The Hardy-Littlewood maximal operator $M$ is defined by $$ M f(x)=\sup_{r>0}|B(x,r)|^{-1}\int_{B(x,r)} |f(y)|dy. $$
There is the following example in Grafakos's "Classical Fourier Analysis" book about this function. $$M \chi_{_{[a,b]}}(x)= \left\{ \begin{array}{ll} \frac{b-a}{2|x-b|} & x\leq a \\ 1 & a< x< b \\ \frac{b-a}{2|x-a|} & x\geq b \\ \end{array} \right. $$
It was written "it is clear" for the case "$M \chi_{_{[a,b]}}(x)=1$ when $x\in (a,b)$". Unfortunately i can't understand this fact. Can anyone write the proof of this fact more explicitly?
My other question: is that fact is true also for n-dimension, i.e., $M \chi_{_{B(z,t)}}(x)= 1$, when $x\in B(z,t)$.
Just recall that we have $$ \dfrac{1}{|B(x,r)|}\int_{B(x,r)} |f(y)|dy \leq \sup_{B(x,r)} |f(y)|\dfrac{1}{|B(x,r)|}\int_{B(x,r)} dy = \sup_{B(x,r)} |f(y)|\leq \sup_{\mathbb{R}^n}|f(y)|. $$
This holds for every $r>0$ and $x\in \mathbb{R}^n$ so that $M f(x)\leq \sup_{\mathbb{R}^n} |f(y)|$. Taking $f=\chi_{B(z,s)}$ gives that the maximal function applied to this $f$ is bounded above by 1. On the other hand if you pick $x\in B(z,s)$ and $0<r<s-|x-z|$, then $$ 1= \dfrac{1}{|B(x,r)|}\int_{B(x,r)} \chi_{B(z,s)}(y)dy\leq M \chi_{B(z,s)}(x). $$ Combining this implies that $M \chi_{B(z,s)}(x)=1$ for every $x\in B(z,s)$.