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!