# Pointwise inequality between a function and its fractional maximal function

Let $0\le\alpha<n$. The fractional maximal operator $M_{\alpha}$ is defined by $$M_{\alpha} f(x)=\sup_{t>0}\frac{t^{\alpha}}{|B(x,t)|}\int _{B(x,t)} |f(y)|dy,$$ $B(x,r)$ is the ball in $\mathbb{R}^n$ and $|B(x,r)|$ is its Lebesgue measure.

If $\alpha=0$, we have the classical Hardy-Littlewood maximal operator.

By the help of Lebesgue differentiation theorem we can show that (see, for example I want to show that $|f(x)|\le(Mf)(x)|$ at every Lebesgue point of $f$ if $f\in L^1(R^k)$)

$$|f(x)|\le Mf(x),\quad \text{a.e. } x\in\mathbb{R}^n.$$

I wonder that is there an analogue of this inequality for the fractional maximal operator?

The fractional maximal function $M_\alpha$ does not provide any pointwise control of $f$. For example, let $f(x) = \log^+|x|$ (in one dimension), then $M_\alpha f(0)$ is $$\sup_{0<t\le 1} \frac{t^\alpha}{2t} \int_{-t}^t -\log x\,dx = \sup_{0<t\le 1} t^\alpha(1-\log t) < \infty$$ (one can compute the maximum explicitly, but it suffices to note that $t^\alpha(1-\log t)$ has a finite limit at $0$.) Meanwhile, $|f(0)|=\infty$.