Maximal Function Estimate Suppose $\psi$ is a rapidly decreasing function; i.e. for all $N>0$ there exists a constant $C_{N}$ such that $\left|\psi(x)\right|\leq C_{N}(1+\left|x\right|)^{-N}$. Define a family of functions $\left\{\psi_{j,k}\right\}_{j,k\in\mathbb{Z}}$ by $\psi_{j,k}(x)=2^{j/2}\psi(2^{j}x-k)$. For each $j,k\in\mathbb{Z}$, let $D(j,k)=\lfloor{2+\left|2^{j}x-k\right|}\rfloor$, where $x\in\mathbb{R}$ is fixed.
In a paper that I'm reading, the author states that the following inequality is a consequence of the rapid decrease of $\psi$ and standard estimates on approximations to the identity:
$$\left|a_{jk}\right|=\left|\int_{\mathbb{R}}f(y)\overline{\psi_{j,k}(y)}dy\right|\leq C2^{-j/2}D(j,k)Mf(x), \ \forall f\in L^{p} (1<p<\infty)$$
where $C$ is some constant which depends only on $\psi$.
I have tried proving this estimate by using the rapid decrease hypothesis to get a continuous, integrable majorant of $\psi_{jk}$ and then using (Lebesgue-Stieltjes) integration by parts, but with no luck. For one thing, I'm not sure how to pick up a factor of $Mf(x)$ instead of $Mf(0)$. Any suggestions on how to proceed would be appreciated.
 A: So this is not as nice as I would have liked. I'm hoping somebody will come up with something cleaner.
Set $\delta:=(1+\left|2^{j}x-k\right|)$ for $j,k\in\mathbb{Z}$. Without loss of generality, assume $f\geq 0$. Observe that
$$\left|a_{jk}\right|\leq\int_{\left|2^{j}y-k\right|\leq\delta}f(y)\left|\psi_{jk}(y)\right|dy+\sum_{k=0}^{\infty}\int_{2^{k}\delta\leq\left|2^{j}y-k\right|<2^{k+1}\delta}f(y)\left|\psi_{jk}(y)\right|dy$$
By the triangle inequality, $\left|2^{j}(y-x)\right|\leq 2\delta$, whence the first term is majorized by
$$2^{j/2}\left\|\psi\right\|_{\infty}\int_{\left|y-x\right|\leq 2^{-j+1}\delta}f(y)dy\leq2^{-j/2+2}\delta\left\|\psi\right\|_{\infty}Mf(x)$$
Using the hypothesis that $\left|\psi(y)\right|\leq C_{N}(1+\left|y\right|)^{-N}$, we see that the general term in the series is majorized by
$$C_{N}2^{j/2}(1+2^{k}\delta)^{-N}\int_{\left|y-x\right|\leq 2^{k-j+2}\delta}f(y)dy\leq C_{N}2^{j/2}(1+2^{k}\delta)^{-N}2^{k-j+3}\delta Mf(x)$$
Take $N>1$ above. Since $\delta\geq 1$, we see that the series is majorized by
$$C_{N}2^{-j/2+3}\delta Mf(x)\sum_{k=0}^{\infty}(2^{N-1})^{-k}$$
Clearly, $\delta\leq\lfloor{2+\left|2^{j}x-k\right|}\rfloor$. Taking
$$C=\max\left\{4\left\|\psi\right\|_{\infty}, 8C_{N}\sum_{k=0}^{\infty}(2^{N-1})^{-k}\right\}$$
completes the proof.
