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During my graduate studies I've been told that pseudo-differential operators with symbols in $S^0=S^0_{1,0}$ (the simplest class) are bounded $L^2 \to L^2$, and also $L^p \to L^p$ for all $p \in (1, \infty)$. I have also heard that such pseudo-differential operators need not be bounded $L^{\infty} \to L^{\infty}$. But no counter examples were presented to me and I can't find any counter example in any book I came across. The Hilbert transform (in Fourier space $\widehat{Hf}(\xi)=i\operatorname{sgn}(\xi) \;\hat f (\xi)$) is unbounded $L^{\infty} \to L^{\infty}$ but strictly speaking it is not a standard pseudo-differential operator.

What are the canonical examples of zeroth order pseudo-differential operators that are not bounded $L^{\infty} \to L^{\infty}$?

If there is a "standard" reference dealing with this problem I would be glad to know where to read it, but any relevant reference would be great!

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    $\begingroup$ Can't you just take a smooth version of $\textrm{sgn}(\xi)$? The modification can of course be done by a compactly supported (in $\xi$) perturbation, so this will be bounded on $L^{\infty}$ and thus you still have a counterexample. $\endgroup$ Sep 12, 2018 at 18:07
  • $\begingroup$ This is true and answers the question... I should have seen it before! However a better question is the following: is there a more general class of pseudo-differential operators unbounded on $L^{\infty}$? $\endgroup$
    – user94415
    Sep 13, 2018 at 6:37

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