TITLE: Fourier transform of signum function
QUESTION [6 upvotes]: If we treat fourier transform as an operator on $L^1(\mathbb{R})$, then its image under fourier transform is the set of continuous functions which will vanish at infinity.
It is well known that the fourier transform of signum function is
$$\mathcal{F} (sgn)(u) =\frac{2}{ui}. $$
I know that signum function is not integrable over the real line. So, to evaluate its fourier transform, one can use limiting argument, say a sequence of functions that converges to signum function, because fourier transform is a bounded linear operator, and hence is continuous.
What puzzles me is that when using continuity, don't we need to ensure that the fourier transform is defined on the limiting function?
In this case, fourier transform of signum function is not defined due to reason given above. If this is the case, how would one obtain formula of the fourier transform of signum function?

REPLY [4 votes]: Let $\phi(x)$ be a Schwartz function and $\displaystyle \psi(x)=\text{PV}\left(\frac1{ix}\right)$.
Using the distributional relationships $\langle d',\phi\rangle =-\langle d,\phi'\rangle $ for any tempered distribution $d$ and $\text{sgn}'(x)=2\delta(x)$, we have
$$\begin{align}
\langle  \mathscr{F}\{\text{sgn}\},\phi \rangle &=\langle \text{sgn},\mathscr{F}\{\phi\}\rangle\\\\
&=\langle \text{sgn},\left(\mathscr{F}\{\phi \psi\}\right)'\rangle\\\\
& =-\langle(\text{sgn})', \mathscr{F}\{\phi \psi\}\rangle\\\\
& =-2\langle \delta, \mathscr{F}\{\phi\psi\}\rangle\\\\
&=2i\text{PV}\left(\int_{-\infty}^\infty \frac{\phi(x)}{x}\,dx\right)
\end{align}$$
Hence, we can write the distributional relationship
$$\bbox[5px,border:2px solid #C0A000]{\mathscr{F}\{\text{sgn}\}(x)=\text{PV}\left(\frac{2i}{x}\right)}$$