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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. :(

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You have taken the Fourier transform of $x^n$ as a distribution, but why do you treat the Hilbert transform of $x^n$ as a function[thus it is divergent]? –  Yimin Feb 15 '13 at 17:39
is $isgn$ actually $sign$? –  rschwieb Feb 15 '13 at 17:45
What is ‘$ w $’, by the way? –  Haskell Curry Feb 15 '13 at 17:46
@rschwieb $i$ is correct: see, e.g., here. –  user53153 Feb 15 '13 at 18:18
@5pm Oh :) I thought it might be a simple transposition. Good thing I didn't change it! –  rschwieb Feb 15 '13 at 19:08

1 Answer 1

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.

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