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. 



$L^p$ estimates and functions with positive Fourier transform
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.


