Suppose I have a distribution, $T_\varphi\in \mathcal{\mathcal{D}}'(\mathbb{R^n})$, which can be represented by a locally-integrable function, $\varphi:\mathbb{R}^n\to \mathbb{R}$. i.e., for any test function, $f\in \mathcal{D}(\mathbb{R}^n)$, $T_\varphi$ defines a linear mapping $T_\varphi:f\to\mathbb{R}$ via,
\begin{align} \langle T_\varphi,f\rangle=\int_{\mathbb{R}^n}d^{n}y\;\varphi(x)f(x). \end{align}
Further suppose that the wavefront set of $T_\varphi$ is such that the pointwise product, $\varphi(x)\varphi(x)$, is well-defined as a distribution everywhere except for a finite number of isolated points (i.e., "Hormander's condition" is satisfied everywhere except isolated points--see Thm 13.). For simplicity, let's suppose $\varphi^2(x)$ is everywhere distributionally well-defined except at $x=0$. In which case, I think the convolution of this distribution with itself,
\begin{align} (\varphi\star \varphi) (x)=\int_{\mathbb{R}^n} d^4y \;\varphi(y)\varphi(x-y), \end{align}
should be well-defined and finite everywhere except at $x=0$.
Question: I suspect Hormander's condition is sufficient to guarantee the convolution $(\varphi\star\varphi)(x)$ is finite everywhere $x\ne 0$; however, I would like to further determine where this convolution is smooth. Are there well-known conditions for determining where self-convolutions of distributions will be smooth?
Edit: For the integral of the convolution to be convergent, I probably also need to impose fall-off conditions on the distribution $\varphi(x)$. So, suppose this has also been imposed--e.g., I don't anticipate any issue with specializing to tempered distributions.