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On page 111 of Stein the maximal function defined as $\mathcal M_0 f(x) = \sup_{t>0} |f*\Phi_t(x)|$ (on page 106), where $\Phi$ is a smooth function supported in the unit ball about the origin with $\int \Phi \ne 0$, $\Phi_t(x) = t^{-n} \Phi(x/t)$. I don't understand how did he derive the inequality: $$\mathcal{M}_0 \sum_{far} b_k(x) \le c\alpha \sum_{far} \frac{l_k^{n+1}}{(l_k+|x-x_k|)^{n+1}}$$

I mean shouldn't the denominator be $|x-x_k|^{n+1}$, why did it change to $(l_k+|x-x_k|)^{n+1}$?

Thanks in advance.

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Recall that $Q_{k}$ is a cube with center $x_{k}$ and side length $l_{k}$, and $Q_{k}^{*}$ is the "concentric" enlargement of $Q_{k}$, where we expand the side length by a factor of $a^{*}$ (see the footnote on pg. 102). If $x\notin Q_{k}^{*}$, then

$$\left|x-x_{k}\right|\geq\dfrac{a^{*}l_{k}}{2},$$ which implies that

$$\left|x-x_{k}\right|+l_{k}\leq\left(1+\dfrac{2}{a^{*}}\right)\left|x-x_{k}\right| \tag{1}$$

and therefore

$$\left|x-x_{k}\right|^{-n-1}\leq\underbrace{\left(1+\dfrac{2}{a^{*}}\right)^{-n-1}}_{c=c(n)}\left(l_{k}+\left|x-x_{k}\right|\right)^{-n-1}$$

If you accept the estimate $$(\mathcal{M}b_{k})(x)\lesssim\alpha\dfrac{l_{k}^{n+1}}{\left|x-x_{k}\right|^{n+1}} \tag{2}$$

for a cube $Q_{k}^{*}$ which is far from the cube $Q_{m}\ni x$, then sublinearity of $\mathcal{M}_{0}$ together with (1) and (2) yield

$$\left(\mathcal{M}_{0}\sum_{\text{far}}b_{k}\right)(x)\lesssim c\alpha\sum_{\text{far}}\dfrac{l_{k}^{n+1}}{(l_{k}+\left|x-x_{k}\right|)^{n+1}}$$

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  • $\begingroup$ BTW Matt, on the same page 111 I am not sure I understand why does: $\sum_{near}|c_k|\mathcal M_0 \eta_k (x) \le c\alpha \frac{l_m^{n+1}}{(l_m +|x-x_m|)^{n+1}}$, I do see that by $(22)$ we get $|c_k| \le c\alpha$, but I don't see why does $\sum_{near} \mathcal M_0 \eta_k(x) \le \frac{l_m^{n+1}}{(l_m +|x-x_m|)^{n+1}}$, do you see why is that? $\endgroup$ Oct 17 '15 at 11:41
  • $\begingroup$ Matt, did you read my last comment? $\endgroup$ Oct 31 '15 at 12:59
  • $\begingroup$ Yes, at first look, it wasn't obvious to me why the inequality holds, and I don't have the time to read through the section right now. $\endgroup$ Oct 31 '15 at 13:07

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