The Hardy-Littlewood maximal operator
$$Mf(x)=\sup\_{x\in B}\frac1{\vert B\vert}\int\_B\vert f(y)\vert dy$$
where the supremum is taken over all balls $B\subset\mathbb{R}^n$ which contain $x$. 


It is well-known that $M$ is both strong (for $p>1$) and weak-type (for $p\geq1$) integral operator; see for example [this paper and references therein](http://www.emis.de/journals/AASF/Vol28/cruz.pdf). It's a very nice operator.


Now, replace $M$ by a weighted-Maximal operator
$$M\_{\mu}f(x)=\sup\_{x\in B}\frac1{\mu(B)}\int\_B\vert f(y)\vert \,d\mu(y)$$
where the Gaussian measure $d\mu(y)=e^{-\vert y\vert^2/2}dy$ takes over the Lebesgue measure.


Things become bad. I believe the following should be true. Any idea for a proof or reference?



> 
> The operator $M\_{\mu}$ in not of weak-type $(p,p)$ for any $p\geq1$.
> 
> 
> 



is this weighted-maximal function unbounded?
It is bounded with constant depending only on $n$ and $p$ (apparently we get strong estimates). This is a result of Forzani et al (see [here](http://www.ams.org/journals/proc/2002-130-01/S0002-9939-01-06156-1/S0002-9939-01-06156-1.pdf)) which states the following:


**Theorem (L.Forzani, R. Scotto,
P. Sjogren & W. Urbina).**  

$M$ is a bounded operator on $L^p(d\gamma)
$ for $p > 1$, that is, there exists a
constant $C\_{n, p}$ such that for $f\in L^p(d\gamma)$


$$\|Mf\|\_{L^p(d\gamma)}\leq C\_{n,p}\|f\|\_{L^p(d\gamma)}.$$
(their measure is simply the Gaussian $d\gamma=e^{-|x|^2}dx$ with no factor 1/2 which doesn't bother anyone) 


