# A simple question about the Hardy-Littlewood maximal function

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:)

• Wasn't this answered in your previous question? – Anthony Quas Oct 6 '16 at 18:09
• Hi Anthony, the example given by Wong in the previous question is not in $L^\infty$. – Mr.right Oct 6 '16 at 18:11

$$f(x) := \sum_{m=2}^\infty 1_{[m, m + \frac{1}{m \log^2 m}]}.$$
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).
• More generally, let $a_n$ be a sequence of positive numbers such that $\sum a_n$ converges but $\sum a_n\log n$ diverges. Then $f(x)=\sum 1_{[n,n+a_n]}$ is a (counter)example. This example is just the superposition of (countably) many rescaled versions of the "nonintegrable at infinity" scenario that the OP had mentioned, and the scales become smaller and smaller so that even the ball of radius 1 looks not quite different from the entire space on these tiny scales. – Fan Zheng Oct 6 '16 at 18:31
• Thanks Terry. I considered this function before but I didn't know how to show $M_\phi f$ is not in $L^1$ after some attempts. Could you give more hints? – Mr.right Oct 6 '16 at 18:32
• Lower bound the $L^1$ norm of $M_\phi 1_{[m, m+\frac{1}{m \log^2 m}]}$ on $[m,m+1]$. – Terry Tao Oct 6 '16 at 18:35
• More precisely, between $[m + \frac{1}{m \log^2 m}, m+1]$, you have that $M_\phi f$ is bounded below by something that looks like $\frac{1}{(x-m)m \log^2 m}$. // Incidentally, this example also suggests some sort of hierarchy where $M_\phi$ only loses a log at infinity (similar to how it only loses a log locally). Is there a way to make this precise? – Willie Wong Oct 6 '16 at 18:37
• @WillieWong Could it be related to the scaling symmetry of $M_\phi$? – Fan Zheng Oct 6 '16 at 18:43