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The inequality seems to be simple but I could not find the right limits of integration.

$$\sup_{\delta>0} |f*K_{\delta}|(x)\leq c f^*(x)$$

Where is some positive constant, $f$ is integrable, $K_\delta$ is an approximation of the identity and $f^*$ is the Hardy-Littlewood maximal function of $f$.

An approximation of the identity is family of Kernel satisfying:

I)$\int_{\mathbb{R}^n}K_{\delta}(x)dx = 1$;

II)$|K_{\delta}(x)|\leq A\delta^{-n}$;

III)$|K_{\delta}(x)|\leq A\delta /|x|^{n+1}$.

And the maximal function is the non-centered.

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You should include what definition of the approximation of identity you are using. Some people consider them to be the same as compactly supported mollifiers, others allow $K_\delta$ to be a scaling family of any $L^1$ function with mass 1, while in other cases the family $K_\delta$ doesn't even have to be scaling: it can be any set of functions indexed by $\mathbb{R}_+$ each with mass 1, uniformly bounded in $L^1$, and has $\lim_{\delta\to 0} \int_{\epsilon < |x|} |K_\delta|dx = 0$ for any $\epsilon > 0$. –  Willie Wong Feb 28 '12 at 8:52
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up vote 4 down vote accepted

It is enough to assume just II and III. Define $L_{\delta}(x) = \min(\delta^{-n}, {\delta \over |x|^{n+1}})$. Since $|K_{\delta}(x)| \leq L_{\delta}(x)$, you have $|f \ast K_{\delta}(x)| \leq |f| \ast L_{\delta}(x)$ and thus it suffices to show your statement for $L_{\delta}(x)$ in place of $K_{\delta}(x)$.

Observe that $$|f\ast L_{\delta}(x)| \leq \int_{{\mathbb R}^n} |f(x - y)|L_{\delta}(y)\,dy$$ $$\leq \int_{|y| \leq \delta} |f(x - y)|L_{\delta}(y)\,dy + \sum_{k = 0}^{\infty}\int_{2^k\delta \leq |y| \leq 2^{k+1}\delta}|f(x - y)|L_{\delta}(y)\,dy$$ In the first term $L_{\delta} = \delta^{-n}$ and thus the term is bounded by $cf^*(x)$. In the $k$th term of the sum, $L_{\delta}(y) \leq C{\delta \over (2^{k}\delta)^{n+1}} = C2^{-k} (2^k\delta)^{-n}$. Thus this term is bounded by $C'2^{-k}f^*(x)$. Adding over all $k$ gives a bound of $$cf^*(x) + C'\sum_{k=0}^{\infty}2^{-k}f^*(x) = C''f^*(x)$$ This is independent of $\delta$ so your statement follows.

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