Show that $|\Phi_t \ast f| \leq AMf(x)$ for all $x \in \mathbb{R}^d$

Question

Let $d \geq 1$. Let $\{\Phi_t\}_{t > 0}$ be a collection of functions from $\mathbb{R}^{d}$ to $\mathbb{R}$ satisfying the following conditions for some constants $c_0, c_1, c_2 \in \mathbb{R}$, all $x \in \mathbb{R}^{d}$ and $t > 0$.

1. $\int_{\mathbb{R}^{d}} \Phi_t = c_0$,
2. $|\Phi_t(x)| \leq c_1 t^{-d}$,
3. $|\Phi_t(x)| \leq c_2t/|x|^{d + 1}$.

Show that there exists $A > 0$ such that $|(\Phi_t \ast f)(x)| \leq AMf(x)$ for all $x \in \mathbb{R}^{d}$ and $f \in L^1(\mathbb{R}^d)$.

Here $M$ denotes the Hardy-Littlewood maximal operator defined as

$Mf(x) = \sup\limits_{r > 0} \frac{1}{|B_r(x)|}\int_{B_r(x)}|f(y)|dy$

My work so far:

We have \begin{align*}|(\Phi_t \ast f)(x)| &= \left|\int_{\mathbb{R}^{d}}\Phi_t(y)f(x - y)dy\right| \\ &\leq \int_{\mathbb{R}^{d}} |\Phi_t(y)||f(x - y)|dy \end{align*} One thing from here that I can do is plit the integral into the region where $|y| > 1$ and $|y| \leq 1$ and then use (2) and (3) to get the following bound

$ |(\Phi_t \ast f)(x)| \leq (c_2t + c_1t^{-d})||f||_{L^1}. $

However I've now lost my dependence on $x$, which is a little troubling so I think this is not the way to go.

Another idea I had was to maybe aim for some kind of contradiction and suppose $Mf(x) < |(\Phi_t \ast f)(x)|$ for some $x$. However, this direction seems difficult as well. I would rather have a hint than a full solution.

Answer 1

Fix $x\in\mathbb R^d$ and $t>0$. Let $A_j=B(0,2^{j+1}t)\setminus B(0,2^jt)$ for $j\in\mathbb N$.

Note first that $$\int_{B(0,t)}|\Phi_t(y)||f(x-y)|\,dy\leq\int_{B(0,t)}c_1t^{-d}|f(x-y)|\,dy\leq c_1|B_1|Mf(x),$$ where $|B_1|$ is the volume of the unit ball in $\mathbb R^n$. Also, for any $j\in\mathbb N$, if $y\in A_j$ then $|y|>2^jt$, therefore $$|\Phi_t(y)|\leq\frac{c_2t}{|y|^{d+1}}\leq\frac{c_2}{2^{j(d+1)}t^d},$$ which implies that \begin{align*}\int_{A_j}|\Phi_t(y)||f(x-y)|\,dy&\leq\frac{c_2}{2^{j(d+1)}t^d}\int_{A_j}|f(x-y)|\,dy\leq \frac{c_2}{2^{j(d+1)}t^d}\int_{B(0,2^{j+1}t)}|f(x-y)|\,dy\\&\leq\frac{c_2}{2^{j(d+1)}t^d}|B(0,2^{j+1}t)|Mf(x)\leq c_22^{d-j}|B_1|Mf(x).\end{align*} Then, \begin{align*}|(\Phi_t*f)(x)|&\leq\int_{\mathbb R^d}|\Phi_t(y)||f(x-y)|\,dy\\ &\leq\int_{B(0,t)}|\Phi_t(y)||f(x-y)|\,dy+\sum_{j=0}^{\infty}\int_{A_j}|\Phi_t(y)||f(x-y)|\,dy\\&\leq c_1|B_1|Mf(x)+c_22^d|B_1|Mf(x)\cdot\sum_{j=0}^{\infty}2^{-j},\end{align*} so the inequality holds for $A=(c_1+c_22^{d+1})|B_1|$.