Week 13 -- BP3, 4, 5, 6

[Victor Ivrii]

$\renewcommand{\Re}{\operatorname{Re}}$
There are several connected bonus problems:
a.
Let $(x\pm i0)^{\nu}= \lim _{\varepsilon\to +0} (x\pm \varepsilon i)^{\nu}$. Prove, it exists for $\Re\nu >-1$.
b. Using equality   
\begin{equation}
[(x\pm \varepsilon i)^{\nu}]'= \nu (x\pm \varepsilon i)^{\nu-1},
\label{eq-11.1.12}
\end{equation}
prove that these limits exist (in the sense of distributions) for  $ \nu \ne -1,-2,\ldots$.
c. Using
\begin{equation}
[\ln (x\pm \varepsilon i) ]'=
 (x\pm \varepsilon i)^{-1},
 \label{eq-11.1.13}
\end{equation}
prove that these limits exist (in the sense of distributions) for  $\nu =-1$. Then, using (\ref{eq-11.1.12}), prove that these limits exist (in the sense of distributions) for $\nu=-2,-3,\ldots$.
d. Prove that
\begin{align}
&(x-i0)^{-1} + (x+i0)^{-1}=2 x^{-1},
\label{eq-11.1.14}\\
&(x-i0)^{-1}- (x+i0)^{-1}=2\pi i \delta(x)
\label{eq-11.1.15}
\end{align}
with integral understood in the sense of principal value, use (\ref{eq-11.1.13}) and 
\begin{align}
&\ln (x+i0)+\ln (x-i0)= 2\ln |x|,\\
&\ln (x+i0)-\ln (x-i0)= -2\pi i\theta(-x).
\end{align}
You need to justify these formulae.

[Andrew Hardy]

a) Consider
$$ (x\pm i0)^{\nu}= f'$$
Now via definitions $$ (f',\theta) = - (f,\theta') $$ and then f is defined as
 $$ \frac{1}{\nu+1}(x \pm i0)^{\nu+1}  = f $$ This  exists for  $ \Re\nu >-1 $. Therefore the limit exists for this condition.
b) Extending the proof from part a, define $$ (x\pm i0)^{\nu}= f^{(n)}$$ then with the same rules of distribution derivatives I can write $$  [(x\pm \varepsilon i)^{\nu}]^{(n)}= (\nu ! )(x\pm \varepsilon i)^{\nu-n}$$ which is then defined for corresponding $ \Re\nu \neq -n $ where  $ n = 2, 3,\ldots $

[Victor Ivrii]

One does not need derivatives to prove that the limit (in $L^1$) $\lim \varepsilon\to +0 (x\pm i\varepsilon)^\nu $ exists and equals
$$
|x|^\nu \left\{\begin{aligned}
&1 &&x>0,\\
&e^{\pm i\pi \nu} && x<0,
\end{aligned}\right.
$$
provided $\Re\nu >-1$ because then this function belongs to $L^1$. The fin starts from (b)

[Zhongnan Wu]

I think there is a type of question c). Should be [ln(x+epsilon i)]', here the power of v is 1 and should not be shown.
Also for the second half part of this question, should we use the equality from b) rather than from a)?

[Jingxuan Zhang]

Zhongnan,
I think c) is alright as the way it stands. After all \eqref{eq-11.1.12} is deduced from \eqref{eq-11.1.13} so that is also fine...

[Victor Ivrii]

Andrew, you right with powers but losing numerical factor.
Zhongnan  is right (copy-paste-not-clening-up error). I corrected

[Andrew Hardy]

Dr. Ivrii? Are Part a and b correct now? In Gel'fand's "Normalized Functions" he has what seems the same argument, but rather does the integrals explicitly. I can go that in depth if I haven't earned the points yet. I'm stilling looking for c and d.

[Andrew Hardy]

d) *from Gel'fand*
Equations 5 and 6 come from the natural log of  the generalized expression
$$ (x+iy)^{\nu} =  \exp[\nu \log(x+iy) + iArg(x+iy)]
\tag{A}$$  which is just the representation on the complex plane
Then we can find
$$ (x+i0)^{-\nu} = x^{-\nu} - \frac{i\pi(-1)^{\nu-1}}{(\nu -1)!}\delta^ {(nu -1)}(x)
\tag{B}$$
$$ (x-i0)^{-\nu} = x^{-\nu} + \frac{i\pi(-1)^{\nu-1}}{(\nu -1)!}\delta^ {(nu -1)}(x) \tag{C}$$
and by combining these and setting $ \nu =1 $ we arrive at
$$ (x-i0)^{-1} + (x+i0)^{-1}=2 x^{-1}  \tag{D}$$
$$(x-i0)^{-1}- (x+i0)^{-1}=2\pi i \delta(x)  \tag{E}$$

[Victor Ivrii]

I added tags to comment

(A): multiple errors: it should be $=\exp \bigl[\nu \bigr( \log|x+iy|+i\arg (x+iy)\bigr)\bigr]$. First factor $\nu$ is applied to everything second, we do not allow $(x+iy)$ to cross $(-\infty,0]$; so angle is defined uniquely and it rans from $-\pi$ to $\pi$.
Anyway: $[(x\pm i \varepsilon)^\nu]' =\nu (x\pm i \varepsilon)^{\nu-1}$ from complex variables and here $\nu$ could be even complex.

As $\varepsilon\to +0$, there is a limit in $\mathscr{D}'$ (even in $L^1_{loc}$) of $(x\pm i \varepsilon)^\nu$ provided $\Re\nu >-1$; but then there exists a limit of its derivative, i.e. $\nu (x\pm i \varepsilon)^{nu-1}$ and we can divide by $\nu\ne 0$. So we defined $(x\pm i \varepsilon)^{\nu}$ as long as $\Re \nu >-2$ and $\nu \ne -1$.

Repeating, we define $(x\pm i \varepsilon)^{\nu}$ as long as $\Re \nu >-3$ and $\nu \ne -1,-2$. ... and so on... as $\nu\ne -1,-2,\ldots$.

Remark. To mitigate the latter restriction, $f_\nu ^\pm :=\frac{(x\pm i0)^\nu}{\Gamma(\nu+1)}$ could be considered where $\Gamma$ is Euler's $\Gamma$-function; it has simple poles at $0,-1,-2,\ldots$ and $\Gamma(\nu+1)=\nu\Gamma(\nu)$.

(B), (C) are out of the window: they do not follow; also what is $x^{-\nu}$ for $x<0$ and $\nu\notin\mathbb{Z}$? What is $\delta^{\nu-1}$ for $\nu\ne 1,2,\ldots$? We can define those but a posteriori.

I suggested a simple way: look at $\log (x\pm i0)$ as $x>0$ and $x<0$; obviously $\log (x\pm i0)=\log|x|$ as $x>0$ and $\log (x\pm i0)=\log|x|\pm i\pi$ as $x<0$. In other words $\log (x\pm i0)=\log |x| \pm i\pi \theta(-x)$. Differentiating in $\mathscr{D}'$ we get
$$
(x\pm i0)^{-1}= (\log |x|)' +\pm i \pi (\theta (-x))'=x^{-1} \mp i\pi \delta(x)
$$
where (see other of bonus problems, http://forum.math.toronto.edu/index.php?topic=1167.0) $(\log |x|)' =x^{-1}$ in vp sence, and
$(\theta (-x))'=-\delta(x)$.

PS Gelfand--Shilov 1--6 (+coauthors in higher volumes) is a truly remarkable book, but IMHO, sometimes they go too far. F.e. considering F.T. of distributions not in $\mathscr{S}'$ they get distributions over some classes of the entire analytic functions , in particular they get $\delta (x-c)$ with any $c\in \mathbb{C}$ (which is definitely a perversion). Unfortunately, G.Shilov died too young (at 58) and I never met him.  I.Gelfand was a great mathematician.