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Sep 13, 2018 at 6:37 comment added user94415 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 $L^{\infty}$?
Sep 12, 2018 at 18:07 comment added Christian Remling 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.
Sep 12, 2018 at 14:47 history asked user94415 CC BY-SA 4.0