Let be $f$ a $2\pi-$periodic function and $\hat{f}(k)=\frac{1}{2\pi}\int\_0^{2\pi}f(x)e^{-ikx}dx$. Consider the operator:
\begin{equation}
Tf(x)=\sum\_{k\in\mathbb{Z}}sign(k)\ \hat{f}(k)\ e^{ikx}.
\end{equation} 
I would like to know if the operator $T:L^p(0,2\pi)\rightarrow L^p(0,2\pi)$ is bounded.



About the boundedness of a multiplication operator.
When $p=2$, boundedness is a triviality and it is the only trivial case. It is not true for $p=1$ nor for $p=\infty$, although the Fourier multiplier $sign(D\_x)$ sends $L^1$ into $L^1\_w$ and the Marcinkiewicz interpolation theorem implies boundedness in $L^p$ for all $p\in]1,+\infty[$.


The operator $sign(D\_x)$ is is a particular case of the wider class of singular integrals, extensively studied by Calderon and Zygmund, later by Hörmander, Stein & Fefferman. They are defined via a simple condition on their kernels, easily proven $L^2$ bounded, with the property that they send $L^1$ into $L^1\_w$. Again Marcinkiewicz Theorem allows to finish the job of proving boundedness in $L^p$ for all $p\in]1,+\infty[$.


To give a simple class of example would be to consider Fourier multiplier $F(D\_x)$ where $F$ is an homogeneous function of degree 0 which is smooth outside of the origin. Note that it works as well in any dimension and that the so-called Hörmander-Mihlin multiplier Theorem allows to weaken significantly the smoothness assumption.


