# If a function is radial, then its Hardy-Littlewood maximal function is radial as well

I'm looking for a proof of the following statement:

$$f\in L(\mathbb R^n)\ \text{ radial } \implies f^* \ \text{ radial}$$

where $f^*$ is the Hardy-Littlewood maximal function defined by:

$$f^*(x)= \sup_{\{B\, :\, x\in B\}}\frac{1}{|B|}\int_B{|f(y)|dy}$$

I'm pretty sure that it is necessary using the invariance of the Lebesgue measure under rotations; but, unfortunately, I have some problem with the notations.

• I'm pretty sure that it necessary using the invariance of the Lebesgue-mesaure under rotations; but, unfortunately, i have some problem with the notations – Nicolò Oct 25 '14 at 22:06

To avoid losing track of notation, focus on two points $x$ and $x'$ with $|x|=|x'|$. The goal is to prove $f(x)=f(x')$. By symmetry it suffices to show $f(x)\le f(x')$. Since we are dealing with supremum, it's good to have an epsilon of room: I will show that $f(x)-\epsilon \le f(x')$ for every $\epsilon>0$.
Pick a ball $B$ containing $x$ such that $$\frac{1}{|B|}\int_B{|f(y)|dy}>f(x)-\epsilon \tag1$$ Let $O$ be an orthogonal matrix such that $Ox'=x$. Substitute $y = Oy'$ in the integral (1): $$\int_B{|f(y)|dy} = \int_{B'}{|f(y')|dy'}$$ where $B'=O^{-1}B$. (This relies on the change of variables formula, and $|\det O|=1$.)
Since $B'$ contains $x'$, it follows that $$f(x') \ge \int_{B'}{|f(y')|dy'}$$ which leads to the desired conclusion.