It is well known that the convolution of a $L^1$ function and a Schwartz function is also in $L^1$, by Young's inequality for convolution. Let $f\in L^1(\mathbb{R}^n)$ and $\phi\in S(\mathbb{R}^n)$, where $\phi$ is nonnegative, radial, and radially decreasing. Set $\phi_\epsilon(x)=\epsilon^{-n}\phi(x/\epsilon)$. My question is, whether the supremum of this convolution $${\rm sup}_{\epsilon>0}|f\ast \phi_\epsilon|\in L^1(\mathbb{R}^n)\ ?$$

Remark: We denote the Hardy-Littlewood maximal function of $f$ by $Mf$. Recall that $$ {\rm sup}_{\epsilon>0}|f\ast \phi_\epsilon|(x)\le Mf(x)\int_{\mathbb{R}^n} \phi dx.$$ The Hardy-Littlewood maximal theorem tells us that if $f\in L^p(\mathbb{R}^n)$, $1<p\le \infty$, then $ Mf\in L^p(\mathbb{R}^n),$ while if $f\in L^1(\mathbb{R}^n)$, then $Mf\notin L^1(\mathbb{R}^n)$, whenever $f\ne 0$ on some positive measure set. Therefore, we get $$f\in L^p(\mathbb{R}^n),\ 1<p\le \infty\Rightarrow{\rm sup}_{\epsilon>0}|f\ast \phi_\epsilon|\in L^p(\mathbb{R}^n).$$ However, we don't know whether it still holds for $p=1$.

  • $\begingroup$ Which operator is $M$? Please add this info in the question, it helps. $\endgroup$ – T. Amdeberhan Sep 18 '16 at 14:25
  • $\begingroup$ @T.Amdeberhan Done. $\endgroup$ – Mr.right Sep 18 '16 at 14:26
  • 1
    $\begingroup$ It is well known that the Hardy-Littlewood maximal inequality is not of strong type (1,1), by the fact that $Mf(x) \geq 1/|x|^{n}$. $\endgroup$ – Asaf Sep 18 '16 at 19:20
  • $\begingroup$ $Mf$ is "weak $(1,1)$" on $L_1$. $\endgroup$ – T. Amdeberhan Sep 18 '16 at 20:57
  • $\begingroup$ Probably related: math.stackexchange.com/questions/1930461/…. $\endgroup$ – PhoemueX Sep 19 '16 at 12:47

It is not difficult to see that a necessary (but not sufficient) condition is $\int f=0$.

The set of functions $f\in L^1(\mathbb{R}^n)$ such that this kind of maximal function that you defined is still in $L^1$ is known as the real Hardy space $H^1(\mathbb{R}^n)$ (not to be confused with the Sobolev space $W^{1,2}(\mathbb{R}^n)$, which is usually denoted in the same way). In particular, the space of such good functions is independent of the choice of $\phi$ (and actually you can take any function in the Schwartz space satisfying $\int\phi\neq 0$, with no need to require it to be nonnegative, radial, decreasing etc).

There are several equivalent characterizations of this space, which happens to be the "correct" replacement of $L^1(\mathbb{R}^n)$ for the purposes of harmonic analysis: notably singular integral operators satisfying reasonable assumptions map $H^1(\mathbb{R}^n)$ into itself (which as you probably know happens to hold for $L^p$ with $p\in(1,\infty)$ and to barely fail for $p=1$, since you just get a weak (1,1) estimate).

A very good reference for the basic theory concerning this space is the third chapter in the book Harmonic Analysis: Real-Variable Methods, Orthogonality, and Oscillatory Integrals, by Stein.


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.