Hilbert transform and Fourier transform Assume the following relationship between the Hilbert and Fourier transforms:
$$
\mathcal{H}(f) = {\mathcal{F}^{-1}}(-i ~ \text{sgn}(\cdot) \cdot \mathcal{F}(f)),
$$
where $ \displaystyle [\mathcal{H}(f)](x) \stackrel{\text{def}}{=} \text{p.v.} \frac{1}{\pi} \int_{- \infty}^{\infty} \frac{f(t)}{x - t} ~ d{x} $.
What happens when $ f(x)$ is a distribution? We know that the Fourier transform exists for distributions, but what about the Hilbert transform?
For example, take $ f(x) = x^{n} $. Its Fourier transform exists as the $ n $-th derivative of the delta function $ \delta(x) $. However, the integral
$$
\text{p.v.} \frac{1}{\pi} \int_{- \infty}^{\infty} \frac{x^{n}}{x - t} ~ d{x}
$$
is divergent. :(
 A: The Hilbert transform is anti-self-adjoint. Therefore, it is natural to define it on distribution by passing $\mathcal H$ to the test functions, similar to "pass the hat" definition of the Fourier transform. In fact, the Wikipedia article already says this. 
Since the stated relation between $\mathcal{F}$ and $\mathcal H$ holds for test functions, the duality-based definition implies that it holds for distributions as well. 
A: The multiplication by "sign" (with or without a constant "$i$" in front) does not map Schwartz functions to Schwartz functions, so that the otherwise-appealing notion of defining an automorphism on (tempered?) distributions by defining it on (Schwartz?) test functions does not succeed here.
Although in general it is ... very risky... to try to "multiply" distributions, a more delicate approach does allow multiplication when the "singular sets" are disjoint.
In the case at hand, multiplication of a distribution by "sign" would make sense, for example, if the singular support does not contain $0$. Thus, a tempered distribution whose Fourier transform has singular support not including $0$ has a Hilbert transform.
