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If we treat fourier transform as an operator on $L^1(\mathbb{R})$, then its image under fourier transform is the set of continuous functions which will vanish at infinity.

It is well known that the fourier transform of signum function is $$\mathcal{F} (sgn)(u) =\frac{2}{ui}. $$ I know that signum function is not integrable over the real line. So, to evaluate its fourier transform, one can use limiting argument, say a sequence of functions that converges to signum function, because fourier transform is a bounded linear operator, and hence is continuous.

What puzzles me is that when using continuity, don't we need to ensure that the fourier transform is defined on the limiting function?

In this case, fourier transform of signum function is not defined due to reason given above. If this is the case, how would one obtain formula of the fourier transform of signum function?

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  • $\begingroup$ The image of $L^1$ by Fourier transform is included in the set of continuous functions vanishing at infinity, but different from it! $\endgroup$
    – LL 3.14
    Jun 19, 2020 at 23:07

2 Answers 2

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1. BOTTOM LINE UP FRONT: Treat it as a distribution.

Since the signum function is not integrable on $\mathbb{R}$, it may be useful to view it as a tempered distribution.

Such "generalized functions" are bounded linear functionals on a class of very-well-behaved functions called Schwartz functions. One of Laurent Schwartz's achievements was finding a collection $\mathcal{S}$ of functions on $\mathbb{R}^n$ such that the set of Fourier transforms of these functions is $\mathcal{S}$ itself. That put the original functions and their Fourier transforms on equal footing.


2. Fourier transform of a distribution

Why is this useful? It means that every tempered distribution has a Fourier transform that is also a tempered distribution. It also provides some useful notation to derive expressions and properties of the Fourier transform of a known tempered distribution.

Given any distribution $\mathsf{T}$ we write the result of applying it to a Schwartz function $\varphi$ as $\left<\mathsf{T},\varphi\right>$, but it is to be understood that this is not an inner product of two objects of the same kind. The Fourier transform of the distribution $\mathsf{T}$ is the distribution $\widehat{\mathsf{T}}$ for which \begin{equation} \left<\widehat{\mathsf{T}},\varphi\right> = \left<\mathsf{T},\widehat{\varphi}\right> \end{equation} for every $\varphi\in\mathcal{S}$, where $\widehat{\varphi}$ is the Fourier transform of $\varphi$. Since $\varphi\in\mathcal{S}$, $\widehat{\varphi}\in\mathcal{S}$, too.
3. Fourier transform of signum

How is this related to the signum function? If $\mathsf{T}$ is the signum function viewed as a distribution, then \begin{equation} \left<\mathsf{T},\varphi\right> = \int\textrm{sgn}(x)\varphi(x)dx. \end{equation} The Fourier transform of this distribution satisfies (or is defined by) \begin{equation} \begin{split} \left<\widehat{\mathsf{T}},\varphi\right> &=~ \left<\mathsf{T},\widehat{\varphi}\right>\\ &=~ \int\textrm{sgn}(x)\widehat{\varphi}(x)dx\\ &=~ -\int_{-\infty}^{0}\widehat{\varphi}(x)dx + \int_{0}^{\infty}\widehat{\varphi}(x)dx. \end{split} \end{equation}
4. Changing order of integration

Let's consider the integral for positive reals. The very good behavior of $\varphi$ allows changing orders of integration in many, many situations. \begin{equation} \begin{split} \int_{0}^{\infty}\widehat{\varphi}(x)dx &=~ \int_{0}^{\infty}\left[\int\varphi(k)e^{-ixk}dk\right]dx\\ &=~ \lim_{R\to\infty}\int_{0}^{R}\left[\int\varphi(k)e^{-ixk}dk\right]dx\\ &=~ \lim_{R\to\infty}\int\left[\int_{0}^{R}e^{-ixk}dx\right]\varphi(k)dk \end{split} \end{equation} We do something very similar for the negative reals. \begin{equation} \begin{split} -\int_{-\infty}^{0}\widehat{\varphi}(x)dx &=~ -\int_{-\infty}^{0}\left[\int\varphi(k)e^{-ixk}dk\right]dx\\ &=~\lim_{R\to\infty}-\int_{-R}^{0}\left[\int\varphi(k)e^{-ixk}dk\right]dx\\ &=~ \lim_{R\to\infty}-\int\left[\int_{-R}^{0}e^{-ixk}dx\right]\varphi(k)dk \end{split} \end{equation}

We now address the sum of the $R$-dependent integrals. \begin{equation} \begin{split} \int_{0}^{R}e^{-ikx}dx - \int_{-R}^{0}e^{-ikx}dx &=~ \left.\frac{e^{-ikx}}{-ik}\right|_{x=0}^{x=R} - \left.\frac{e^{-ikx}}{-ik}\right|_{x=-R}^{x=0}\\ &=~ \frac{1 - e^{-ikR}}{-ik} - \frac{e^{ikR} - 1}{-ik}\\ &=~ \frac{e^{ikR} + e^{-ikR}}{ik} - \frac{2}{ik} \end{split} \end{equation}


5. Singularity at $k = 0$; Riemann-Lebesgue Lemma

The $k$ in the denominator will be a problem at $k=0$. But we know that the original integrals coverge. We must consider the new one as the limit of integrals from $\epsilon$ to $\infty$ and from $-\infty$ to $-\epsilon$.

\begin{equation} \int_{|k|>\epsilon}\frac{e^{ikR} + e^{-ikR}}{ik}\varphi(k)dk = \int 1_{\{k:|k|>\epsilon\}}(k)\frac{\varphi(k)}{ik}\left(e^{ikR} + e^{-ikR}\right)dk \end{equation} For each $\epsilon >0$, the function $1_{\{k:|k|>\epsilon\}}(k)\frac{\varphi(k)}{ik}$ is in $L^1(\mathbb{R})$, so this integral is that function's Fourier transform evaluated at $\omega = R$ plus the same Fourier transform evaluated at $\omega = -R$. The Riemann-Lebesgue Lemma shows that if $f\in L^1(\mathbb{R})$, then $\lim_{|R|\to\infty}\widehat{f}(R) = 0$. Hence, these $R$-dependent terms vanish as $R\to\infty$.

It is worth noting that this shows that we must take the $R$-limit first and then take the $\epsilon$-limit. The opposite order would not work.


6. Cauchy Principal Value

We are left with \begin{equation} \lim_{\epsilon\to 0}2i\int_{|k|>\epsilon}\frac{\varphi(k)}{k}dk. \end{equation} This is the Cauchy Principal Value of this integral. This shows that we must interpret the Fourier transform of the signum function very carefully, but we can do it in the sene of distributions: if $\textrm{sgn}(x)$ is the signum of $x$, then \begin{equation} \widehat{\textrm{sgn}}(k) = 2i~\mathsf{PV}\left(\frac{1}{k}\right). \end{equation}
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  • $\begingroup$ Hi Joe. You beat me to this way forward. I had written up to the result in your Section 3 when I noticed your very well-written post (+1). The analysis reminds me of the development in This Answer. Your post is of exceptional value and I wish I could give more than one up vote. $\endgroup$
    – Mark Viola
    Jun 19, 2020 at 20:31
  • $\begingroup$ Joe, I took the liberty of posting a fairly compact development using bracket notation that relies on some distributional relationships. It is, therefore, a bit different in kind from yours and serves as a complementary way forward. Let me know your thoughts. $\endgroup$
    – Mark Viola
    Jun 19, 2020 at 20:51
  • $\begingroup$ @mark-viola: You are too kind. If I had not remembered the Riemann-Lebesgue Lemma, I would still be worrying about the terms that vanish as $R\to\infty$. Using that result felt like cheating, but I had spent enough time on it that I decided not to seek a more direct method. $\endgroup$
    – Joe Mack
    Jun 19, 2020 at 20:51
  • $\begingroup$ Thank you my friend! Much appreciated. I took for granted several well-known relationships that require some background in the theory of generalized functions. Once equipped with that background, then all proceeds nicely. I really like your post inasmuch as you explain everything along the way for a reader who is less familiar with tempered distributions and the Fourier Transform. $\endgroup$
    – Mark Viola
    Jun 19, 2020 at 20:56
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    $\begingroup$ Just to be clear, I am talking about going from $\lim_{R \to \infty} \lim_{\epsilon \to 0^+}$ to $\lim_{\epsilon \to 0^+} \lim_{R \to \infty}$, not about changing $\int_0^R \int_{\mathbb R}$ to $\int_{\mathbb R} \int_0^R$. The latter is straightforward because $\iint_{\mathbb R \times [0, R]} |\phi(k) e^{-i k x}| \, d(k, x)$ is finite. For the former, since we know that the single limits exist, it would suffice to show uniform convergence of one of the single limits. $\endgroup$
    – Maxim
    Feb 1, 2021 at 18:14
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Let $\phi(x)$ be a Schwartz function and $\displaystyle \psi(x)=\text{PV}\left(\frac1{ix}\right)$.

Using the distributional relationships $\langle d',\phi\rangle =-\langle d,\phi'\rangle $ for any tempered distribution $d$ and $\text{sgn}'(x)=2\delta(x)$, we have

$$\begin{align} \langle \mathscr{F}\{\text{sgn}\},\phi \rangle &=\langle \text{sgn},\mathscr{F}\{\phi\}\rangle\\\\ &=\langle \text{sgn},\left(\mathscr{F}\{\phi \psi\}\right)'\rangle\\\\ & =-\langle(\text{sgn})', \mathscr{F}\{\phi \psi\}\rangle\\\\ & =-2\langle \delta, \mathscr{F}\{\phi\psi\}\rangle\\\\ &=2i\text{PV}\left(\int_{-\infty}^\infty \frac{\phi(x)}{x}\,dx\right) \end{align}$$

Hence, we can write the distributional relationship

$$\bbox[5px,border:2px solid #C0A000]{\mathscr{F}\{\text{sgn}\}(x)=\text{PV}\left(\frac{2i}{x}\right)}$$



ADDENDUM TO ORIGNAL POST:

Here we present an approach very similar to that of @JoeMack.

For any $\delta>0$ we have

$$\begin{align} \langle \mathscr{F}\{\text{sgn}\},\phi\rangle &=\langle \text{sgn}, \mathscr{F}\{\phi\}\rangle\\\\ &=\int_0^\infty \int_{-\infty}^\infty \phi(k)e^{ikx}\,dk\,dx-\int_{-\infty}^0 \int_{-\infty}^\infty \phi(k)e^{ikx}\,dk\,dx\\\\ &=2i\int_0^\infty \int_{-\infty}^\infty \phi(k)\sin(kx)\,dk\,dx\\\\ &=\lim_{L\to \infty}\int_{-\infty}^\infty \phi(k)\frac{1-\cos(kL)}{k}\,dk\\\\ &=\lim_{L\to \infty}2i\left(\int_{|k|\ge \delta}\phi(k)\frac{1-\cos(kL)}{k}\,dk+\int_{|k|\le \delta}\phi(k)\frac{1-\cos(kL)}{k}\,dk\right)\\\\ &=2i\int_{|k|\ge \delta}\frac{\phi(k)}{k}\,dk+O(\delta)\tag1 \end{align}$$

Letting $\delta\to 0$ on the right-hand side of $(1)$, we find that

$$\langle \mathscr{F}\{\text{sgn}\},\phi\rangle=\lim_{\delta\to 0}\int_{|k|\ge \delta}\frac{2i\phi(k)}{k}\,dk\tag2$$

The right-hand side of $(2)$ is the Cauchy Principal Value of $\int_{-\infty}^\infty \frac{2i\phi(k)}{k}\,dk$. Therefore, we find that in distribution

$$\mathscr{F}\{\text{sgn}\}(x)=\text{PV}\left(\frac{2i}x\right)$$

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  • $\begingroup$ That is pretty slick; I like it. $\endgroup$
    – Joe Mack
    Jun 19, 2020 at 20:53
  • $\begingroup$ Thank you Joe! Much appreciative. $\endgroup$
    – Mark Viola
    Jun 19, 2020 at 20:56
  • $\begingroup$ Strictly speaking, $\mathcal F[\phi \psi]$ is not a Schwartz function, $(f, \mathcal F[\phi \psi])$ is not well-defined. $\endgroup$
    – Maxim
    Jan 31, 2021 at 15:37
  • $\begingroup$ @Maxim How would you propose to salvage the development? $\endgroup$
    – Mark Viola
    Jan 31, 2021 at 16:08
  • $\begingroup$ I don't think we can get $\mathcal F[\operatorname {sgn}]$ (or the solution of $x f = 1$) in a purely algebraic way. If we already know that $\mathcal F[f]$ is the limit of $\int_{|k| < R} f(k) e^{i x k} dk$ if $\int_{\mathbb R} |f(k)| (1 + |k|)^{-m} dk$ is finite for some $m \geq 0$, then we have $$F(x) = \int_{|k| < R} e^{i x k} \operatorname {sgn} k \, dk = \frac {2 i (1 - \cos R x)} x, \\ (F, \phi) = 2 i \int_{\mathbb R^+} \frac {\phi(x) - \phi(-x)} x dx - 2 i \int_{\mathbb R^+} \frac {\phi(x) - \phi(-x)} x \cos R x \, dx.$$ The first term is $2 i (\mathcal P(1/x), \phi)$. $\endgroup$
    – Maxim
    Feb 2, 2021 at 18:36

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