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?