Hardy Space Cancellation Condition I have been reading the chapter on Hardy spaces in Stein's Harmonic Analysis book, and I am having a lot of trouble figuring something out.  
The setting here is $\mathbb{R}^n.$  Let $f \in L^q$ be compactly supported, and let $\phi$ be a Schwartz function.  As usual, define $\phi_t(x) = t^{-n} \phi(t^{-1} x).$  Define the maximal function $M_{\phi}f(x)$ to be $\sup_{t > 0} | \phi_t * f(x)|.$  Stein claims that if we assume that $\int f = 0$, then $M_{\phi}(f)$ is less than or equal to $c | x|^{-n-1}$ for large $x$.  He says that the smoothness of $\phi$ and cancellation condition on $f$ are very important here.  
I have not been able to figure out why this is true.  I assume you have to integrate by parts and then use the fact that the gradient of $\phi$ is decreasing really quickly, but I can't seem to get the details to work out correctly.  Can someone please explain why we have this decay bound on $M_{\phi}(f)$?
 A: Let $B$ be the support of $f$. Pick $c_{b} \in B$. By the cancellation condition on $f$, we have
$$M_{\phi}(f)(y)=\sup_{t>0} \left|\int_{y+B} \left(\frac{1}{t^n}\phi\left(\frac{x}{t}\right)-\frac{1}{t^n}\phi\left(\frac{y+b_{0}}{t}\right)\right)f(y-x)\,  dx\right|$$
Using the mean value theorem, we get
$$\frac{1}{t^n}\phi\left(\frac{x}{t}\right)-\frac{1}{t^n}\phi\left(\frac{y+b_{0}}{t}\right)=\frac{1}{t^{n+1}}\phi'\left(\frac{x_{b}}{t}\right)$$
for some $x_{b} \in y+B$.
Since $\phi$ is Schwartz function, we have 
$$|\phi'(x)| \leq \frac{C}{1+|x|^{n+1}}$$
Now we can prove what we wanted.
$$\begin{split}
|M_{\phi}(f)(y)| &= \sup_{t>0}\left|\int_{y+B}\frac{1}{t^{n+1}}\phi'\left(\frac{x_{b}}{t}\right)f(y-x) \, dx\right| \\
& \leq \|f\|_{L^q}\sup_{t>0}\left(\int_{y+B}\left|\frac{1}{t^{n+1}}\phi\left(\frac{x_{b}}{t}\right)\right|^{p} \, dx\right)^{1/p} \\
& \leq \|f\|_{L^q}\sup_{t>0}\left(\int_{y+B}\left|\frac{1}{t^{n+1}+x_{b}^{n+1}}\right|^p \, dx\right)^{1/p} \\
& =\|f\|_{L^q} \left(\int_{y+B}\frac{1}{x_{b}^{np+p}}\, dx\right)^{1/p} \\ 
&\approx \|f\|_{L^{q}} \,m(B)^{1/p}\, \frac{1}{(y+c_{b})^{n+1}} \approx \frac{C}{y^{n+1}}
\end{split}$$
