My question concerns the sharp maximal operator, mapping a locally integrable function $f\colon\mathbb{R}^{n}\to\mathbb{R}$ to $f^{\sharp}\colon \mathbb{R}^{n}\to\mathbb{R}$, where $$f^{\sharp}(x):=\sup_{Q\ni x}\frac{1}{|Q|}\int_{Q}|f-(f)_{Q}|dy,$$ the $\sup$ running over all cubes with sides parallel to the coordinate axes and containing $x$.
Similar to the Hardy-Littlewood maximal operator which maps $L^{1}$ to weak-$L^{1}$, but not $L^{1}$ to $L^{1}$, I guess that the sharp maximal operator does not map $L^{1}$ to $L^{1}$. The standard example for this fact with respect to the Hardy Littlewood maximal operator, namely, the indicator function of the unit cube in $\mathbb{R}^{n}$ does not seem to serve as a counterexample in this situation.
My question: Is there a similar "cheap" counterexample which shows that the sharp maximal operator does not map $L^{1}$ to $L^{1}$ - I would appreciate this very much! I've tried to find a solution to this issue in the web, but all references do not deal with this, so that I guess that it's either entirely trivial and I don't see it - or one does have a bound $\|f^{\sharp}\|_{L^{1}}\leq C \|f\|_{L^{1}}$.
However, referring to Variable Lebesgue Spaces of Cruz-Uribe & Fiorenza (Birkhaeuser 2013), p. 230, one has $\|f\|_{L^{p}}\leq c\|f^{\sharp}\|_{L^{p}}$ for all $f\in L^{p}$, $1\leq p<\infty$. Since $f^{\sharp}\leq C Mf$ with $Mf$ the Hardy-Littlewood maximal function pointwisely, one would obtain $\|f\|_{L^{p}}\leq c \|f^{\sharp}\|_{L^{p}}\leq C \|Mf\|_{L^{p}}\leq C'\|f\|_{L^{p}}$ provided $1<p<\infty$ which implies that the norms $\|\cdot\|_{L^{p}}$ and $\|(\cdot)^{\sharp}\|_{L^{p}}$ are equivalent, but somehow I cannot image that this remains valid for $p=1$ (which would be the case if one had a bound of $f^{\sharp}$ in $L^{1}$-norm against the $L^{1}$-norm of $f$).
Consider $n=1$ and $$ f\left(x\right)=\begin{cases} 0, & x\leq-1,\\ -1, & -1<x\leq0,\\ 1, & 0<x\leq1,\\ 0, & x>1. \end{cases} $$ For $x\in\mathbb{R}$ with $\left|x\right|\geq1$, we then have $x\in\left[-\left|x\right|,\left|x\right|\right]=:Q_{x}$ with $Q_{x}\supset\left[-1,1\right]$ which implies $\left(f\right)_{Q_{x}}=\left(f\right)_{\left[-1,1\right]}=0$ and thus \begin{eqnarray*} f^{\#}\left(x\right) & \geq & \frac{1}{\left|Q_{x}\right|}\int_{Q_{x}}\left|f\left(y\right)\right|\,{\rm d}y\\ & = & \frac{1}{2\left|x\right|}\cdot\int_{-1}^{1}\left|f\left(y\right)\right|\,{\rm d}y\\ & = & \frac{1}{\left|x\right|}, \end{eqnarray*} which easily implies $$ \int_{\mathbb{R}}f^{\#}\left(x\right)\,{\rm d}x\geq\int_{1}^{\infty}f^{\#}\left(x\right)\,{\rm d}x\geq\int_{1}^{\infty}\frac{1}{x}\,{\rm d}x=\infty $$ and hence $f^{\#}\notin L^{1}$, although clearly $f\in L^{1}$.
I leave the modification for $n>1$ to you.