Given a measurable function $f : \mathbb{R}^n \to \mathbb{C}$, we can define the Hardy-Littlewood maximal operator $M$ by $$Mf (x) = \sup_{0 < r < \infty} \mu(B(x, r))^{-1} \int_{B(x, r)} | f | \mathrm{d} \lambda ,$$ where $\lambda$ here denotes Lebesgue measure. This function can be shown to be measurable for all locally integrable $f$. The classical maximal inequalities say that there exist constants $C_p \in (0, \infty)$ for $p \in [1, \infty]$ such that \begin{align*} \lambda \left( \left\{ x \in \mathbb{R}^n : M f (x) > t \right\} \right) & \leq \frac{C_1 \|f\|_1}{t} & \left( \forall f \in L^1 , t > 0 \right) , \\ \| Mf \|_p & \leq C_p \|f\|_p & \left( \forall p \in (1, \infty], f \in L^p \right) . \end{align*} My question is this:
Is there a nice characterization of the functions $f$ for which $M f \in L^\infty$, and in particular are there any such $f$ which aren't $L^\infty$?
That $M$ maps $L^\infty$ functions to $L^\infty$ functions follows from the strong type $(\infty, \infty)$ inequality, but more directly we could just note that $\lambda(B(x, r))^{-1} \int_{B(x, r)} | f | \mathrm{d} \lambda \leq \lambda(B(x, r))^{-1} \int_{B(x, r)} \| f \|_\infty \mathrm{d} \lambda = \|f\|_\infty$. Therefore $L^\infty \subseteq M^{-1} L^\infty$, but I'm unsure as of yet if the opposite containment holds, or how to construct an unbounded element of $M^{-1} L^\infty$ if it exists.
Any help is appreciated!
If $f$ is not in $L^\infty$ but in some other $L^p$ with $p < \infty$ then $Mf$ cannot be in $L^\infty$. Indeed $f$ is certainly $L^1_{loc}$ and for $L^1_{loc}$ functions it holds that almost every point is a Lebesgue point, i.e. you can choose a representative $\hat{f}$ of $f$ such that $\hat{f} (x) = \lim_{r \to 0^+} \lambda(B(x,r))^{-1} \int_{B(x,r)} f(y) d \lambda (y)$ where $\lambda$ is the Lebesgue measure (by the way, you used a $\mu$ for the definition of the maximal function, is this wanted? As far as I know you should work with the same measure you're integrating with). But then if $f \notin L^\infty$ there exists, for every $n$, a set of positive measure $A_n$ such that $ \hat{f}|_{A_n} \geq n$. By definition of the Hardy-Littlewood maximal function, $Mf (x)\geq \hat{f} (x) \geq n$ for almost every $x$ in $A_n$, and so $Mf \notin L^\infty$.
Have I understood correctly what you were asking? I mean that if the maximal function is in $L^\infty$ then also $f$ must be there, simply by the fact that in general $Mf \geq f$ almost everywhere due to the Lebesgue points property I mentioned.