Consider the following maximal function : in dimension $n$ consider $B(0,1)\subset \mathbb{R}^n$ the unit ball, if $f\in L^1(B(0,1))$, $\alpha\geq 0$ : $$ M_\alpha f : x \mapsto \sup \left\{ \frac{1}{r^n} \left( \int_{B(x,r)} |f| \right)^{1+\alpha} :r>0, B(x,r)\subset B(0,1) \right\} $$ Where $B(x,r) = \{ y\in \mathbb{R}^n : |x-y|<r\}$.
If $\alpha=0$, then $M_0 : L^1 \to L^{(1,\infty)}$ is bounded, where $L^{(1,\infty)}$ is the weak $L^1$ space. If $\alpha>0$, we still have a bound of the form $$ \| M_\alpha f\|_{L^{(1,\infty)}} \leq C_{n,\alpha} \|f\|_{L^1}^{1+\alpha} $$ The question is : can we improve this inequality if $\alpha>0$ ? More precisely, can we find a strict subspace $X\subset L^{(1,\infty)}$ such that $M_\alpha : L^1 \to X$ is bounded ?
I look for a norm $\|\cdot \|_X$ on $X$ which is strictly better than $\|\cdot \|_{L^{(1,\infty)}}$ in the sense that there exists $C>0$ satisfying $\forall f\in X,\ \|f\|_{L^{(1,\infty)}(B(0,1))} \leq C\|f\|_X$, and that $L^{(1,\infty)}\setminus X$ is not empty.
And $M_\alpha : L^1\to X$ is bounded means that there exists $C>0$ such that $\forall f\in L^1(B(0,1)),\ \|M_\alpha f\|_X \leq C\|f\|_{L^1}$, or maybe with a power $\forall f\in L^1(B(0,1)),\ \|M_\alpha f\|_X \leq C\|f\|_{L^1}^{1+\alpha}$, to have a homogenous inequality.