I am struggling to know whether the product function $fg$, where $f$ is in $L^2$ and $g$ is in $C^{\infty}_0$ belong to hardy space $H^1$. $fg$ has compact support but I can't figure out how I can try to find a atomic decomposition with cancellation property. Any suggestion will be greatly appreciated.
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1$\begingroup$ An $H^1$ function has zero mean, but there's no reason for $fg$ to have this property. $\endgroup$– Christian RemlingCommented May 21, 2016 at 5:45
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$\begingroup$ If by Hardy space you mean the complex Hardy space then obviously no, since $fg$ has no reason to be holomorphic; if you mean the real Hardy space then obviously yes, since locally the maximal function belongs to $L^2$, thus also to $L^1$. $\endgroup$– Alexander ShamovCommented May 21, 2016 at 6:34
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1$\begingroup$ Dear Alexander Shamov, thank you very much! I meant real hardy space $H^1(\mathbb{R}^n)$ Could you please clarify what maximal function do you use here? It seems there are several equivalent definitions for $H^1(\mathbb{R}^n)$ $\endgroup$– Dulguun DarkhanCommented May 21, 2016 at 15:50
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$\begingroup$ @AlexanderShamov: This is incorrect; you ignore the necessary (and, here, also sufficient) condition I mentioned in my comment for a function to belong to the real Hardy space (if you want to look at the maximal function, then this will not decay fast enough as $x\to\pm\infty$ unless $fg$ has zero mean). See here also: en.wikipedia.org/wiki/Hardy_space#Real_Hardy_spaces_for_Rn $\endgroup$– Christian RemlingCommented May 21, 2016 at 19:06
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$\begingroup$ @DulguunDarkhan: en.wikipedia.org/wiki/Hardy%E2%80%93Littlewood_maximal_function. Christian is right - there is a local version and a homogeneous version; if you need the latter one, it imposes the zero mean condition. $\endgroup$– Alexander ShamovCommented May 21, 2016 at 20:40
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This is obvious not sure, aמ obvious concentrate example is : The step function $f(x)$ of closed interval $[0,1]$,
It can be expressed as $\ f(x) = g(x)h(x)\\$, with $g(x) \in C^{\infty} (\mathbb{R})$ is a cut function, and $h(x)=f(x) \in L^2(\mathbb{R})$. But $f(x)$ itself is not in $H^1(\mathbb{R})$, because $f'(x) =\delta(1) - \delta(0)$, which is not in $L^2(\mathbb{R})$, in general, all step function in $\mathbb{R}^n$ are also counter examples.
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$\begingroup$ I am not really convinced. $fg$ is product not convolution. What do you mean by f' =delta(1) - delta(0). Is it derivative? Why f is not in H1(R) because of f' =delta(1) - delta(0). $\endgroup$ Commented May 21, 2016 at 5:05
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$\begingroup$ sorry!i misunderstanding the question at first,i lisp it and misunderstanding the space $H1$ to $W1,1$,but i should point out in the hardy space case this is also not true,the point is Hp(T) is the closed subspace of Lp(T) and f contain in Hp(T) iff f contain in Lp(T) and for any n negative integer ,the fourior series f(n)=0. and the fourior series of fg is convolution of f's and g's,so Hp is not closed under $\endgroup$– Hu xiyuCommented May 21, 2016 at 5:22