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