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I am wondering how to imagine the Hardy space on $\mathcal{H}^1(\mathbb{R}^n)$ and in particular what sort of functions are in $L^1(\mathbb{R}^n)\backslash\mathcal{H}^1(\mathbb{R}^n)$. Furthermore, is it possible to find explicit examples of functions in $\partial (L^1(\mathbb{R}^n)\backslash\mathcal{H}^1(\mathbb{R}^n))$?

$\mathcal{H}^p(\mathbb{R}^n)=L^p(\mathbb{R}^n)$ for $p>1$, but what is the "problem"/ difference when $p=1$ ( or $p\leq 1)$?

Thanks!

David
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1 Answers1

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Let $f(x) \in L^1(\mathbb R^n)$ satisfy $\int_{\mathbb R^n} f(x) \, dx = C \ne 0$. If you are using the maximal characterization of $H^1$, then $$ Mf(x) := \sup_{r>0} \frac1{w_nr^n} \int_{y\in B(x,r)} f(y) \, dy \ge \frac1{2^nw_n|x|^n} \int_{B(x,2|x|)} f(y) \, dy \sim \frac C{|2x|^n} \text{ as $x \to \infty$}.$$ Here $w_n = |B(0,1)|$. Hence $Mf \notin H^1$.

The problem is that the inequality $\|Mf\|_p \le c_p \|f\|_p$ is only true if $p>1$.

Also $\partial(L^1\setminus H^1) = L^1$ because $L^1\setminus H^1$ is dense in $L^1$. This isn't really a meaningful question, because $L^1$ and $H^1$ have different topologies.

Stephen Montgomery-Smith
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  • Thanks for your answer. So $\mathcal{H}^1$ is quite a "small" space... – David Dec 14 '13 at 14:48
  • Professor Montgomery-Smith: Although any $f\in L^{1}(\mathbb{R}^{n})$ which satisfies $\int_{\mathbb{R}^{n}}f\neq 0$ clearly doesn't belong to $H^{1}(\mathbb{R}^{n})$ by the moment requirement, I don't see why your post answers the OP's question. It's my understanding that $H^{1}(\mathbb{R}^{n})$ isn't characterized by the Hardy-Littlewood maximal function as $Mf\notin L^{1}(\mathbb{R}^{n})$ if $f\neq 0$ a.e. – Matt Rosenzweig Nov 05 '15 at 16:49
  • @MattRosenzweig I think you are correct. Perhaps I should have used $Mf(x) \sup_{t>0} |P_t*f(x)|$ where $P_t$ is the Poisson kernel, or some other suitably smooth sequence of functions satisfying $P_t(x) = |t|^n P_1(t x)$. – Stephen Montgomery-Smith Nov 06 '15 at 04:10