Let $\rho(x,t)>0$ be a Holder-continuous function on $\mathbb R^d\times(0,\infty)$. I know that:
- there exists (in weak sense) $\partial_t\rho\in L^1_\text{loc}(\mathbb R^d\times(0,\infty))$;
- there exists $\sigma\in L^1(\mathbb R^d)\cap C(\mathbb R^d)$ such that $$\int_{K} |\rho(x,t)-\sigma(x)|^p\,d x\to 0\quad \text{as}\ t\to0$$ for every $p\geq1$, $K$ compact subset of $\mathbb R^d$.
Extending $\rho(\cdot,t)$ by $\sigma$ for $t\leq 0$, can I say that there exists $\partial_t\rho\in L^1_\text{loc}(\mathbb R^d\times(-\infty,\infty))$ ? Of course the value of this derivative should be zero for $t<0$. The problem could be at $t=0$.
I'll give a rough sketch.
The answer to this is yes, and it basically boils down to the fact that $\sigma$ will be a trace in the functional analytic sense: Show that for every $\phi\in C_c^\infty(\mathbb{R}^{n+1})$ it holds $$ \int_{\mathbb{R}^n} \sigma(x) \phi(x,0)\, dx = -\int_{\mathbb{R}^{n+1}_+} [\phi(x,t) \partial_t\rho(x,t)+ \rho(x,t) \partial_t \phi(x,t)]\, dx dt. $$ Then the analogous thing will be true in the lower half space $\mathbb{R}^{n+1}_-$ for the function $\rho(x,-t)=\sigma(x)$ for $t>0$ (to define a trace in this setting only requires a derivative in the last variable).
Now use these identities to show that your extension of $\rho$ does have a weak derivative in $t$.