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I have tried to solve the following exercise but I don't know if I have missed anything.

Show that the Hilbert Transform $Hf = \mathcal{F}^{-1}(\operatorname{sgn}(\xi)\hat{f}(\xi))$ is a pseudo-differential operator of order 0 in $\mathbb{R}$.

My thoughts:

$\mathcal{F}^{-1}(sgn(\xi)\hat{f}(\xi)) = \frac{1}{2\pi}\int_{-\infty}^{\infty}sgn(\xi)\hat{f}(\xi)e^{-ix\xi}d\xi = \frac{1}{2\pi}\int_{-\infty}^{0}-\hat{f}(\xi)e^{-ix\xi}d\xi + \frac{1}{2\pi}\int_{0}^{\infty}\hat{f}(\xi)e^{-ix\xi}d\xi = g(x)$

Now $f(x) = \frac{1}{2\pi}\int_{-\infty}^{\infty}\hat{f}(\xi)e^{-ix\xi}d\xi$ and $g(x) - f(x) = -\frac{1}{\pi}\int_{-\infty}^{0} \hat{f}(\xi)e^{-ix\xi}d\xi = h(x)$.

Hence, $Hf = g(x) = f(x) + h(x)$, so $H$ must be of order 0.

Is this reasonable or completely wrong?

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    $\begingroup$ I'm just a novice on this, but does it make sense to say that an operator is a pseudodifferential operator without specifying the class of the symbol? Also, what function space does $f$ belong to? It seems to me that the question asks you to show first that H is a PsiDO, and then that it is of order 0. $\endgroup$ – Student tea May 30 '14 at 22:51
  • $\begingroup$ Yeah, you are probably right, The book is not really clear in many aspects, Ill have another look. $\endgroup$ – El_Loco May 31 '14 at 8:22
  • $\begingroup$ The sign in the exponential for the inverse Fourier transform is wrong. $\endgroup$ – JamalS May 31 '14 at 20:09
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    $\begingroup$ What do you mean? There are different conventions for the sign of the exponential and where to put $2\pi$ $\endgroup$ – El_Loco May 31 '14 at 22:32

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