Fdtxjr

There are four common versions of Hardy-Littlewood Maximal operator $M(f)$: centered/uncentered + ball/cube.
I noticed that the continuity of $M(f)$ depends on the version. For example, let $f$ be the characteristic function of the interval $[0,1]$. Then the centered version of $M(f)$ is not continuous while the uncentered version is continuous. (ball and cube are the same in 1-dim). My question is: for each version of $M(f)$, under what conditions for $f$ is $M(f)$ continuous?

Here is a partial answer:

Let $Mf(x)=sup_r>0frac1lambda(B(x,r))int_B(x,r)|f(y)|dy$ be the (centered) Hardy Littlewood Maximal Function.

Proposition: Let $fin L^1_loc(mathbbR^n)$. Then $Mf$ is lower semicontinuous, and in particular, measurable.

Proof: Lower semicontinuity means $E(t)=x: Mf(x)>t$ is open for all $tin mathbbR$. Let $xin E(t)$. Then $Mf(x)>t$, so there is an $r>0$ such that $Mf(x)geqfrac1lambda(B(x,r))int_B(x,r)|f(y)|dy>t$. If we choose $s>r$ small enough, then $$Mf(x)geqfrac1lambda(B(x,r))int_B(x,r)|f(y)|dy>frac1lambda(B(x,s))int_B(x,r)|f(y)|dy>t$$Then, for $z$ such that $|z-x|+r<s$, we will have $B(x,r)subset B(z,s)$, so $$Mf(z)geqfrac1lambda(B(z,s))int_B(z,r)|f(y)|dy>frac1lambda(B(x,s))int_B(x,r)|f(y)|dy>t$$ Thus $zin E(t)$, and the set of $z$ is open, so $E(t)$ is open.

Note: This is much easier to prove for the uncentered Hardy Littlewood Function.

Now, we are reduced to showing when $Mf$ is upper semicontinuous, since if a function is lower and upper semicontinuous somewhere, it is continuous. Searching hasn't gotten me much in terms of if and only if, but there is this result:

Proposition: If $fin L^1_loc(mathbbR^n)$ is continuous at some point $x$, then $Mf$ is also continuous at $x$.

The proof is found in this article: Lemma 7.3. A slightly different requirement is the following:

Proposition: If $fin L^1_loc(mathbbR^n)$ is upper semicontinuous at $x$ and $f(x)leq Mf(x)$, then $Mf$ is continuous at $x$.

The proof is in this paper (Lemma 3.4). Although $chi_[0,1]$ is upper semicontinuous everywhere, $Mchi_[0,1]$ is not continuous at $0$ or $1$. The theorem holds because $$chi_[0,1](0)=1>Mchi_[0,1](0)=frac12$$This provides a counterexample showing that we need the second condition.

Hope this helps as a partial answer!

In Folland's Real Analysis : Modern Techniques and Their Applications, it is proved that $A_rf(x)$ is jointly continuous in $r$ and $x$. Hence, because the preimage of any open set is open under $A_rf$ and that you can write $Hf^-1(a,infty)=cup_r>0 (A_r|f|)^-1(a,infty)$, you have that $Hf$ is lower semicontinuous. Same applies to prove upper semicontinuity.