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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.

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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.

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