# is this weighted-maximal function unbounded?

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. 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 $p\geq1$.

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) which states the following:
$M$ 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)
• Thanks, this paper is helpful. It might also be worthy noting that $M_{\mu}$ is not $L_p\rightarrow L_p$ bounded when $p=1$, **not even weak-type $(1,1)$ unlike its classical counterpart. This is one of the results in the references of the paper you cite. – T. Amdeberhan Dec 23 '16 at 2:56