Let $f\in L^1(\mathbb{R}^n)$. It is well known that the Hardy-Littlewood maximal function $Mf\notin L^1(\mathbb{R}^n)$ (if $f \ne 0$ a.e.), though there is a weak-type (1,1) bound for this maximal operator. More explicitly, we have a lower bound estimate $ Mf(x)\ge C|x|^{-n}$ (see e.g. https://math.stackexchange.com/questions/252595/lower-bound-for-the-hardy-littlewood-maximal-function-implies-it-is-not-integrab?rq=1) We consider a "local" maximal function defined by $$(M_\phi f)(x)= \sup_{0<\epsilon<1}|\phi_\epsilon \ast f|(x),$$ which is similar to the classical Hardy-Littlewood maximal function : $$(Mf)(x)= \sup_{\epsilon>0}|(\chi_B)_\epsilon \ast f|(x),$$
where $\phi\in S(\mathbb{R}^n)$ is some nonnegative, radial, radially decreasing Schwartz function and $\phi_\epsilon(x)=\epsilon^{-n}\phi(x/\epsilon)$. The word "local" means that the supremum is only taken on small $\epsilon\in(0,1)$.
It has been shown in How much do we know about this "local" Hardy-Littlewood maximal function? that $M_\phi f\notin L^1(\mathbb{R}^n)$ for general $f\in L^1(\mathbb{R}^n)$, though if $f$ is some "good function"(e.g. compactly supported bounded functions) we can easily see that $M_\phi f$ is integrable.
My question is: if we assume that $f\in L^1(\mathbb{R}^n)\cap L^\infty(\mathbb{R}^n)$, do we have $M_\phi f\in L^1(\mathbb{R}^n)$?
Remark: Since $f\in L^p$ for all $p>1$, by Hardy–Littlewood maximal inequality, we have $Mf\in L^p$, for all $p>1$. Then $M_\phi f\in L^p,\ p>1$, since $M_\phi f\le Mf$.
My attempt: Since $\phi$ is a Schwartz fucntion, I decompose the integral into three parts: $$\epsilon^{-n}\int |f(y)|(1+|x-y|/\epsilon)^{-N}dy=\int_{|y|\le|x|/10}+\int_{|y|\ge 10|x|}+\int_{|x|/10<|y|<10|x|}.$$
Since $f$ is bounded, we see that $M_\phi f$ is also bounded. So we only consider $|x|\gg1$. We see that the first two integrals are both bounded by $C\epsilon^{N-n}|x|^{-N}$, so we only need to consider the third one. If $f$ has good decay (e.g. $|f(y)|\le C|y|^{-\alpha}, \alpha>n$), we see that the third integral also has "good dacay", which makes $M_\phi f\in L^1(\mathbb{R}^n)$. But I don't know what to do for general $f$. Moreover, I don't know whether there is any simple counterexample for it. Any comments are welcome:)
Best Answer
No. Say we are in one dimension. Consider the function
$$ f(x) := \sum_{m=2}^\infty 1_{[m, m + \frac{1}{m \log^2 m}]}.$$
this function is (barely) in $L^1$ (because $\sum_{m=2}^\infty \frac{1}{m \log^2 m}$ converges), but $M_\phi f$ is (barely) outside of $L^1$ (basically because $\sum_{m=2}^\infty \frac{1}{m \log m}$ diverges).