I want to show that the Hilbert Transform satisfies the Mikhlin Multiplier Theorem (i.e. that it satisfies the following. $m: \mathbb{R}^{n}\backslash \{0\} \rightarrow \mathbb{C}$ satisfies $|\partial^{\alpha}m(\xi)| \leq C|\xi|^{-|\alpha|} $ for $\alpha$ a multi-index with $|\alpha| \leq n+2, \xi \neq 0$ and $C>0$ a constant. Then for all $1 < p < \infty $ there exists a $B > 0$ such that $||m(D)f||_{L^{p}} \leq B||f||_{L^{p}}, f \in \mathcal{S}(\mathbb{R}^{n})$).
Attempt:
Let $m(\xi) = -isgn(\xi)$. Letting $m(D)$ equal the Hilbert transform the result is obviously true for $\alpha = 0$. Then I get a bit confused how to proceed. If I start taking derivatives of $m(\xi)$ it starts to get messy and involving dirac-delta distribution which I don't think is the way to proceed?