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Problem: Find the Fourier transform of the tempered distributions $F_1(f) = \int_0^{\infty} f(x)xdx$, and $F_2(f) = (\frac{d^n}{dx^n} \delta)(f)$, where $f$ is assumed to be a Schwartz function.

Attempt:

$$(1)\,\begin{align*}\mathcal{F}F_1(f) &= \int_0^{\infty} \widehat f(x)x\,dx\\ &= \int_0^{\infty} \int_{-\infty}^{\infty}f(y)e^{-2 \pi i yx} x\,dy\,dx\\ &= \int_0^{\infty} 0 + \frac{1}{2\pi i}\int_{-\infty}^{\infty} f'(y)e^{-2 \pi i xy}\,dy\,dx \end{align*}$$

where the last step follows from integration by parts and the fact the $f$ is a Schwartz function. Now, I am not sure how to proceed from here to reach a significant new tempered distribution. Moreover, the integrand doesn't seem to belong to $L^1(\mathbb{R} \times \mathbb{R_{\gt 0}})$, so I can't even use Fubini... $$(2)\begin{align*} \mathcal{F}F_2(f) &= \int_{-\infty}^{\infty} \widehat{(f^{(n)})}(x)\,dx\\ &= \widehat {(f^{(n)})}(0)\\ &= \int_{-\infty}^{\infty} f(y) \frac{e^{-2 \pi i xy}}{-2 \pi i y}\,dy \end{align*}$$

At this point, this seems like a complex analysis problem... I am not sure how to proceed with the integral though as my background in complex analysis is weak.

Any help is appreciated!

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    $\begingroup$ For 2) show that $\hat\delta \equiv 1$ and that the Fourier transform of derivatives tempered distributions obey the same law as for Schwartz functions. $\endgroup$
    – md2perpe
    Mar 7, 2018 at 21:37
  • $\begingroup$ For 1) ... Do you know the Fourier transform of the Heaviside function $$H(x) = \begin{cases}1 & (x>0) \\ 0 & (x<0)\end{cases}$$? $\endgroup$
    – md2perpe
    Mar 7, 2018 at 22:34
  • $\begingroup$ No we did not cover that $\endgroup$
    – Longti
    Mar 7, 2018 at 23:16
  • $\begingroup$ @md2perpe I have looked up Fourier transform of Heaviside function. But I have a question first. Can we even apply Fubini to the integral? Because I cannot see how the integrand $|f'(y)e^{2\pi i yx}| = |f'(y)|$ can ever lie in $L^1(\mathbb{R}\times \mathbb{R_{\gt 0}})$... $\endgroup$
    – Longti
    Mar 8, 2018 at 16:34
  • $\begingroup$ No, you can't. You will have to use other ways. For example, we know that $H' = \delta$. Therefore, taking the Fourier transform, we have $2\pi i y \hat H = 1$ so $$\hat H = \frac{1}{2\pi i} \mathrm{pv}\frac{1}{y} + C \, \delta(y)$$ for some constant $C$. $\endgroup$
    – md2perpe
    Mar 8, 2018 at 20:31

1 Answer 1

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The definition of Fourier transform that we are using: $$\hat{f}(y) = \int f(x) \, e^{-i \, 2\pi \, xy} \, dx$$

The inverse is $$f(x) = \int \hat{f}(y) \, e^{i \, 2\pi \, xy} \, dy$$ and $$\hat{\hat{f}}(t) = f(-t).$$


First we note that $\hat{\delta}(y) = 1,$ that $\widehat{f'} = i \, 2\pi \, y \, \hat{f}(y)$ and that $\widehat{xf}(y) = \frac{i}{2\pi} \hat{f}'(y).$ Also note that the property of being an even (odd) function is preserved under the Fourier transform.

We can use this directly for $F_2$: $$\widehat{F_2}(y) = \widehat{\delta^{(n)}}(y) = (i \, 2\pi \, y)^n \hat{\delta}(y) = (i \, 2\pi \, y)^n.$$

For $F_1$ we note that $F_1(x) = x H(x),$ where $H$ is the Heaviside step function, having derivative $H' = \delta.$ Therefore, $\widehat{F_1}(y) = \frac{i}{2\pi} \hat{H}(y).$ But what is $\hat{H}$?

We have $H' = \delta$ so $i \, 2\pi \, y \, \hat{H}(y) = 1,$ i.e. $\hat{H} = \frac{1}{i \, 2\pi \, y} + C \, \delta(y)$ for some constant $C.$ But we also have $H(x) = \frac12 \theta(x) + \frac12,$ where $\theta$ is the sign function. Thus we have $$\frac{1}{i \, 2\pi \, y} + C \, \delta(y) = \frac12 \hat{\theta}(y) + \frac12 \hat{1}(y).$$

Now it's important to note that the first term on both sides is an odd function while the second term is an even function. We must therefore have $$ \frac{1}{i \, 2\pi \, y} = \frac12 \hat{\theta}(y) \quad \text{and} \quad C \, \delta(y) = \frac12 \hat{1}(y). $$

But $\hat{1}(y) = \hat{\hat{\delta}}(y) = \delta(-y) = \delta(y).$ Therefore we must have $C = \frac12$ and thus $$\hat{H} = \frac{1}{i \, 2\pi \, y} + \frac12 \, \delta(y)$$ and therefore $$\widehat{x H}(y) = \frac{i}{2\pi} \frac{d}{dy} \left( \frac{1}{i \, 2\pi \, y} + \frac12 \, \delta(y) \right) = \frac{i}{2\pi} \left( -\frac{1}{i \, 2\pi \, y^2} + \frac12 \, \delta'(y) \right) \\ = -\frac{1}{(2\pi)^2 y^2} + \frac{i}{4\pi} \delta'(y) $$

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  • $\begingroup$ Thanks a lot, I know where I had gone wrong now. $\endgroup$
    – Longti
    Mar 8, 2018 at 23:59

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