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Let $f\in C^\infty_0(\mathbb{R})$ with $\operatorname{supp}f\subset(0,\infty)$. I would like to prove that

$$\mathcal{F}^{-1}\left\{\frac{i}{2t}\hat{f}(t)\right\}=-\int_{-\infty}^tf(s)\,ds,\qquad t\in\mathbb{R}\qquad(\star)$$

where $c>0$ is a constant, using rigorous distribution theory

"Unrigorous" Proof:

The approach I considered is to consider the fact that the Fourier transform of the Heaviside function is given by

$$\hat{H}(\omega)=\frac{1}{2}\delta(\omega)-\frac{i}{2\pi t}.$$

Hence, we may write

$$ \begin{aligned} \mathcal{F}^{-1}\left\{\frac{i}{2t}\hat{f}(t)\right\}&=\pi\mathcal{F}^{-1}\left\{\frac{i}{2\pi t}\hat{f}(t)\right\} \\ &=-\pi\mathcal{F}^{-1}\{\hat{H}(\omega)\hat{f}(t)\}\qquad(\omega\ne 0) \end{aligned}$$

where we have used that $0\notin\operatorname{supp}f$ and $\operatorname{supp}\delta=\{0\}$.

Then $(\star)$ follows via an application of the convolution theorem.

Now, my question is:

Can we rigorously prove $(\star)$ in the sense of distributions

Jason Born
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  • You mean $\mathcal{F}^{-1}[P.V.(\frac{\hat{f}(t)}{t})+ c \delta(t)]$. How do you prove rigorously that $\hat{H}(x) = P.V.(\frac{1}{2 i \pi x}) + \frac{\delta}{2}$ ? The proof of $\widehat{H \ast f}$ won't be very different. If you like the distributions, then show that the kernel of the operator : $T \mapsto x T$ defined by $$\langle x T,\varphi \rangle = \langle T,x\varphi \rangle$$ is the subspace generated by $\delta$ – reuns May 26 '17 at 22:05
  • @user1952009 Where did you derive $\mathcal{F}^{-1}\left[\operatorname{P.V.}\left(\frac{\hat{f}(t)}{t}\right)+c\delta(t)\right]$ from? – Jason Born May 27 '17 at 08:43
  • By the convolution theorem for example : what is $\hat{f} \hat{H}$ ? – reuns May 27 '17 at 20:58
  • Also your question and your level is unclear. Do you know how to show that (in the sense of distributions) $x T = 0 \implies T = c \delta$ ? (where $xT = 0$ means $$\forall \varphi \in C^\infty_c, \qquad 0=\langle xT,\varphi \rangle = \langle T,x\varphi \rangle $$ – reuns May 27 '17 at 21:00
  • @user1952009 Well, $\hat{f}\hat{H}=\widehat{f\ast H}$. And if $\langle T,x\varphi\rangle=0$ for all $\varphi\in C^\infty_c$, then the proof that $T=c\delta$ follows from the fact that $x\mapsto x$ is $C^\infty$ and we must have $\operatorname{supp} T\subset{x:x=0}\implies\operatorname{supp}T={0}\implies T=c\delta$. – Jason Born May 27 '17 at 21:19
  • @user1952009 My question, in the plainest language, is: deducing $(\star)$ in the sense of distributions. – Jason Born May 27 '17 at 21:20
  • Take a sequence $\phi_n \in C^\infty_c$ such that $\phi_n \to T$ in the sense of tempered distributions. Then $\varphi \ast \phi_n \to T \ast \varphi$. Since the Fourier transform is a continuous linear operator (in the sense of tempered distributions) we obtain $$\mathcal{F}[\varphi \ast T] = \mathcal{F}[ \varphi \ast (\lim_{n \to \infty} \phi_n)] = \lim_{n \to \infty}\mathcal{F}[ \varphi \ast \phi_n)]= \lim_{n \to \infty}\hat{\varphi} \hat{\phi_n} = \hat{\varphi} \hat{T}$$ Conclude by taking $T = H$ – reuns May 27 '17 at 21:34
  • @user1952009 Perhaps I didn't make my question clear. I would like to prove the identity $(\star)$. You have merely proven the convolution theorem. – Jason Born May 28 '17 at 18:17
  • @JasonBorn: You write "where we have used that $0 \notin \operatorname{supp} f$ and $\operatorname{supp}\delta = {0}$." How does this make $\hat f \delta$ disappear? $\operatorname{supp}\hat f$ doesn't exclude $0$. – md2perpe May 30 '17 at 21:19

2 Answers2

1

Using $\hat{H}(\xi ) = pv.(\frac{1}{2 i\pi \xi})+\frac{1}{2} \delta(\xi)$ and the convolution theorem $$\forall \phi,f\in S(\mathbb{R}), \qquad \langle pv.(\frac{\hat{f}(\xi)}{2 i\pi \xi})+\frac{\hat{f}(0)}{2} \delta(\xi), \hat{\phi}\rangle = \langle \hat{H}, \hat{f}\hat{\phi}\rangle = \langle H, f \ast \phi\rangle =\langle H \ast f, \phi\rangle $$ Qed. if $F(x) =H \ast f(x)= \int_{-\infty}^x f(y)dy$ then $$ \hat{F}(\xi) = pv.(\frac{\hat{f}(\xi)}{2 i\pi \xi})+\frac{\hat{f}(0)}{2} \delta(\xi)$$

reuns
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  • It is not clear what convention the OP is using. I used the $\xi$ convention $\int_{-\infty}^\infty f(x) e^{-2i \pi \xi x}dx$ – reuns Jun 09 '17 at 13:57
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As definition of the Fourier transform we take $$\hat\phi(\xi) = \int \phi(x) e^{-i\xi x} \, dx$$

Then we have $\hat\delta = 1$ since $\langle \hat\delta(t), \phi(t) \rangle = \langle \delta(t), \hat\phi(t) \rangle = \hat\phi(0) = \int \phi(x) 1 \, dx = \langle 1(t), \phi(t) \rangle.$

Now, $H = \frac12 (1 + \theta),$ where $\theta(x) = -1$ when $x<0$, and $\theta(x) = +1$ when $x>0$.

For the constant part we have $\hat 1(t) = \hat{\hat\delta}(t) = 2\pi \, \delta(-t) = 2\pi \, \delta(t).$

For $\theta$ we have $\theta' = 2\delta$ so $it \hat\theta(t) = \widehat{\theta'}(t) = \hat\delta(t) = 1(t).$ Thus $\hat\theta(t) = -i \, \operatorname{pv}\cfrac{1}{t}$.

Summarizing we get $$\hat H(t) = \frac12 (\hat 1 + \hat\theta) = \frac12 \left(2\pi\,\delta(t) - i\,\operatorname{pv}\cfrac{1}{t}\right)$$

If we instead define the Fourier transform as $$\hat\phi(\xi) = \frac{1}{2\pi} \int \phi(x) e^{-i\xi x} \, dx$$ we get $$\hat H(t) = \frac12 \left(\delta(t) - i \frac{1}{2\pi} \,\operatorname{pv}\cfrac{1}{t}\right)$$

There's a small difference from the formula that Jason Born had. Have I made a mistake or is it an error in Jason's formula?

md2perpe
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  • there are many errors in Jason's formulas. It is not clear how you know $it \hat{\theta} = 1 \implies \hat{\theta} = P.V.(\frac{1}{it})$. You should show first that $\forall \varphi \in C^\infty_c, \langle t S,\varphi \rangle=\langle S,t\varphi \rangle = 0 \implies S = c \delta , c \in \mathbb{C}$ – reuns May 28 '17 at 02:50
  • $\operatorname{pv}\frac{1}{x}$ is by definition the odd solution to $xu = 1$ (https://en.wikipedia.org/wiki/Cauchy_principal_value#More_general_definitions). My calculations are on a level above such fundamental proofs as showing $tu(t) = 0 \implies u(t) = c\delta$. – md2perpe May 28 '17 at 07:06
  • Well not really, and this is the core of the proof. $p.v. (\frac{1}{x})$ is by definition the distribution $$\forall \varphi \in \mathbb{C}^\infty_c, \qquad\langle p.v. (\frac{1}{x}), \varphi \rangle = \langle 1, \frac{\varphi-\varphi(0)\phi}{x} \rangle $$ where $\phi \in \in \mathbb{C}^\infty_c$ is even and $\phi(0) = 1$. But that doesn't tell us that $i x \hat{\theta} = 1 \implies\hat{\theta} = p.v.(\frac{1}{ix})$ if $\hat{\theta}$ is odd – reuns May 28 '17 at 07:12
  • It seems more common to define $\langle \operatorname{p.v.}\frac1x, \phi(x) \rangle = \lim_{\epsilon\to 0} \int_{|x|>\epsilon} \frac1x \phi(x),dx$. See for example https://en.wikipedia.org/wiki/Cauchy_principal_value, https://math.stackexchange.com/questions/162931/principal-value-as-distribution-written-as-integral-over-singularity, https://math.stackexchange.com/questions/245211/principal-value-of-1-x-equivalence-of-two-definitions, http://www.ueltschi.org/teaching/2012-MA433/distributions.pdf – md2perpe May 28 '17 at 17:11
  • Yes it is the same, but I wanted to say $x T = 1$ means that $T=pv.(1/x)+S$ where $xS = 0$, which reduces to $S = c\delta$ 3 lines latter – reuns May 29 '17 at 13:33
  • Okey, I agree, but since $\theta$ is odd, so is $\hat\theta$ which makes the $c\delta$ part disappear. Or I could have concluded just from $it \hat\theta(t) = 1$ that $\hat\theta(t) = \operatorname{pv}\frac1t + c\delta(t)$. But then I would have the problem to determine $c$. – md2perpe May 30 '17 at 09:32
  • @md2perpe $c=\pi$, but you're not answering my question. I do not ask to determine the Fourier transform of the Heaviside function. – Jason Born May 30 '17 at 14:57
  • Sorry, @JasonBorn, I misunderstood your question. I realized that a few days ago. – md2perpe May 30 '17 at 16:30
  • @JasonBorn He answered your question, because the proof works exactly the same with $f , \hat{f}$ instead of $\delta,1$. – reuns May 30 '17 at 16:54
  • @user1952009 Which part? The part where we show $\mathcal{F}f=\hat{f}$, or when the constant part of $H$, i.e. $1$, is the double Fourier transform of the delta distribution (which is expressed as a function here, contrary to what I asked for in my question)? – Jason Born May 31 '17 at 15:32
  • @JasonBorn The part where $F(t) = \int_{-\infty}^t f(s)ds$ is the inverse Fourier transform of $$\displaystyle P.V.(\frac{\hat{f}(\omega)}{2i \pi \omega} )+ \frac{F(\infty)}{2} \delta(\omega)$$ – reuns May 31 '17 at 20:28
  • @user1952009 Thanks, but that doesn't address the fundamental point of this question: How do I go from

    $$\frac{ic}{2\omega}\hat{f}(\omega)\leadsto\operatorname{P.V.}\left(\frac{\hat{f}(\omega)}{2i\pi\omega}\right)$$

     – Jason Born Jun 04 '17 at 09:24
  • @user1952009 Oh, sorry. I think I understand. I have to use the theorem that $\exists c\in\mathbb{C}$ such that $\hat{f}=c\delta+\frac{1}{2\pi i}\operatorname{P.V.}\left(\frac{\hat{f}}{\omega}\right)$ – Jason Born Jun 04 '17 at 09:34