For $f\in L^1(\mathbb{R})$, let us denote the restricted maximal function $Mf$ as $$(Mf)(x)=\sup_{0<t<1}\frac{1}{2t}\int_{x-t}^{x+t}|f(z)|\,dz.$$ I would like to show that $$M(f*g)\le M(f)*M(g)$$ for all $f, g\in L^1(\mathbb{R})$.
I computed $$M(f*g)(x) =\sup_{0<t<1}\frac{1}{2t}\int_{x-t}^{x+t} \left|\int_{\mathbb{R}} f(z-y)g(y)\,dy \right|\,dz\le \sup_{0<t<1}\frac{1}{2t}\int_{x-t}^{x+t} \int_{\mathbb{R}} |f(z-y)||g(y)|\,dy\,dz.$$ A change of variables shows that $$M(f*g)(x) \le \sup_{0<t<1} \frac{1}{2t}\int_{\mathbb{R}} \int_{-t}^{t} |f(z+x-y)||g(y)|\,dz\,dy= \frac{1}{2}\sup_{0<t<1} \int_{\mathbb{R}} \int_{-1}^{1} |f(tz+x-y)||g(y)|\,dz\,dy$$ On the other hand, we have that \begin{aligned} (Mf)*(Mg)(x) &= \int_{\mathbb{R}}Mf(x-y)Mg(y)\,dy \\ &= \int_{\mathbb{R}} \left(\sup_{0<t<1}\frac{1}{2t}\int_{x-y-t}^{x-y+t}|f(z)|\,dz\right)\left(\sup_{0<t'<1}\frac{1}{2t'}\int_{y-t'}^{y+t'}|g(z')|\,dz'\right)\,dy \\ &= \sup_{0<t, t'<1} \frac{1}{4tt'} \int_{\mathbb{R}}\int_{-t}^{t}\int_{y-t'}^{y+t'} |f(z+x-y)| |g(z')|\,dz'\,dz\,dy. \end{aligned} It is starting to look promising but I am not sure where to go from here.