Let $f \in L^1 (R)$. By $M f$ be the restricted maximal function defined by
$$(M f )(x) = \sup_{0 <t<1} \frac{1}{2t} \int_{x-t}^{x+t} |f(z)| dz.$$ Show that $M(f * g) \leq (M f) * (M g)$ for all $f, g \in L^1(R)$
My Proof
\begin{align} M(f * g)(x) &= \sup_{0<t<1} \frac{1}{2t} \int_{B(x,t)} |(f*g)(z)| dz\\ &= \sup_{0<t<1} \frac{1}{2t} \int_{B(x,t)} \left|\int_\mathbb{R} f(z-y)g(y) dy\right| dz\\ &\leq \int |g(y)| \left( \frac{1}{2t} \int_{B(x,t)} |f(z-y)|dz\right)dy\\ &\leq \int |g(y)| Mf(x-y)dy=(Mf*g)(x)\\ \end{align}
Similarly, we have
$$M(f * g)(x) \leq (f*Mg)(x).$$
Edit
We consider $M (M(f*g)) \leq M(Mf * g)$, and the right side becomes
\begin{align} M (Mf * g)(x) &= \sup_{0<t<1} \frac{1}{2t} \int_{B(x,t)} \left| \int Mf(y) g(x-y) dy \right| dx\\ & \leq \int |Mf(y)| \left( \sup_{0<t<1} \frac{1}{2t} \int_{B(x-y,t)} |g(x-2y)|dx \right) dy\\ & \leq 2 \int |Mf(2y) Mg(x-2y)| dy\\ &= (Mf * Mg)(x) \end{align}