I believe that the function $f$ below has a Hardy-Littlewood maximal function that is finite everywhere but unbounded. The definition of the Hardy-Littlewood maximal function I am using is: $f^{*}(x) = sup_{t>0} \frac{1}{2t} \int^{x+t}_{x-t} |f|$ where $f$ is a Lebesgue-measurable function.
Define $f:\mathbb{R} \rightarrow [0,\infty)$ to be the function that is zero everywhere except on $(1,1+1), (10,10+0.5),(100,100+0.25),(1000,1000+0.125),…$ and so on. On $(10^k,10^k + 2^{-k})$ define $f$ to be the constant $log(k)$. Denote $f$'s Hardy-Littlewood maximal function by $f^*$. Clearly for $x \in (10^k,10^k + 2^{-k})$ $f^*$ is at least $log(k)$ (just take $t$ small enough to stay on $(10^k,10^k + 2^{-k})$), so $f^{*}$ is unbounded.
I also believe $f^{*}(x)$ is always finite because the $(10^k, 10^k + 2^{-k})$ are quite far apart and the area under $f$ in each of these interval is only $2^{-k}log(k)$ so $ \int^{x+t}_{x-t} |f|$ grows much much slower than $\frac{1}{2t}$. For any $x$ and $t$, $t$ has to be increased by a factor of 10 to increase $\int^{x+t}_{x-t} |f|$ by a very small amount.
Am I correct?
Best Answer
It's true that with $f$ so-defined, $f^*$ is unbounded and finite almost everywhere. As you showed, $f^*$ is unbounded, and because the $f$ you gave is integrable, $f^* < \infty$ a.e. (This is a general result that follows by the Lebesgue differentiation theorem, but I believe you can give a direct proof with the function you defined here.)
Proof. If $x$ does not belong to the support of $f$, then there exists $\eta > 0$ so that $x$ is a distance $\eta$ from $\mathrm{supp}(f)$, so $f^*(x)\le (1/2\eta)\int_{\mathbb R}f$. If $x\in (10^k, 10^k + 2^{-k})$, then there exists $\delta > 0$ so that $f^*(x) = \sup_{t>\delta}\frac{1}{2t}\int_{x-t}^{x+t}f$ (because for $\delta$ sufficiently small, and $0<t\le\delta$, $\frac{1}{2t}\int_{x-t}^{x+t}f = \log k$). Therefore, $f^*(x)\le (1/2\delta) \int_{\mathbb R} f$. The $f$ you defined is integrable, so this is finite. This covers all $x$ except for a set of measure $0$, so $f^*<\infty$ a.e.
(Note that each $\eta,\delta$ above depends on the choice of $x$, but that's inconsequential.)