The "local" Hardy-Littlewood maximal function is given by $$(M_\phi f)(x)= \sup_{0<\epsilon<1}|\phi_\epsilon \ast f|(x),$$ which is similar to the classical Hardy-Littlewood maximal function : $$(Mf)(x)= \sup_{\epsilon>0}|(\chi_B)_\epsilon \ast f|(x),$$ where $\phi\in S(\mathbb{R}^n)$ is some nonnegative, radial, radially decreasing Schwartz function and $\phi_\epsilon(x)=\epsilon^{-n}\phi(x/\epsilon)$.

The word "local" means that the supremum is only taken on small $\epsilon\in(0,1)$. It seems that this maximal function should have better properties than the classical one. For example, it is well known that if $f\in L^1$, then $Mf\notin L^1$ in general. However, we can easily check that for some $f\in L^1$, we do have $M_\phi f\in L^1$(e.g. $f(x)=\phi(x)=e^{-|x|^2}$). Are there any references(e.g. books, papers,...) on this "local" maximal function? Thanks!

  • $\begingroup$ I'm sure you know this, but since you only refer to the classical $M$, it's worth mentioning that $\sup_{\epsilon>0} |\phi_t * f(x)| \leq \|\phi\|_1Mf(x)$. I can't grasp how the supremum taken on small epsilon changes the facts, but maybe in the notes and further results of Chapter 2 of Duoandikoetxea's Fourier Analysis book you can find something useful. $\endgroup$ – Silvia Ghinassi Sep 23 '16 at 17:42
  • $\begingroup$ @SilviaGhinassi Thanks. Something may be different if the supremum is only taken on small $\epsilon$. For example, if $f\in L^1$, one can show that $Mf(x)\ge C|x|^{-n}, |x|>>1$ by taking $\epsilon\approx |x|$, e.g. math.stackexchange.com/questions/252595/… But this inequality can not hold if if the supremum is only taken on small $\epsilon$. $\endgroup$ – Mr.right Sep 23 '16 at 18:05

The failure of the maximal function to be a bounded map in $L^1$ is not just an issue of "failure at infinity".

Consider the case of $n = 1$. Let $f$ be a non-negative function that is compactly supported, and smooth away from the origin. Near the origin suppose that $$ f(x) = \frac{1}{|x| (\ln |x|)^2} $$ This function can be easily checked to be in $L^1$, as we can integrate, for $y> 0$ close to zero, $$ \int_0^y f(x) ~\mathrm{d}x = - \frac{1}{\ln y} $$

If you apply your local maximal function to $f$ (with the positivity assumption, WLOG, that $\phi(y) \geq c > 0$ whenever $|y| \leq 1$), you have that for $y\in (0,\epsilon)$ that $$ M_\phi f(y) \gtrsim - \frac{1}{y \ln y} $$ which implies that $M_\phi f(y)$ is not integrable near the origin.

The local properties of the maximal function (the classical one, but it applies almost identically to your version with the Schwartz weight) is remarked upon in Section 5.2, Chapter 1 of Stein's Singular Integrals and Differentiability Properties of Functions, where you can find the assertion $$ Mf [\log (2 + Mf)]^k \in L^1_{loc} \iff |f| [\log (2 + |f|)]^{k+1} \in L^1_{loc} $$ for any non-negative integer $k$.

  • $\begingroup$ Thanks! Your answer is very helpful. BTW, since $\frac1{|x|(\ln |x|)^2}$ is not integrable near $x=1$, you may need to multiply it by a characteristic function, e.g. $f(x)=\frac{\chi_{(0,1/2)}}{|x|(\ln |x|)^2}$. $\endgroup$ – Mr.right Sep 24 '16 at 3:49
  • $\begingroup$ @Mr.right: I wrote "Let $f$ be a non-negative function that is compactly supported, and smooth away from the origin. Near the origin suppose..." $\endgroup$ – Willie Wong Sep 25 '16 at 13:19
  • $\begingroup$ Thanks. I have one more question. If $f\in L^1\cap L^\infty$, do we have $M_\phi f\in L^1$? (Clearly, $M_\phi f\in L^p,\ p>1$). More generally, how much do we know about $f$ when $M_\phi f\in L^1$? $\endgroup$ – Mr.right Sep 26 '16 at 19:45

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