Let $\mu\_{n}$ be the unit measure over $S^{n-1}$,and consider the convolution operator$$Tf=\mu\_{n}\ast f,\quad f\in \mathcal{S}$$
then,it's well-known that T can be extend to a bounded operator on $L^{1}$.My question is whether it's bounded from $H^{1}$ to $H^{1}$ ?


The question is equivallent to say that whether $m(\xi)=\hat{\mu\_{n}}$ is a $H^1$ multiplier(I also care about $\frac{d}{dr}m$,where $r=|\xi|$)or not.The theorem related to this is that(the most convenient one I know so far) if $\mathcal{F}^{-1}m$ has compact support,and $$|m(\xi)|\leq (1+|\xi|)^{-b},\quad b>0$$
then m is a $H^p$ multiplier,where $\frac{1}{p}-\frac{1}{2}=\frac{b}{n}$.In our case $p=1$,$b=\frac{n}{2}$.On the other hand,we also know that $\hat{\mu\_{n}}=J\_{\frac{n}{2}-1}(|\xi|)|\xi|^{-\frac{n}{2}+1}$,so $\hat{\mu\_{n}}\sim |\xi|^{-\frac{n-1}{2}}$ when $|\xi|$ large.the problem is that this theorem cann't be applied since we need the decay of $|\xi|^{-\frac{n}{2}} $,and I don't know how to do with it.So are there other mathods to prove and disprove it?



Convolution operators defined by compactly supported distribtion
The Hardy space $H^1(\mathbb R^n)$ is equal to 
{$u\in L^1(\mathbb R^n),\forall j, R\_ju\in L^1(\mathbb R^n)$},
where
$R\_j$ are the Riesz operators (Fourier multiplier $\xi\_j/\vert\xi\vert$)
and the following norm is equivalent to the $H^1$ norm:
$$
\Vert u \Vert\_{H^1}=\Vert u \Vert\_{L^1}+\sum\_{1\le j\le n}\Vert R\_ju \Vert\_{L^1}.
$$
Let $T$ be a Fourier multiplier (thus commuting with the $R\_j$) bounded on $L^1$ : then for $u\in H^1$
$$
\Vert Tu \Vert\_{H^1}=\Vert Tu \Vert\_{L^1}+\sum\_{1\le j\le n}\Vert R\_jTu \Vert\_{L^1}
=\Vert Tu \Vert\_{L^1}+\sum\_{1\le j\le n}\Vert TR\_ju \Vert\_{L^1}\lesssim
\Vert u \Vert\_{L^1}+\sum\_{1\le j\le n}\Vert R\_ju \Vert\_{L^1},
$$
and the last term is $\Vert u \Vert\_{H^1}$, proving the $H^1$ boundedness.


