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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 $p=1$.

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  • $\begingroup$ 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
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    $\begingroup$ 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
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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.

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