Extend the result that the Hilbert transform is of weak-type $latex L^1$ to any operator of the form $latex Tf = K*f$ where $latex K$ satisfies:
$latex \hat K\in L^\infty(\R)$ (say, taking the Fourier transform in the $latex L^2$ sense);
there exists a constant $latex A>0$ such that $latex |K'(x)| \le A/x^2$ for any $latex x\in\R, \; x\not=0$.
Show that we can replace (2) above by the condition 2'. There exists a constant $latex A>0$ such that $latex \displaystyle \int_{|x|\ge 2|t|} |K(x-t) - K(x)| dx \le A$ for any $latex t\in\R$.
(Chebyshev Inequality) If $latex f\in L^p$, for some $latex 1 < p <\infty$, then $latex |\{x : |f(x)|>\alpha\}| < \dfrac{1}{\alpha^p} ||f||_{L^p}^p$
Prove that we can substitute the condition $latex \hat K\in L^\infty$ with the boundedness of $latex T$ on any $latex L^q(\R), \; q>1$, that is, 1'. there exists a constant $latex M>0$ such that $latex ||Tf||_{L^q} \le M||f||_{L^q}$.
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