Let $latex \{\Phi_t\}_{t>0}$ be a collection that satisfies, for some constants $latex c_0, c_1, c_2$, and all $latex x\in\R^d$ and $latex t>0$,
$latex \int \Phi_t = c_0$,
$latex |\Phi_t(x)| \le c_1 t^{-d}$,
$latex |\Phi_t(x)| \le c_2 t/|x|^{d+1}$.
Then there exists $latex A>0$ such that $latex |\Phi_t*f(x)|\le A Mf(x)$, for all $latex x\in\R^d$, and thus $latex \Phi_t*f(x)\to f(x)$ as $latex t\to 0$ for almost every $latex x\in\R^d.$
(Needs complex analysis) $latex \displaystyle \hat Q _y(\xi) = \lim_{N\to\infty}\frac{1}{\pi} \int_{-N}^N \frac{x e^{-2\pi ix\xi}}{x^2+y^2} dx = -i\text{sgn}\xi e^{-2\pi y |\xi|}$. (Hint: Use the residue theorem from complex analysis.)
(Needs complex analysis) Give another proof that the Hilbert transform is bounded on $latex L^2$ by considering the Cauchy integral $latex \displaystyle Cf(z) = \frac{1}{i\pi} \int_{-\infty}^\infty \frac{f(t)}{t-z} dt$.
If $latex f\in C_c(\R)$, $latex Cf(z) = O(1/|z|)$.
If $latex F(z) = Cf(z)$, then $latex \displaystyle \int_{-\infty}^\infty F(x)^2 dx = 0$
Calculate $latex \Re (F(x)^2)$, and conclude $latex ||Hf||_{L^2} = ||f||_{L^2}.$
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