The Fourier Transform of the Principal Value Distribution And Some Others
In this article, we will essentially solve exercise 19 of chapter 9 of Folland’s “Real Analysis: Modern Techniques and Their Applications.” That is, we wish to solve:
Folland’s Challenge: On $\mathbb R,$ let $F_0=\text{PV}(1/x).$ Also, for $\epsilon>0,$ let $F_\epsilon=x(x^2+\epsilon^2)^{-1},$ $G_\epsilon=(x\pm i\epsilon)^{-1},$ and $S_\epsilon=e^{2\pi\epsilon|x|}\text{sgn}(x).$
$\lim_{\epsilon\to 0}F_\epsilon=F_0$ in the $\text{weak}^*$ topology of $\mathscr{S}’$
$\lim_{\epsilon\to 0}G_\epsilon=F_0\mp \pi i\delta$
$\hat{S}_\epsilon=(\pi i)^{-1}F_\epsilon$ and hence $\widehat{\text{sgn}}=(\pi i)^{-1}F_0$
From (3) it follows that $\hat{F}_0=-\pi i\text{sgn}.$ Prove this directly by showing that $F_0=\lim_{\epsilon\to 0,N\to\infty}H_{\epsilon,N},$ where $H_{\epsilon, N}(x)=x^{-1}$ if $\epsilon<|x|<N$ and $H_{\epsilon,N}(x)=0$ otherwise.
Compute $\hat{\chi}_{(0,\infty)}$ in two ways: (i) by writing $\chi_{(0,\infty)}=\frac{1}{2}\text{sgn}+\frac{1}{2}$ and using (3), (ii) by writing $\chi_{(0,\infty)}(x)=\lim_{\epsilon\to 0}e^{-\epsilon x}\chi_{(0,\infty)}$ and using (2)
I particularly liked this exercise, because once I could solve this (after quite some difficulty), I could solve pretty much every other problem in that section of chapter 9. Since the proof uses many common tools useful for distribution theory (by virtue of simply being long as hell), it is very instructive in solidifying ones understanding of the chapter. Of course, I would advise the reader to attempt the problem by oneself first!
The Solution
Proof of Part I
Fix some $a>0.$ Then, for all $\epsilon\neq 0,$ $\int_{|x|\le a}\frac{x}{x^2+\epsilon^2}dx=0$ since the integrand is odd (integrability follows because the function is bounded and the domain of integration $[-a,a]$ has finite measure).
So, for any $\phi\in\mathscr{S},$ we have $$\int_{|x|\le a}\frac{x\phi(x)}{x^2+\epsilon^2}=\int_{|x|\le a}\frac{x}{x^2+\epsilon^2}\big[\phi(x)-\phi(0)\big]dx\text{ ($\star_1$)}$$
Since $\phi\in\mathscr{S},$ we know that $P\cdot D^\alpha\phi$ is bounded for every polynomial $P$ and multi-index $\alpha$ (by the discussion in $\S$7.3, Grandpa Rudin). In particular, there exists some $A\ge 0$ such that $\sup_{x\in\mathbb{R}}|\phi'(x)|\le A$ and some $B\ge 0$ such that $\sup_{x\in\mathbb{R}}|x\phi(x)|\le B.$
Note that by the mean value theorem, $$\left|\frac{x}{x^2+\epsilon^2}\big[\phi(x)-\phi(0)\big]\right|\le \frac{|x|}{x^2+\epsilon^2}\Big(|x|\sup_{y\in\mathbb{R}}|\phi'(y)|\Big)\le \frac{Ax^2}{x^2+\epsilon^2}\le A\text{ (since $x^2+\epsilon^2\ge x^2$)}$$
Furthermore, we have $\int_{|x|\le a}Adx=2Aa<\infty.$ For a fixed $x\neq 0,$ $\lim_{\epsilon\to 0}\frac{x}{x^2+\epsilon^2}\big[\phi(x)-\phi(0)\big]dx=\frac{\phi(x)-\phi(0)}{x}$ and hence, we can apply the dominated convergence theorem to conclude that $$\lim_{\epsilon\to 0}\int_{|x|\le a}\frac{x\big[\phi(x)-\phi(0)\big]}{x^2+\epsilon^2}dx=\int_{x\le a}\frac{\phi(x)-\phi(0)}{x}dx\text{ ($\star_2$)}$$
On the other hand, we have $$\left|\frac{x}{x^2+\epsilon^2}\phi(x)\right|\le \left|\frac{x}{x^2}\phi(x)\right|=\left|\frac{\phi(x)}{x}\right|\text{ (since $x^2+\epsilon^2\ge x^2$)}$$
with the last function being integrable in $\{x\in\mathbb{R}:|x|>a\}$ since $$\int_{|x|>a}\left|\frac{\phi(x)}{x}\right|dx\le \int_{|x|>a}\frac{B}{x^2}dx=2B\int_{a}^\infty\frac{1}{x^2}dx=\frac{2B}{a}<\infty$$
For a fixed $x\neq 0,$ $\lim_{\epsilon\to 0}\frac{x}{x^2+\epsilon^2}\phi(x)=\frac{\phi(x)}{x}$ so, again, using the dominated convergence theorem, $$\lim_{\epsilon\to 0}\int_{|x|>a}\frac{x}{x^2+\epsilon^2}\phi(x)dx=\int_{|x|>a}\frac{\phi(x)}{x}dx\text{ ($\star_3$)}$$
Combining $(\star_1),(\star_2),$ and $(\star_3)$ we get $$\lim_{\epsilon\to 0}\langle F_\epsilon,\phi\rangle=\lim_{\epsilon\to 0}\int \frac{x}{x^2+\epsilon^2}\phi(x)dx=\lim_{\epsilon\to 0}\int_{|x|\le a} \frac{x}{x^2+\epsilon^2}\phi(x)dx+\lim_{\epsilon\to 0}\int_{|x|>a} \frac{x}{x^2+\epsilon^2}\phi(x)dx$$
$$=\int_{|x|\le a}\frac{\phi(x)-\phi(0)}{x}dx+\int_{|x|>a}\frac{\phi(x)}{x}dx$$
Now, note that for any $t$ such that $0<t<a,$ we have $$\int_{|x|>t}\frac{\phi(x)}{x}dx=\int_{t<|x|\le a}\frac{\phi(x)}{x}dx+\int_{|x|>a}\frac{\phi(x)}{x}dx$$
Since $\int_{t<|x|\le a}\frac{1}{x}dx=0,$ we have $$\int_{t<|x|\le a}\frac{\phi(x)}{x}dx=\int_{t<|x|\le a}\frac{\phi(x)-\phi(0)}{x}dx$$
We have already shown that $x\mapsto \chi_{[-a,a]}(x)\frac{\phi(x)-\phi(0)}{x}$ is $L^1$ (a consequence of the dominated convergence theorem in $(\star_2)$). So again we can apply the dominated convergence theorem to get $$\lim_{t\to 0^+}\int_{t<|x|\le a}\frac{\phi(x)-\phi(0)}{x}dx=\int_{|x|\le a}\frac{\phi(x)-\phi(0)}{x}dx$$
Combing this with our previous result, $$\langle\text{PV(1/x)},\phi\rangle=\lim_{t\to 0^+}\int_{|x|>t}\frac{\phi(x)}{x}dx=\int_{|x|\le a}\frac{\phi(x)-\phi(0)}{x}dx+\int_{|x|>a}\frac{\phi(x)}{x}dx=\lim_{\epsilon\to 0}\langle F_\epsilon,\phi\rangle$$
Since $\phi\in\mathscr{S}$ is arbitrary, we conclude that $F_\epsilon\longrightarrow F_0$ in $\mathscr{S}'$ as $\epsilon\longrightarrow 0.$
Proof of Part II
Consider any $\epsilon>0.$ For any $\phi\in \mathscr{S},$ we have $A=\sup_{x\in \mathbb{R}}|\phi(x)|<\infty.$ Now, $x\mapsto \frac{\epsilon\phi(x)}{x^2+\epsilon^2}$ is integrable, as by the monotone convergence theorem, we have $$\int \left|\frac{\epsilon\phi(x)}{x^2+\epsilon^2}\right|dx=\lim_{M\to\infty}\int_{-M}^M \left|\frac{\epsilon\phi(x)}{x^2+\epsilon^2}\right|dx\le A\lim_{M\to\infty}\int_{-M}^M \left|\frac{\epsilon}{x^2+\epsilon^2}\right|dx$$$$=A\lim_{M\to\infty}\left[\tan^{-1}\left(M/\epsilon\right)-\tan^{-1}\left(-M/\epsilon\right)\right]=A\pi<\infty$$
So, using the dominated convergence theorem and integration by parts (since the integrands are atleast $C^1$), $$\int \frac{\epsilon\phi(x)}{x^2+\epsilon^2}dx=\lim_{M\to\infty}\int_{-M}^M\frac{\epsilon\phi(x)}{x^2+\epsilon^2}dx=\lim_{M\to\infty}\left[\phi(x)\tan^{-1}(x/\epsilon)\Big|_{-M}^M-\int_{-M}^M\phi'(x)\tan^{-1}(x/\epsilon)dx\right]$$
Now, $\lim_{M\to\infty}\phi(\pm M)=0$ since $\mathscr{S}\subset C_0$ (by Reimann Lebesgue lemma and the fact that Fourier transform is a bijection on $\mathscr{S}$) and $\lim_{M\to\infty}\tan^{-1}(\pm M/\epsilon)=\pm \pi/2.$ Thus, $\lim_{M\to\infty}\phi(\pm M)\tan^{-1}(\pm M/\epsilon)=0.$
Furthermore, $\phi'\in L^1(\mathbb{R})$ since $x\mapsto (1+|x|)^2\phi'(x)$ is bounded as $\phi\in\mathscr{S}$ and $x\mapsto \frac{1}{(1+|x|)^2}\in L^1(\mathbb{R}).$ Since $x\mapsto |\tan^{-1}(x)|$ is uniformly bounded by $\pi/2,$ $x\mapsto \phi(x)\tan^{-1}(x/\epsilon)$ is also an $L^1$ function and we can apply the dominated convergence theorem to conclude $$\lim_{M\to\infty}\int_{-M}^M\phi'(x)\tan^{-1}(x/\epsilon)dx=\int \phi'(x)\tan^{-1}(x/\epsilon)dx$$
Hence, we have $$\int \frac{\epsilon\phi(x)}{x^2+\epsilon^2}dx=-\int \phi'(x)\tan^{-1}(x/\epsilon)dx$$
Again, using the dominated convergence theorem (since $|\phi'(x)\tan^{-1}(x/\epsilon)|\le \pi|\phi'(x)|/2$ and $\phi'\in L^1$), $$\lim_{\epsilon\to 0^+}\int \frac{\epsilon\phi(x)}{x^2+\epsilon^2}dx=-\frac{\pi}{2}\int \phi'(x)\text{sgn}(x)dx=\pi\phi(0)$$
To get the last equality, we again applied dominated convergence theorem along with the fundamental theorem of calculus: $$\int_0^\infty \phi'(x)dx=\lim_{M\to\infty}\int_0^M\phi'(x)dx=\lim_{M\to\infty}\big[\phi(M)-\phi(0)\big]=-\phi(0)$$ and similarly on $\int_{-\infty}^0-\phi'(x)dx.$ Hence, using part (a), $$\lim_{\epsilon\to 0^+}\langle G_\epsilon^{\pm},\phi\rangle=\lim_{\epsilon\to 0^+}\int \frac{\phi(x)}{x\pm i\epsilon}dx=\lim_{\epsilon\to 0^+}\int \frac{x\mp i\epsilon}{x^2 +\epsilon^2}\phi(x)dx=\lim_{\epsilon\to 0^+}\langle F_\epsilon,\phi\rangle\mp i\lim_{\epsilon\to 0^+}\int\frac{\epsilon\phi(x)}{x^2+\epsilon^2}dx$$$$=\langle F_0,\phi\rangle\mp i\pi\langle\delta,\phi\rangle=\langle F_0\mp \pi i\delta,\phi\rangle$$
Since $\phi\in\mathscr{S}$ is arbitrary, we have our result.
Proof of Part III
We know from calc-2 class, that the following indefinite integral evaluates as (using integration by parts twice or by direcly checking the derivative of the left hand side) $$\int e^{ax}\sin(bx)dx=\frac{e^{ax}}{a^2+b^2}[a\sin(bx)-b\cos(bx)]+c$$
Using this, we have for any $\xi\in\mathbb{R},$ $$\hat{S}_\epsilon(\xi)=\int e^{-2\pi \epsilon |x|}\text{sgn}(x)e^{-2\pi i \xi x}dx=\int_0^\infty e^{-2\pi \epsilon |x|}e^{-2\pi i \xi x}dx-\int_{-\infty}^0 e^{-2\pi \epsilon |x|}e^{-2\pi i \xi x}dx$$$$=\int^{\infty}_0 \Big[e^{-2\pi i \xi x}-e^{2\pi i \xi x}\Big]e^{-2\pi \epsilon x}dx=\int^{\infty}_0 \Big[-2i\sin(2\pi\xi x)\Big]e^{-2\pi \epsilon x}dx$$
Hence, using the fundamental theorem of calculus and dominated convergence theorem, $$\hat{S}_\epsilon(\xi)=-2i\frac{2\pi \xi}{(2\pi \epsilon)^2+(2\pi\xi)^2}=\frac{-i}{\pi}\frac{\xi}{\xi^2+\epsilon^2}=\frac{(i\pi)^{-1}}{\xi^2+\epsilon^2}=(i\pi)^{-1}F_\epsilon(\xi)$$
This follows from the fact that $\lim_{M\to\infty}e^{-2\pi \epsilon M}=0.$ Hence, $\hat{S}_\epsilon=(i\pi)^{-1}F_\epsilon.$ Now, for any $\phi\in\mathscr{S},$ $$\lim_{\epsilon\to 0^+}\langle S_\epsilon,\phi\rangle=\lim_{\epsilon\to 0^+}\int e^{-2\pi \epsilon|x|}\text{sgn}(x)\phi(x)dx=\int \text{sgn}(x)\phi(x)dx=\langle\text{sgn},\phi\rangle$$
using the dominated convergence theorem since $|e^{-2\pi \epsilon |x|}\text{sgn(x)}\phi(x)|\le |\phi(x)|$ for all $x\in\mathbb{R}$ and $\phi\in L^1(\mathbb{R})$ (as argued before using ideas from $\S$7.3, Grandpa Rudin). Since $\phi\in\mathscr{S}$ is arbitrary, $S_\epsilon\longrightarrow\text{sgn}$ in $\mathscr{S}'$ as $\epsilon\longrightarrow 0^+.$
So, using Theorem 7.15 (Grandpa Rudin) and part (a), $$\widehat{\text{sgn}}=\lim_{\epsilon\to 0^+}\hat{S}_\epsilon=\lim_{\epsilon\to 0^+}(i\pi)^{-1}F_\epsilon=(i\pi)^{-1}\lim_{\epsilon\to 0^+}F_\epsilon=(i\pi)^{-1}F_0\text{ (with convergence and equality in $\mathscr{S}'$)}$$
Proof of Part IV
We have already proved in part (a) that for any $a>0$ and $\phi\in\mathscr{S},$
$$\langle \text{PV}(1/x),\phi\rangle=\int_{|x|\le a}\frac{\phi(x)-\phi(0)}{x}dx+\int_{|x|>a}\frac{\phi(x)}xdx\text{ for all }\phi\in\mathscr{S}\text{ ($\star$)}$$
where both integrals are absolutely convergent. Now using dominated convergence theorem (and the fact that $\int_{t<|x|\le a}\frac{1}{x}dx=0$), we have $$\lim_{t\to 0^+}\int_{t<|x|\le a}\frac{\phi(x)}{x}dx=\lim_{t\to 0^+}\int_{t<|x|\le a}\frac{\phi(x)-\phi(0)}{x}dx=\int_{|x|\le a}\frac{\phi(x)-\phi(0)}{x}dx$$$$\text{ and }\lim_{N\to\infty}\int_{a<|x|< N}\frac{\phi(x)}{x}dx=\int_{a<|x|}\frac{\phi(x)}{x}dx$$
So there exists $t_0>0$ such that $\left|\int_{t<|x|\le a}\frac{\phi(x)}{x}dx-\int_{|x|\le a}\frac{\phi(x)-\phi(0)}{x}dx\right|<\frac{\epsilon}{2}$ for all $0<t\le t_0$ and an $N_0$ such that $\left|\int_{a<|x|\le N}\frac{\phi(x)}{x}dx-\int_{a<|x|}\frac{\phi(x)}{x}dx\right|<\frac{\epsilon}{2}$ for all $N\ge N_0.$ Hence, for any $t\le t_0,N\ge N_0,$ we have $$\left|\int_{t<|x|<N}\frac{\phi(x)}xdx-\langle \text{PV}(1/x),\phi\rangle\right|\le \left|\int_{t<|x|\le a}\frac{\phi(x)}{x}dx-\int_{|x|\le a}\frac{\phi(x)-\phi(0)}{x}dx\right|$$$$+\left|\int_{a<|x|< N}\frac{\phi(x)}{x}dx-\int_{a<|x|}\frac{\phi(x)}{x}dx\right|<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon\text{ (using $(\star)$)}$$
Since $\phi\in\mathscr{S}$ is arbitrary, we conclude that $F_0=\lim_{t\to 0^+,N\to\infty} H_{t,N}.$
We first note that for any $t>0$ and $N$ we have using Tonelli's theorem $$\int_{t<|\xi|<N,-\infty<|x|<\infty}\left|\frac{\phi(x)e^{-2\pi i\xi x}}{\xi}\right|d(x,\xi)=\int_{t<|\xi|<N}\int\left|\frac{\phi(x)e^{-2\pi i\xi x}}{\xi}\right|dxd\xi$$$$\le \int_{t<|\xi|<N}\frac{1}{|\xi|}\int\left|\phi(x)\right|dxd\xi=2||\phi||_{L^1}\int_{t}^N\frac{1}{\xi}d\xi=2||\phi||_{L^1}\ln\left(\frac{N}{t}\right)<\infty$$
So, we can apply Fubini's theorem to get $$\int_{t<|\xi|<N}\frac{\hat{\phi(\xi)}}{\xi}d\xi=\int_{t<|\xi|<N}\int\frac{\phi(x)e^{-2\pi i\xi x}}{\xi} dxd\xi=\int\int_{t<|\xi|<N}\frac{\phi(x)e^{-2\pi i\xi x}}{\xi} d\xi dx$$$$=\int\int_{t<|\xi|<N}\phi(x)\frac{\cos(2\pi \xi x)}{\xi} d\xi dx-i\int\int_{t<|\xi|<N}\phi(x)\frac{\sin(2\pi \xi x)}{\xi} d\xi dx$$
For any $x,$ we have $\int_{t<|\xi|<N}\frac{\cos(2\pi \xi x)}{\xi} d\xi dx=0$ since the integrand is odd. So the first integral above evaluates to $0.$ Now, via a change of variables ($y=2\pi \xi x$ and $y=-2\pi \xi x$ in the first and second integral respectively) and the fact that $\xi\mapsto \sin(2\pi\xi x)/\xi$ is odd for all $x,$ $$\int\int_{t<|\xi|<N}\phi(x)\frac{\sin(2\pi \xi x)}{\xi} d\xi dx=\int_0^\infty2\phi(x)\left(\int_{2\pi t x}^{2\pi N x}\frac{\sin(y)}{y} dy\right)dx-\int_{-\infty}^02\phi(x)\left(\int_{-2\pi t x}^{-2\pi N x}\frac{\sin(y)}{y} dy\right)dx$$
It is easy to show that $\left|\int_a^b\frac{\sin(x)}{x}dx\right|\le 8$ for all $a\ge b\ge 0.$ Read this previous article for details. Now, $$\left|2\phi(x)\left(\int_{2\pi t x}^{2\pi N x}\frac{\sin(y)}{y} dy\right)\right|\le 8|\phi(x)|\text{ for all $x\ge 0$ }\text{ and }\phi\in L^1$$
Hence, using dominated convergence theorem, $$\lim_{t\to 0^+\atop N\to\infty}\int_0^\infty2\phi(x)\left(\int_{2\pi t x}^{2\pi N x}\frac{\sin(y)}{y} dy\right)dx=\int_0^\infty\lim_{t\to 0^+\atop N\to\infty}\left(2\phi(x)\left(\int_{2\pi t x}^{2\pi N x}\frac{\sin(y)}{y} dy\right)\right)dx=\pi\int_0^\infty \phi(0)dx$$
This is because for any $x>0,$ $\int_{2\pi t x}^{2\pi N x}\frac{\sin(y)}{y} dy$ converges pointwise to $\frac{\pi}{2}$ as $t\to 0^+,$ $N\to \infty$ ($\dagger$). Similarly, $$\lim_{t\to 0^+\atop N\to\infty}\int_{-\infty}^02\phi(x)\left(\int_{-2\pi t x}^{-2\pi N x}\frac{\sin(y)}{y} dy\right)dx=\pi\int_{-\infty}^0\phi(x)dx$$
Combining all these facts, we have $$\langle \hat{F}_0,\phi\rangle=\langle F_0,\hat{\phi}\rangle=\lim_{t\to 0^+\atop N\to\infty}\int_{t<|\xi|<N}\frac{\hat{\phi(\xi)}}{\xi}d\xi=-\pi i\left[\int_{0}^\infty \phi(x)dx-\int_{-\infty}^0 \phi(x)dx\right]=-\pi i\int \phi(x)\text{sgn}(x)dx$$
Since $\phi$ is arbitrary, we conclude that $\hat{F}_0=-\pi i\text{ }\text{sgn}.$
$(\dagger)$ We used the fact that $\lim_{a\to 0^+\atop b\to\infty}\int_a^b\sin(x)/xdx=\pi/2.$ To show this, first consider $f_b:(x,y)\mapsto e^{-xy}\sin(x)\chi_{(0,b)^2}(x,y).$ Since $|f|\le 1$ and non-zero only in $(0,b)^2$ it must be integrable. Hence, by Fubini's theorem, we have $$\int f_b(x,y)d(x,y)=\int_0^b\int_0^be^{-xy}\sin(x)dydx=\int_0^b\left(\frac{\sin(x)}{x}-\frac{e^{-xb}\sin(x)}{x}\right)dx$$
However, $$\left|\int_0^b\frac{e^{-xb}\sin(x)}{x}dx\right|\le \int_0^b\left|e^{-xb}\frac{\sin(x)}{x}dx\right|\le \int_0^b e^{-xb}dx=\frac{1-e^{-b^2}}{b}$$
Again using Fubini's theorem and the first line in the proof of (c), $$\int f_b(x,y)d(x,y)=\int_0^b\int_0^be^{-xy}\sin(x)dxdy=\int_0^b \left(\frac{1}{y^2+1}-e^{-by}\frac{y\sin(b)+\cos(b)}{y^2+1}\right)dy$$
However, $$\left|\int_0^be^{-by}\frac{y\sin(b)+\cos(b)}{y^2+1}dy\right|\le 2\int_0^be^{-by}dy=2\frac{1-e^{-b^2}}{b}$$ since $$\left|\frac{y\sin(b)+\cos(b)}{y^2+1}\right|\le \frac{|y|}{y^2+1}+\frac{1}{y^2+1}\le 1\text{ as }|y|\le y^2\text{ if }|y|\ge 1\text{ so }|y|\le y^2+1\text{ in either case}$$
Since $\lim_{b\to\infty}\frac{1-e^{-b^2}}{b}=0,$ we get $$\lim_{b\to\infty}\int_0^b\frac{\sin(x)}{x}dx=\lim_{b\to\infty}\int f_b(x,y)d(x,y)=\lim_{b\to\infty}\int_0^b\frac{1}{1+y^2}dy=\lim_{b\to\infty}\big[\tan^{-1}(b)-\tan^{-1}(0)\big]=\frac{\pi}{2}$$
In other words, for any $\epsilon>0,$ there exists a $b_0$ such that $\left|\int_0^b\frac{\sin(x)}{x}dx-\frac{\pi}{2}\right|<\frac{\epsilon}{2}$ for all $b\ge b_0.$
Then for any $b\ge b_0$ and $0<a\le \frac{\epsilon}{2},$ since $|\sin(x)/x|\le 1$ for all $x\neq 0,$ we have $$\left|\int_a^b\frac{\sin(x)}{x}dx-\frac{\pi}{2}\right|\le \left|\int_0^b\frac{\sin(x)}{x}dx-\frac{\pi}{2}\right|+\left|\int_0^a\frac{\sin(x)}{x}dx\right|<\frac{\epsilon}{2}+\int_0^adx=\frac{\epsilon}{2}+a<\epsilon$$
This justifies $(\dagger).$
Proof of Part V
Since $\chi_{(0,\infty)}=\frac{1}{2}\text{sgn}(x)+\frac{1}{2},$ we have for any $\phi\in\mathscr{S},$ $$\langle \widehat{\chi_{(0,\infty)}},\phi\rangle=\langle \chi_{(0,\infty)},\hat{\phi}\rangle=\frac{1}{2}\langle \text{sgn},\hat{\phi}\rangle+\frac{1}{2}\langle x\mapsto 1,\hat{\phi}\rangle=\frac{1}{2}\langle \hat{\text{sgn}},\phi\rangle+\frac{1}{2}\langle x\mapsto 1,\hat{\phi}\rangle$$$$=\frac{1}{2}\langle (i\pi)^{-1}F_0,\phi\rangle+\frac{1}{2}\int\hat{\phi(x)}dx=\frac{1}{2}\langle (i\pi)^{-1}F_0,\phi\rangle+\frac{1}{2}\langle \delta,\phi\rangle=\left\langle \frac{1}{2}\Big((i\pi)^{-1}F_0+\delta\Big),\phi\right\rangle$$
Here, we used the fact that $\int\hat{\phi}(x)dx=\widehat{\hat{\phi}}(0)=\phi(-0)=\phi(0)$ using the inversion theorem.
Hence, $\widehat{\chi_{(0,\infty)}}=\frac{1}{2}\Big((i\pi)^{-1}F_0+\delta\Big)$ since the above equality holds for arbitrary $\phi\in\mathscr{S}.$
Alternatively, since $\chi_{(0,\infty)}(x)=\lim_{\epsilon\to 0^+}e^{\epsilon x}\chi_{(0,\infty)}(x)$ pointwise, the limit also exists in $\mathscr{S}'$ (as a result of the dominated convergence theorem: $|e^{-\epsilon x}\phi(x)\chi_{(0,\infty)}(x)|\le |\phi(x)|$ for all $x\in \mathbb{R}$ and $\phi\in L^1(\mathbb{R})$).
Hence, using theorem 7.15, $\hat{f}_\epsilon\longrightarrow \widehat{\chi_{(0,\infty)}}$ in $\mathscr{S}'$ as $\epsilon\longrightarrow 0^+$ where $f_\epsilon(x)=e^{-\epsilon x}\chi_{(0,\infty)}(x)\text{ ($\star$)}$
Now, using lemma 8.25 (Folland), since $f_\epsilon$ is obviously $L^1$ (for any $\epsilon>0$), $$\left\langle \hat{f}_\epsilon,\phi\right\rangle=\int f_\epsilon(x)\hat{\phi}(x)dx=\int \hat{f}_\epsilon(x)\phi(x)dx$$
However, $$\hat{f}_\epsilon(\xi)=\int_0^\infty e^{-\epsilon x}e^{-2\pi i\xi x}dx=\int_0^\infty e^{-2\pi i\xi x-\epsilon x}dx=\frac{1}{2\pi i \xi+\epsilon}=\frac{1}{2\pi i}\frac{1}{\xi-i\Big(\epsilon/(2\pi)\Big)}=\frac{1}{2\pi i}G^{-}_{\big(\epsilon/(2\pi)\big)}(\xi)$$
Hence, (taking limits in $\mathscr{S}'$), $$\lim_{\epsilon\to 0^+}\left\langle \hat{f}_\epsilon,\phi\right\rangle=\frac{1}{2\pi i}\lim_{\epsilon\to 0^+}\left\langle G^{-}_{\big(\epsilon/(2\pi)\big)},\phi\right\rangle=\frac{1}{2\pi i}\langle F_0+\pi i\delta,\phi\rangle=\left\langle \frac{1}{2}\Big((i\pi)^{-1}F_0+\delta\Big),\phi\right\rangle$$
Since this holds for alrbitrary $\phi\in\mathscr{S},$ we conclude that $\hat{f}_\epsilon\longrightarrow \frac{1}{2}\Big((i\pi)^{-1}F_0+\delta\Big)$ in $\mathscr{S}'$ as $\epsilon\longrightarrow 0^+.$ Thus, $(\star)$ gives us the same result as before.
