we do have the relation $\omega\delta'(\omega)=-\delta(\omega)$, so in this sense, your two answers are the same, they are both correct.

P.S. As for your second solution, it is helpful to think that way, and it can give you the correct answer quickly, but this is not how maths works.

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  • $\begingroup$ Thank you for the answer. I see, they're actually equivalent! Could you elaborate on what you mean by "this is not how maths works" please? $\endgroup$
    – xkef
    Commented Dec 10, 2015 at 12:39
  • $\begingroup$ I mean that the integral doesn't converge so it doesn't make sense that you took derivatives on both sides. Besides, this is not how we define the Fourier transformations for distributions. $\endgroup$ Commented Dec 10, 2015 at 23:49
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I thought it might be instructive to present a rigorous solution. To that end we proceed.

Let $\phi\in \mathbb{S}$ and $f(x)=x$. Then, the Fourier Transform of $f$, $\mathscr{F}\{f\}$, is defined in terms of its action on Schwartz functions. Note that we have

$$\begin{align} \langle \mathscr{F}\{f\},\phi \rangle &=\langle f,\mathscr{F}\{\phi\} \rangle\\\\ &=\int_{-\infty}^\infty x \int_{-\infty}^\infty \phi(k)e^{-ikx}\,dk\,dx\\\\ &=\lim_{L\to\infty}\int_{-\infty}^\infty \phi(k)\int_{-L}^L xe^{-ikx}\,dx\,dk\tag{Fubini-Tonelli}\\\\ &=\lim_{L\to\infty}\int_{-\infty}^\infty \phi(k)\left(i\frac{d}{dk}\frac{2\sin(kL)}{k}\right)\,dk\\\\ &=-i2\pi\lim_{L\to\infty}\int_{-\infty}^\infty \phi'(k)\left(\frac{\sin(kL)}{\pi k}\right)\,dk\tag{Integrate by Parts}\\\\ &=-i2\pi \phi'(0)\tag1 \end{align}$$

where we arrived at $(1)$ using the proof in THIS ANSWER.

Therefore, we conclude in distribution that $\mathscr{F}\{f\}(k)=i2\pi \delta'(k)$.

Finally, note that for $f(x)=x$, the distribution $d(x)=f(x)\delta'(x)$ is such that for any $\psi\in C_C^\infty$, we have

$$\begin{align} \langle d,\psi\rangle &=\langle \delta',f\psi\rangle\\\\ &=-f(0)\psi'(0)-f'(0)\psi(0)\\\\ &=-\psi(0) \end{align}$$

Hence, we find that in distribution $x\delta'(x)=-\delta(x)$. However, it is not true that $\delta'(x)=-\delta(x)/x$ since $\delta(x)/x$ is not defined as a distribution..

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  • $\begingroup$ That's seems a lengthy way to get the result, there is no problems to rigorously write $\mathcal F(x) = i\mathcal F(1)' = 2i\pi\delta_0'$ $\endgroup$
    – LL 3.14
    Commented Jan 3 at 23:31
  • $\begingroup$ Of course. But $(i)$ derivations from first principals can be instructive and $(ii)$ you assumed a priori knowledge of the FT of $1$ and the derivative of the FT. $\endgroup$
    – Mark Viola
    Commented Jan 3 at 23:41

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