Verify the integral $latex \displaystyle \int_{\R^d}\frac{dx}{(|x|^2+1)^\frac{d+1}{2}} = \frac{\pi^\frac{d+1}{2}}{\Gamma(\frac{d+1}{2})}.$
Use Minkowski inequality to prove that, if $latex K\in L^1(\R^d)$ and $latex f\in L^p(\R^d)$, then $latex ||K*f||_{L^p}\le ||K||_{L^1}||f||_{L^p}.$
Prove that, if $latex f\in C_c(\R^d)$ and $latex y\in\R^d$, then $latex ||f(\cdot - ty) - f||_{L^p} \to 0$ as $latex t\to 0$.
Let $latex \Phi\in L^1(\R^d)$ with $latex \int \Phi = 1$, and $latex \Phi_t(x) = t^{-d}\Phi(x/t).$ Then
$latex \int \Phi_t = 1$ for all $latex t>0.$
There exists some $latex M>0$ such that $latex \int |\Phi_t| \le M$ for all $latex t > 0.$
For each $latex \delta>0$, $latex \displaystyle \int_{|x|\ge\delta} |\Phi_t(x)|dx \to 0$ as $latex t\to 0.$
If $latex f\in C_c(\R^d)$, then $latex \Phi_t*f(x) \to f(x)$ uniformly.
State conditions on $latex \Phi$ (as in the previous exercise) so that there exists $latex A>0$ such that, for any $latex t>0$ and $latex f\in L^1(\R^d),$ $latex |\Phi_t*f(x)| \le A Mf(x)$.
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