For $f\in\mathcal{S}$ a Schwartz function on $\mathbb{R}^n$ and $m$ a bounded function, define $Tf$ by $\widehat{T f}=m\cdot \widehat{f}$.

Fix $1<p<\infty$, $p\not=2$. Suppose we have proved that there exists a constant $C>0$ such that $$ \|Tf\|_p \le C \|f\|_p$$ holds for all $f\in\mathcal{S}$ satisfying $\widehat{f}\ge 0$.

Is it still possible that $T$ is not $L^p\to L^p$ bounded?

I believe that the answer is yes, but I don't know of an example. I'd also be grateful for any reference where this or similar situations (possibly in the context of other kinds of operators) are discussed.


1 Answer 1


I apologize for answering my own question (and for asking a trivial question). The answer is yes. If $\widehat {f}\ge 0$, then we always have by the triangle inequality, $$\|Tf\|_4=\|\widehat{Tf}*\widehat{Tf}\|_2^{1/2}\le C \|\widehat{f}*\widehat{f}\|_2^{1/2}=C\|f\|_4 $$ so any multiplier that is not bounded on $L^4$ is a counterexample.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.