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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?

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up vote 3 down vote accepted

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.

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A good lesson for me. sometimes the easiest way is getting back to the definition.Thanks very much. – user23078 Aug 29 '12 at 11:50

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