Given $f \in L^1 (\mathbb R^d)$, and $\varepsilon > 0$, define the *minimal function* $m\_\varepsilon f$ by


$$m\_\varepsilon f(x) := \inf\_B \frac1{|B|} \int\_B |f| ,$$


where the infimum is taken over all balls $B$ containing $x$ of radius less than or equal to $\varepsilon$, and the integral is with respect to Lebesgue measure.


**Question:** Is it true that for every $\varepsilon > 0$, there exists a constant $C\_\varepsilon > 0$ depending only on $\varepsilon$ and the dimension $d$ such that for all $f \in L^1(\mathbb R^d)$,


$$\|m\_\varepsilon f\|\_{L^1} \geq C\_\varepsilon \|f\|\_{L^1}\text{?}$$


Further, is it true that the optimal constants $C\_\varepsilon$ converge to $1$ as $\varepsilon \to 0$?


*Remark: By the Lebesgue differentiation theorem we have $|m\_\varepsilon f| \leq |f|$ a.e., and hence $\|m\_\varepsilon f\|\_{L^1} \leq \|f\|\_{L^1}$. Thus if the conjectured result is true then we would have $m\_\varepsilon f \sim\_\varepsilon f$, with scale tending to $1$ as $\varepsilon \to 0$.*



Is the Hardy Littlewood “minimal function” comparable to the original function in $L^1$ norm?
This is not true. Take $\epsilon=1$ and, on the real line, $f\_\delta=\delta^{-1} \chi\_{(0, \delta)}$, so that $\|f\_\delta\|\_1=1$. Then $0 \leq m\_1 f\_\delta \leq \chi\_{(0, \delta)}$, by choosing small balls outside $(0, \delta)$ and balls of radius 1 otherwise. Thus $\|m\_1 f\_\delta\|\_1 \leq \delta$.


