most recent 30 from http://mathoverflow.net 2013-05-26T00:15:35Z http://mathoverflow.net/feeds/question/111799 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/111799/bound-on-hilbert-transform dynamic89 2012-11-08T10:04:18Z 2012-12-21T11:22:00Z

<p>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 $p\in(0,\infty)$.</p> <p>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}</p> <p>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 $g$..</p>

http://mathoverflow.net/questions/111799/bound-on-hilbert-transform/111879#111879 Delio Mugnolo 2012-11-09T09:32:49Z 2012-11-09T09:32:49Z

<p>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 $p\in (1,\infty)$.</p>