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 $0 < \alpha < 1$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.