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## Bound on Hilbert transform

Consider $\widehat{Tf(\xi)}=m(\xi)\hat{f}(\xi)$, where $m(\xi)=(1-\vert\xi\vert)1_{[-1,1]}$, i.e. $T$ is the operation of taking Fourier transform and multiplying with the function $m(\xi)$. I am asked to show that $T$ is bounded on $L^p(\mathbb{R})$ for $p\in(0,\infty)$.

If $p=2$ the question becomes pretty easy. Using the Plaucherel's theorem I can show that the operator $T$ is bounded on $L^2$, \begin{align} \Vert Tf\Vert_2=\Vert\widehat{Tf(\xi)}\Vert_2=\left(\int_\mathbb{R}\left\vert m(\xi)\hat{f}(\xi)\right\vert^2d\xi\right)^{1/2}\le\Vert m\Vert_\infty\Vert\hat{f}\Vert_2=\Vert m\Vert_\infty\Vert f\Vert_2. \end{align}

My question is what do I do for $p\ne2$. I have this theorem: suppose there $f\in L^q$ and $g\in L^r$ and $1/q+1/r=1+1/p$ then $f*g\in L^p$ and $\Vert f*g\Vert_p\le\Vert f\Vert_q\Vert g\Vert_r$. But I don't know how to write the transformation as convolution with some function $g$..

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multiplication in frequency space is the same as convolution in physical space. $m(\xi)$ is continuous and has compact support. Just compute its inverse Fourier transform explicitly and show that it is in $L^1$. Also, the Hilbert transform is usually something quite specific en.wikipedia.org/wiki/Hilbert_transform and is not what you described in your question. – Willie Wong Nov 8 at 11:18
"I am asked to show" - who is asking you to show? What for? – Yemon Choi Nov 8 at 16:34
Smells like homework... – Alain Valette Nov 9 at 9:37

I am not sure this is what is commonly known under the name "Hilbert transform" - which by the way is known to be bounded on $L^p$ iff $p\in (1,\infty)$.