Is the Hardy Littlewood “minimal function” comparable to the original function in $L^1$ norm?

Question

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 $\varepsilon \to 0$.

Answer 1

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 $\|m_1 f_\delta\|_1 \leq \delta$.