About the boundedness of a multiplication operator.

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: $$Tf(x)=\sum_{k\in\mathbb{Z}}sign(k)\ \hat{f}(k)\ e^{ikx}.$$ I would like to know if the operator $T:L^p(0,2\pi)\rightarrow L^p(0,2\pi)$ is bounded.

• This is just (a scalar multiple of) the Hilbert transform on the circle. It maps Lp to itself for 1<p<∞. See: en.wikipedia.org/wiki/Hilbert_transform Feb 25 '13 at 11:09
• Multiplication with a bounded sequence always gives a bounded operator: en.wikipedia.org/wiki/Multiplication_operator Feb 25 '13 at 11:14
• @András, note the operator acts by multiplying the Fourier transform by a bounded sequence (not the function itself!) Feb 25 '13 at 11:29
• @Mark: Thank you. An amateurish mistake... Feb 27 '13 at 7:51

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.