Let $\phi : \mathbb{R}^n \rightarrow \mathbb{R}_{+}$ be a $C^1$ function with $supp(\phi) \subset B(0,1)$ and $\int \phi = 1$. Define $$\phi_t(x) := t^{-n} \phi({x/t})$$ and set $$ M_{\phi} f(x) := \sup_{t > 0} |f \ast \phi_t (x)|$$ for locally integrable $f$. Now take a function $a$ which is an $\infty$-atom, i.e. with support contained in some cube $Q$ of radius (half the length of a side) $r$ and center $x_0$ and with $||a||_{\infty} \leq \frac{1}{|Q|}$ where $|Q|$ denotes the Lebesgue measure of the cube. I need to show that there exists a constant $c$ depending only on $n$ and $\phi$ such that for all $x \not\in 2Q$ (the cube twice as big as $Q$ and with the same center) we have $$ M_{\phi}a(x) \leq \frac{cr}{||x-x_0||^{n+1}}$$.
My attempt: previous point of the exercise shows that $M_{\phi}(f)$ can be bounded from above by the classical centred maximal function of Hardy Littlewood, hence I tried using that, but it didn't work out. If I use the maximal function, then using the fact the the $L^1$ norm of $a$ can be at most $1$ and taking $x$ far away from $x_0$ we can take a ball of radius $||x-x_0|| +r$, then the average over this ball will be something like $\frac{1}{||x-x_0||^{n}}$ which is much larger than the estimate given in the problem, at least for $x$ far away from $x_0$.
It sounds like you're not exploiting the zero mean of atoms and the compact support of $\phi$. First, observe that we can write the cube $Q$ as $Q=r[-1,1]^{n}+x_{0}$. I claim that it suffices to consider the case $r=1$ and $x_{0}=0$. Indeed, the maximal operator $M_{\phi}$ is translation invariant and dilation invariant (check this). So if we have proven the assertion for $\infty$-atoms supported in the cube $[-1,1]^{n}$, we can apply our result to $(\tau^{-x_{0}}a)_{1/r}$ to obtain \begin{align*} M_{\phi}(a)(x)=M_{\phi}(a)\left(r\dfrac{x-x_{0}}{r}+x_{0}\right)&=r^{-n}M_{\phi}(\tau^{-x_{0}}a)_{1/r}\left(\dfrac{x-x_{0}}{r}\right)\\ &\lesssim_{n,\phi}r^{-n}\left|\dfrac{x-x_{0}}{r}\right|^{-n-1}\\ &=\dfrac{r}{\left|x-x_{0}\right|^{n+1}} \end{align*} for all $\left|x\right|\notin 2Q$.
Using the hypothesis that $a$ has mean zero, we have the estimate $$\left|\int_{\mathbb{R}^{n}}a(y)\phi_{t}(x-y)dy\right|=\left|\int_{\mathbb{R}^{n}}a(y)\left[\phi_{t}(x-y)-\phi_{t}(x)\right]dy\right|\leq\int_{\mathbb{R}^{n}}\left|a(y)\right|\left|\phi_{t}(x-y)-\phi_{t}(x)\right|dy$$ Since $\phi$ is $C^{1}$ and compactly supported, we may use the mean value inequality to obtain the estimate for the RHS above $$\leq t^{-n}\int_{\mathbb{R}^{n}}\left|a(y)\right|\left\|\nabla\phi\right\|_{L^{\infty}}t^{-1}\left|y\right|dy$$ Since $\phi$ has compact support, we see that $\left|x-y\right|/t\leq C$, for some $C>0$. Since $a$ is supported in $Q$, the integrand vanishes for $y\notin Q$, and simple geometry show that $\left|x-y\right|\geq\left|x\right|/2$. Whence, for $x$ fixed, the integrand vanishes unless $$t\geq\dfrac{\left|x\right|}{2C}$$ Using this inequality, we conclude the estimate $$\left|(a\ast\phi_{t})(x)\right|\leq\dfrac{\left\|\nabla\phi\right\|_{L^{\infty}}(2C)^{n+1}}{\left|x\right|^{n+1}}\int_{\mathbb{R}^{n}}\left|a(y)\right|dy\leq\dfrac{\left\|\nabla\phi\right\|_{L^{\infty}}(2C)^{n+1}}{\left|x\right|^{n+1}},\quad\forall t>0$$ Taking the supremum of the LHS over all $t>0$, we conclude that $$M_{\phi}a(x)\leq\dfrac{\left\|\nabla\phi\right\|_{L^{\infty}}(2C)^{n+1}}{\left|x\right|^{n+1}}$$