# Hardy-Littlewood maximal function of a Lipschitz function

In a book, it is said that Hardy-Littlewood maximal function of a Lipschitz function is also Lipschitz. How do we prove this?

+) For Hardy-Littlewood maximal function, see: http://en.wikipedia.org/wiki/Hardy%E2%80%93Littlewood_maximal_function

+) For Lipschitz function, see: http://mathworld.wolfram.com/LipschitzFunction.html

One way to do this is to use the subadditivity of the Hardy-Littlewood maximal function: $M(f + g)(x) \leq Mf(x) + Mg(x)$.
This can be used as follows. Suppose $f(x)$ is some Lipschitz function satisfying $|f(x + a) - f(x)| \leq K|a|$. Let $f_a(x) = f(x + a)$. By subadditivity you have $$Mf(x) \leq Mf_a(x) + M(f - f_a)(x)$$ But the Hardy-Littlewood maximal function of any function is bounded by its $L^{\infty}$ norm, so $M(f - f_a)(x) \leq K|a|$. Thus we have $$Mf(x) \leq Mf_a(x) + Ka$$ Similarly, the fact that $Mf_a(x) \leq Mf(x) + M(f_a - f)(x)$ gives that $$Mf_a(x) \leq Mf(x) + Ka$$ Taken together, the last two equations imply that $$|Mf_a(x) - M f(x)| \leq Ka$$ An examination of the defintion of the maximal function reveals that $Mf_a(x) = Mf(x+ a)$, so that we have $$|Mf(x + a) - M f(x)| \leq Ka$$ Thus the maximal function is also Lipschitz, and with the same constant. Notice that the same argument works if you're looking at $\alpha$-Lipschitz functions for some $0 < \alpha < 1$.