zeros of the dyadic maximal function

Question

Recall the definition of the Hardy-Littlewood maximal function $Mf$ (https://en.wikipedia.org/wiki/Hardy%E2%80%93Littlewood_maximal_function). If we replace the balls in the definition by dyadic cubes (cubes with side length of the form $(2^kn, 2^k(n+1))$, $k,n\in\mathbb{Z}$, $n$ may be different for different sides), then we get a so-called dyadic maximal function, denoted by $M_df$. My question is: is there an $f$ so that $M_df$ vanishes in a set of positive measure? (recall that $Mf$ is always positive for non-zero $f$).

Answer 1

Yes. Let $f:\mathbb{R}\rightarrow\mathbb{C}$ be a compactly supported, bounded function which vanishes on the interval $[-1/3,+\infty)$ and $f\equiv 1$ on the interval $[-2,-1]$. I claim that $M_{d}f(x)=0$ for $0\leq x\leq 1/3$.

Let $Q$ be a dyadic cube of generation $k$ containing $x\in [0,1/3]$. Then $Q$ necessarily has left endpoint $\geq 0$. Since $f$ is identically zero for $x\geq 0$, we see that $$\dfrac{1}{\left|Q\right|}\int_{Q}\left|f\right|=0$$ Taking the supremum over all such $Q$, we conclude that $M_{d}f(x)=0$.