Fourier transform in S′

Question

In class we have next definition for Fourier transform in $\mathcal{S}'$: $$\langle \hat{f}, \varphi \rangle = \langle f, \hat{\varphi} \rangle = \frac{1}{\sqrt{2\pi}}\int\limits_{\mathbb{R}}f(y)\int\limits_{\mathbb{R}}\varphi(x)e^{ixy}dx dy$$

I tried to solve through the definition: $$\frac{1}{\sqrt{2\pi}}\int\limits_{\mathbb{R}_{-}}\cos(y)\int\limits_{\mathbb{R}}\varphi(x)e^{ixy}dxdy + \frac{1}{\sqrt{2\pi}}\int\limits_{\mathbb{R}_{+}}y^2\int\limits_{\mathbb{R}}\varphi(x)e^{ixy}dxdy.$$ But I didn't succeed. I also tried to solve using limits in second integral (and then using Fubini's theorem) or convolution, but failed again.

Can you please suggest ideas for a solution?

I don't know this very well but shouldn't you just calculate the ordinary Fourier transform of $f$? — FShrike, Jun 04 '23 at 13:24
$f(x)$ is not a "generalized" function. It's just a function. — jjagmath, Jun 04 '23 at 13:31
@FShrike I think you are right if f is an absolutely integrable function — lilit, Jun 04 '23 at 13:31
@jjagmath we obtain a generalized function $F$ from $f$ by a formula $F(\varphi) = \int\limits_{\mathbb{R}}f(x)\varphi(x)dx$ — lilit, Jun 04 '23 at 13:35
The formula is often difficult to use directly. Instead some laws and special transforms are derived using it, and these are then used. — md2perpe, Jun 04 '23 at 14:39
Is $f$ the function that is $\cos(y)$ for $y<0$ and $y^2$ for $y>0$? — Mark Viola, Jun 04 '23 at 15:11
The question appears to have been edited into such a state that it is no longer clear what is being asked. Please clarify the body of the question and consider a more descriptive title. — Alex Ortiz, Jun 04 '23 at 23:44

Mark Viola · Accepted Answer · 2023-06-04T23:33:44.253

Let $f$ be the function defined as

$$f(x)=\begin{cases}\cos(x)&, x<0\\\\x^2&,x>0 \end{cases}$$

Then, for $\phi\in \mathbb{S}$, we have

$$\begin{align} \langle \mathscr{F}\{f\},\phi\rangle&=\langle f,\mathscr{F}\{\phi\}\rangle\\\\ &=\int_{-\infty}^0 \cos(x)\int_{-\infty}^\infty \phi(k)e^{ikx}\,dk\,dx+\int_0^{\infty} x^2\int_{-\infty}^\infty \phi(k)e^{ikx}\,dk\,dx \\\\ &=-\lim_{L\to\infty}\int_0^L \cos(x)\int_{-\infty}^\infty \phi(k)e^{-ikx}\,dk\,dx\\\\ &+\lim_{M\to \infty}\int_0^M x^2\int_{-\infty}^\infty \phi(k)e^{ikx}\,dk\,dx\\\\ &=-\lim_{L\to\infty}\int_0^L \cos(x)\int_{-\infty}^\infty \phi(k)e^{-ikx}\,dk\,dx\tag{1a}\\\\ &-\lim_{M\to \infty}\int_0^M \int_{-\infty}^\infty \phi''(k)e^{ikx}\,dk\,dx\tag{1b} \end{align}$$

where we used the fact that $x^2e^{ikx}=-\frac{d^2e^{ikx}}{dk^2}$ and then integrated twice by parts to arive at $(1b)$. Next, we will apply the Fubini-Tonelli Theorem.

Inasmuch as $\phi\in \mathbb{S}$, the Fubini-Tonelli theorem guarantees that we can interchange the order of integration in $(1a)$ and $(1b)$. Proceeding we find that

$$\begin{align} \int_0^L \cos(x)\int_{-\infty}^\infty \phi(k)e^{-ikx}\,dk\,dx&=\frac i2\int_{-\infty}^\infty \phi(k) \left(\frac{e^{-i(k+1)L}-1}{k+1}+\frac{e^{-i(k-1)L}-1}{k-1}\right)\,dk\\\\ &=\frac12 \int_{-\infty}^\infty \phi(k) \left(\frac{\sin((k+1)L)}{k+1}+\frac{\sin((k-1)L)}{k+1}\right)\,dk\tag{2a}\\\\ &+\frac i2 \int_{-\infty}^\infty \phi(k)\left(\frac{\cos((k+1)L)-1}{k+1}+\frac{\cos((k-1)L)-1}{k-1}\right)\,dk \tag{2b} \end{align}$$

and

$$\begin{align} \int_0^M \int_{-\infty}^\infty \phi''(k)e^{ikx}\,dk\,dx&= \int_{-\infty}^\infty \phi''(k)\frac{e^{ikM}-1}{ik}\,dk\\\\ &=\int_{-\infty}^\infty \phi''(k)\frac{\sin(kM)}{k}\,dk\tag{3a}\\\\ &-i\int_{-\infty}^\infty \phi''(k)\frac{\cos(kM)-1}{k}\,dk\tag{3b} \end{align}$$

Using the result in THIS ANSWER or THIS ONE, we let $L\to \infty$ in $(2a)$ and $M\to\infty$ in $(3a)$ to reveal

$$\lim_{L\to\infty}\frac12 \int_{-\infty}^\infty \phi(k) \left(\frac{\sin((k+1)L)}{k+1}+\frac{\sin((k-1)L)}{k+1}\right)\,dk=\frac\pi2 \left(\phi(-1)+\phi(1)\right)$$

and

$$\lim_{M\to\infty}\int_{-\infty}^\infty \phi''(k)\frac{\sin(kM)}{k}\,dk=\pi\phi''(0)$$

$$$$

Finally, to evaluate the remaining limits we use the following analysis. Note that

$$\begin{align} \lim_{M\to\infty}\int_{-\infty}^\infty \phi(k)\frac{\cos(kM)-1}{k}\,dk&=\lim_{M\to\infty}\text{PV}\int_{-\infty}^\infty \phi(k)\frac{\cos(kM)}{k}\,dk\\\\ &-\text{PV}\int_{-\infty}^\infty \frac{\phi(k)}{k}\,dk\\\\ &=\lim_{M\to\infty}\text{PV}\int_{|k|<1} \phi(k)\frac{\cos(kM)}{k}\,dk\\\\ &+\lim_{M\to\infty}\text{PV}\int_{|k|>1} \phi(k)\frac{\cos(kM)}{k}\,dk\\\\ &-\text{PV}\int_{-\infty}^\infty \frac{\phi(k)}{k}\,dk\\\\ &=\underbrace{\lim_{M\to\infty}\int_{|k|<1} (\phi(k)-\phi(0))\frac{\cos(kM)}{k}\,dk}_{=0\,\,\text{using the Riemann-Lebesgue Lemma}}\\\\ &+\phi(0)\lim_{M\to\infty}\underbrace{\text{PV}\int_{|k|<1} \frac{\cos(kM)}{k}\,dk}_{=0\,\,\text{since the integrand is odd}}\\\\ &+\underbrace{\lim_{M\to\infty}\text{PV}\int_{|k|>1} \phi(k)\frac{\cos(kM)}{k}\,dk}_{=0\,\,\text{using the Riemann-Lebesgue Lemma}}\\\\ &-\text{PV}\int_{-\infty}^\infty \frac{\phi(k)}{k}\,dk\\\\ &=-\text{PV}\int_{-\infty}^\infty \frac{\phi(k)}{k}\,dk \end{align}$$

Putting it all together, we find that in distribution

$$\mathscr{F}\{f\}=i\text{PV}\left(\frac{2k}{k^2-1}\right)+\frac{\pi}2 \left(\delta(k+1)+\delta(k-1)\right)-\pi \delta''(k)- i \left(\frac2{k^3}\right)_D$$

where $\left(\frac{2}{k^3}\right)_D$ is the distributional derivative of $\text{PV}\left(\frac1k\right)$

Fourier transform in S′

1 Answers1