# multipliers on $H^{1}$

I'm begining to study the hardy space $H^{p}(\mathbb{R}^{n})$. First recall that a $L^{\infty}$ function is called a $H^{1}$ multiplier if the associated operator $T_{m}(f)=\mathcal{F^{-1}}(m\hat{f})(x)$ is a bounded operator on $H^{1}$. I want to know how to prove that $$m_{1}(\xi)=\hat{\mu}(\xi)$$(where $\mu$ is the unit measure on $S^{n-1}$) $$m_{2}(\xi)=|\xi|^{i\beta}$$($\beta$ is a positive real number) are multipliers on $H^{1}$.

I think these are basic examples about $H^{1}$ multipliers, but I can't find its proof right now, so any comments or references are welcome

EDIT：similar to the case on $L^{p}$,there are many types(integral or differential form) of theorems to prove that a particular function belongs to $\mathcal{M}_{p}$(all the multipliers on $H^{p}$).Here is one:

if $m(\xi)\in C^{k}$,where $k=[\frac{n}{2}]+1$,satisfies $$|D^{\alpha}m(\xi)|\leq \frac{C}{(1+|\xi|)^{|\alpha|}},\quad |\alpha|\leq k$$ then m is $H^{1}$ multiplier.

Applying this to $m_{2}(\xi)$,we immediatly get the result.But for $m_{1}(\xi)$,since we have $\hat{\mu}(\xi)=J_{\frac{n}{2}-1}(\xi)|\xi|^{-(\frac{n}{2}-1)}$,and due to the asymptotic behavour of the bessel function,it seems the theorem above doesn't fit for this case.

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– sun Sep 10 '12 at 14:43
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