Some confusing basic inequalities on the Hardy-Littlewood Maximal Function

Question

I am a beginner in harmonic analysis and I already know the definition of Hardy Littlewood Maximum Function. Our next considerations are all focused on $f\in L^1\left( \mathbb{R} ^n \right) $, but we do not guarantee $f\in C\left( \mathbb{R} ^n \right) $. Recall the Hardy Littlewood Maximum Function is $Mf\left( x \right) =\mathop {\mathrm{sup}} \limits_{r>0}\frac{1}{\left| B\left( x,r \right) \right|}\int_{B\left( x,r \right)}{\left| f\left( y \right) \right|\mathrm{d}y}$.

Now, I have some confusion:

1. If $f\in L^{\infty}\left( \mathbb{R} ^n \right) $, which inequality is correct (or both): $\left\| Mf \right\| _{L^{\infty}\left( \mathbb{R} ^n \right)}\geqslant \left\| f \right\| _{L^{\infty}\left( \mathbb{R} ^n \right)}$ or $\left\| Mf \right\| _{L^{\infty}\left( \mathbb{R} ^n \right)}\leqslant \left\| f \right\| _{L^{\infty}\left( \mathbb{R} ^n \right)}$? Then if $f\in L^{\infty}\left( \mathbb{R} ^n \right) $ cannot be guaranteed, which inequality is correct?
2. Can $Mf\geqslant \left| f \right|$ a.e. be obtained without $f\in C\left( \mathbb{R} ^n \right) $?

My question basically comes from this reference: "LECTURE NOTES ON HARMONIC ANALYSIS" by Chengchun Hao. Specifically, we can refer to Theorem 3.2.2. and Proposition 3.2.3. I know they may be trivial, but I cannot fully understand these inequalities.

Answer 1

1. These two quantities are equal always.

2. Yes, this is true for all measurable $f$.

Let's handle 2) first.

For $f$ measurable, if $f\in L^1_{\text{loc}}(\mathbb R^n)$, this follows since almost every $x\in \mathbb R^n$ is a Lebesgue point of $|f|$, i.e. $$|f(x)| =\lim_{r\to 0}\frac{\int_{B(x,r)}|f(y)|dy}{|B(x,r)|}\leq Mf(x)\text{,}$$ and if $f\notin L^1_{\text{loc}}(\mathbb R^n)$, then $Mf(x)=\infty$ everywhere.

Now as for 1), from 2) we immediately have $$\left\| Mf \right\| _{L^{\infty}\left( \mathbb{R} ^n \right)}\geqslant \left\| f \right\| _{L^{\infty}\left( \mathbb{R} ^n \right)}\text{.} $$ However, since $|f(x)|\leq \|f\|_{L^\infty(\mathbb R^n)}$ a.e., then certainly its average on every ball is at most $\|f\|_{L^\infty(\mathbb R^n)}$, regardless of the ball, so we have $$\left\| Mf \right\| _{L^{\infty}\left( \mathbb{R} ^n \right)}\leqslant \left\| f \right\| _{L^{\infty}\left( \mathbb{R} ^n \right)}$$ as well.