Please see this related question for the definition of the grand maximal function and the class of normalised test functions $\mathcal{T}$. I will refer to them in this question.
Let $f\in L^{\infty}(\mathbb{R}^n)$ with compact support in $\mathbb{R}^n$ and $\int_{\mathbb{R}^n}f \ \mathrm{d}x=0$. I would like to show that $f^{\ast}\in L^1(\mathbb{R}^n)$.
So far my efforts have been to try and show that for all $t>0, \phi\in\mathcal{T}$ we have \begin{equation}\tag1 |\phi_t\ast f(x)|\leq\frac{C}{(1+|x|)^{n+1}} \end{equation}for almost all $x\in\mathbb{R}^n$ and we note that the RHS is integrable in $\mathbb{R}^n$. Unfortunately, I didn't get very far. I tried to exploit the fact that $\|D\phi_t\|_{\infty}\leq t^{-n-1}$. Below is my work or rather the idea that I was trying to follow.
Let $t>0, x\in\mathbb{R}^n$ and $\phi\in\mathcal{T}$ be given. Suppose that $t\geq (1+|x|)^\frac{n+1}{n}$. Let $\zeta_t\in \partial B(x, t)$ and note that $\phi_t(\zeta_t-x)=0$. Then \begin{align*} |\phi_t\ast f(x)|&=\left|\int_{\text{supp } f}[\phi_t(x-y)-\phi_t(\zeta_t-x)]f(y)\ \mathrm{d}y\right|\\ &\leq\|f\|_{\infty}\left|\int_{\text{supp } f}\left[ \int_0^1 \frac{d}{d\tau}\phi_{t}\big(\tau (x-y)+(1-\tau)(\zeta_t-x)\big)\ \mathrm{d}\tau\right]\ \mathrm{d}y\right|\\ &\leq 2\|f\|_{\infty}|\text{supp }f|t^{-n}\\ &\leq C(1+|x|)^{-n-1}, \end{align*}which is fine for the case that $t\geq (1+|x|)^{\frac{n+1}{n}}$. I don't know how to prove for the case that $t< (1+|x|)^{\frac{n+1}{n}}$. My first attempt ends here.
My second attempt at a proof is via contradiction as follows:
If we suppose that $(1)$ is false then we must have (for a given $x\in\mathbb{R}^n$) that for all $m\in\mathbb{N}$ there exists a $\phi^m\in\mathcal{T}, t_m>0$ such that \begin{equation} \Psi_m(x)\equiv|\phi^{m}_{t_m}\ast f(x)|>\frac{m}{(1+|x|)^{n+1}} \end{equation} so as $m\rightarrow\infty$ we have $\Psi_m(x)\rightarrow \infty$ pointwise. However, we have that $\Psi_m(x)\leq \|f\|_{\infty}$ for all $m\in\mathbb{N}$ and $x\in\mathbb{R}^n$ since \begin{equation} \Psi_m(x)=\left|\int_{B(0, t_m)}\phi^m_{t_m}(z)f(x-z)\ \mathrm{d}z\right|\leq\|f\|_{\infty}. \end{equation}Thus we end up with a contradiction and hence $(1)$ holds true for some $C>0$ and all $x\in\mathbb{R}^n$.
