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We denote $\mathcal D'(\mathbb R^n)$ the space of distributions, and $\mathcal D(\mathbb R^n)$ the space of smooth, compactly supported functions.

Let $\rho\in \mathcal D'(\mathbb R^n)$ such that for any function $\varphi \in \mathcal D(\mathbb R^n)$, the convolution $\varphi * \rho$ is a function in $L^p$ where $p<1$. For any $\varphi \in \mathcal D(\mathbb R^n)$ and $t>0$, we define $\varphi_t(x)=t^{-n}\varphi(x/t)$.

We suppose that for any $\varphi \in \mathcal D(\mathbb R^n)$ such that $\int_{\mathbb R^n} \varphi=1$, the family of functions $\rho_t=\varphi_t*\rho\in L^p$ is such that $\rho_t \to \tilde \rho$ in $L^p$ as $t\to 0$, where $\tilde \rho \in L^p$.

Since $\varphi_t$ is a sequence converging to $\delta_0$ in $\mathcal E'(\mathbb R^n)$ (the space of distributions with compact support), we also have $\rho_t \to \rho$ in $\mathcal D'(\mathbb R^n)$ as $t\to 0$.

Since $p<1$, in general $\tilde \rho$ does not define a distribution (it is not necessarily locally integrable). However, we would like to say that in some sense, $\tilde \rho = \rho$. This equality is easily deduced (in the sense of distributions) from Hölder's inequality when $p\geq 1$.

I have come across the notion of Hardy spaces $H^p$, which coincide with $L^p$ if $p>1$. In general, I know that for $f \in H^p$, there is an associated function $f_0$ such that for any $\varphi \in \mathcal D'(\mathbb R^n)$ with $\int_{\mathbb R^n} \varphi=1$, $\varphi_t*f(x) \to f_0(x)$ for almost every $x\in \mathbb R^n$.

I suspect then in my problem that $\rho \in H^p$, and that $\tilde \rho$ is an associated function. Is this statement true, or even plausible?

Thank you for your help!

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    $\begingroup$ The reason why $\rho_t\to \rho$ in $\mathcal{D}'$ is that $\varphi_t\to\delta_0$ in $\mathcal{E}'$ (distributions with compact support), not just in $\mathcal{D}'$ ... $\endgroup$ Aug 26, 2015 at 14:28
  • $\begingroup$ You are right, thank you for pointing that out! $\endgroup$
    – Thomas
    Aug 26, 2015 at 15:30

1 Answer 1

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The Hardy space $H^p$ is defined as the space of those $\rho\in\mathcal{S}'$ such that the maximal function $M_\rho(x):=\sup_{t>0}|\rho_t(x)|$ satisfies $\int M_\rho(x)^p\ dx<\infty$. This is stronger than just $\rho_t\to\tilde{\rho}$ in $L^p$, if $p<1$. If it is satisfied, what you can expect is that on open sets where $\tilde{\rho}$ is locally integrable, the restriction of $\rho$ coincides with (the distribution defined by) $\tilde{\rho}$. While I've never seen a proof of this, I'm pretty confident you could fabricate one (using the closed graph theorem, maybe, as $L^p$, while not locally convex, is metrizable and complete).

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