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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.

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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 –  Mark Lewko Feb 25 '13 at 11:09
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Multiplication with a bounded sequence always gives a bounded operator: en.wikipedia.org/wiki/Multiplication_operator –  András Bátkai Feb 25 '13 at 11:14
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@András, note the operator acts by multiplying the Fourier transform by a bounded sequence (not the function itself!) –  Mark Lewko Feb 25 '13 at 11:29
    
@Mark: Thank you. An amateurish mistake... –  András Bátkai Feb 27 '13 at 7:51
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up vote 4 down vote accepted

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.

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