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\begin{align}
\mrm{H}\pars{x} & = \int_{-\infty}^{\infty}{\expo{\ic kx} \over k - \ic 0^{+}}\,
{\dd k \over 2\pi\ic}
\\[5mm]
\hat{\mrm{H}}\pars{k} & = {1 \over k -\ic 0^{+}}\,{1 \over \ic} =
-\ic\bracks{\mrm{P.V.}{1 \over k} + \ic\pi\,\delta\pars{k}} =
-\ic\,\mrm{P.V.}{1 \over k} + \pi\,\delta\pars{k}
\end{align}
1. BOTTOM LINE UP FRONT: Treat it as a distribution.
Since the signum function is not integrable on $\mathbb{R}$, it may be useful to view it as a tempered distribution.
Such "generalized functions" are bounded linear functionals on a class of very-well-behaved functions called Schwartz functions. One of Laurent Schwartz's achievements was finding a collection $\mathcal{S}$ of functions on $\mathbb{R}^n$ such that the set of Fourier transforms of these functions is $\mathcal{S}$ itself. That put the original functions and their Fourier transforms on equal footing.
2. Fourier transform of a distribution
Why is this useful? It means that every tempered distribution has a Fourier transform that is also a tempered distribution. It also provides some useful notation to derive expressions and properties of the Fourier transform of a known tempered distribution.
Given any distribution
$\mathsf{T}$ we write the result of applying it to a Schwartz function
$\varphi$ as
$\left<\mathsf{T},\varphi\right>$, but it is to be understood that this is not an inner product of two objects of the same kind. The Fourier transform of the distribution
$\mathsf{T}$ is the distribution
$\widehat{\mathsf{T}}$ for which
\begin{equation}
\left<\widehat{\mathsf{T}},\varphi\right> = \left<\mathsf{T},\widehat{\varphi}\right>
\end{equation}
for every
$\varphi\in\mathcal{S}$, where
$\widehat{\varphi}$ is the Fourier transform of
$\varphi$. Since
$\varphi\in\mathcal{S}$,
$\widehat{\varphi}\in\mathcal{S}$, too.
3. Fourier transform of signum
How is this related to the signum function? If
$\mathsf{T}$ is the signum function viewed as a distribution, then
\begin{equation}
\left<\mathsf{T},\varphi\right> = \int\textrm{sgn}(x)\varphi(x)dx.
\end{equation}
The Fourier transform of this distribution satisfies (or is defined by)
\begin{equation}
\begin{split}
\left<\widehat{\mathsf{T}},\varphi\right> &=~
\left<\mathsf{T},\widehat{\varphi}\right>\\
&=~ \int\textrm{sgn}(x)\widehat{\varphi}(x)dx\\
&=~ -\int_{-\infty}^{0}\widehat{\varphi}(x)dx + \int_{0}^{\infty}\widehat{\varphi}(x)dx.
\end{split}
\end{equation}
4. Changing order of integration
Let's consider the integral for positive reals. The very good behavior of
$\varphi$ allows changing orders of integration in many, many situations.
\begin{equation}
\begin{split}
\int_{0}^{\infty}\widehat{\varphi}(x)dx
&=~
\int_{0}^{\infty}\left[\int\varphi(k)e^{-ixk}dk\right]dx\\
&=~
\lim_{R\to\infty}\int_{0}^{R}\left[\int\varphi(k)e^{-ixk}dk\right]dx\\
&=~
\lim_{R\to\infty}\int\left[\int_{0}^{R}e^{-ixk}dx\right]\varphi(k)dk
\end{split}
\end{equation}
We do something very similar for the negative reals.
\begin{equation}
\begin{split}
-\int_{-\infty}^{0}\widehat{\varphi}(x)dx
&=~
-\int_{-\infty}^{0}\left[\int\varphi(k)e^{-ixk}dk\right]dx\\
&=~\lim_{R\to\infty}-\int_{-R}^{0}\left[\int\varphi(k)e^{-ixk}dk\right]dx\\
&=~
\lim_{R\to\infty}-\int\left[\int_{-R}^{0}e^{-ixk}dx\right]\varphi(k)dk
\end{split}
\end{equation}
We now address the sum of the $R$-dependent integrals.
\begin{equation}
\begin{split}
\int_{0}^{R}e^{-ikx}dx - \int_{-R}^{0}e^{-ikx}dx
&=~
\left.\frac{e^{-ikx}}{-ik}\right|_{x=0}^{x=R}
-
\left.\frac{e^{-ikx}}{-ik}\right|_{x=-R}^{x=0}\\
&=~
\frac{1 - e^{-ikR}}{-ik}
-
\frac{e^{ikR} - 1}{-ik}\\
&=~
\frac{e^{ikR} + e^{-ikR}}{ik}
-
\frac{2}{ik}
\end{split}
\end{equation}
5. Singularity at $k = 0$; Riemann-Lebesgue Lemma
The
$k$ in the denominator will be a problem at
$k=0$. But we know that the original integrals coverge. We must consider the new one as the limit of integrals from
$\epsilon$ to
$\infty$ and from
$-\infty$ to
$-\epsilon$.
\begin{equation}
\int_{|k|>\epsilon}\frac{e^{ikR} + e^{-ikR}}{ik}\varphi(k)dk
=
\int 1_{\{k:|k|>\epsilon\}}(k)\frac{\varphi(k)}{ik}\left(e^{ikR} + e^{-ikR}\right)dk
\end{equation}
For each $\epsilon >0$, the function $1_{\{k:|k|>\epsilon\}}(k)\frac{\varphi(k)}{ik}$ is in $L^1(\mathbb{R})$, so this integral is that function's Fourier transform evaluated at $\omega = R$ plus the same Fourier transform evaluated at $\omega = -R$. The Riemann-Lebesgue Lemma shows that if $f\in L^1(\mathbb{R})$, then $\lim_{|R|\to\infty}\widehat{f}(R) = 0$. Hence, these $R$-dependent terms vanish as $R\to\infty$.
It is worth noting that this shows that we must take the $R$-limit first and then take the $\epsilon$-limit. The opposite order would not work.
6. Cauchy Principal Value
We are left with
\begin{equation}
\lim_{\epsilon\to 0}2i\int_{|k|>\epsilon}\frac{\varphi(k)}{k}dk.
\end{equation}
This is the
Cauchy Principal Value of this integral. This shows that we must interpret the Fourier transform of the signum function very carefully, but we
can do it in the sene of distributions: if
$\textrm{sgn}(x)$ is the signum of
$x$, then
\begin{equation}
\widehat{\textrm{sgn}}(k) = 2i~\mathsf{PV}\left(\frac{1}{k}\right).
\end{equation}
Best Answer
I suppose you are considering $T$ as an operator from $L^{1}(\mathbb R^{n})$ into the space $C_0(\mathbb R^{n})$ of continuous functions vanishing at $\infty$ with the sup norm. The norm of $T$ (which is $(2\pi)^{-N/2}$) is attained at the function $(2\pi)^{-N/2} e^{-\|x\|^{2} /2}$.