Question

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.

Answer 1

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.