We consider the equation $$ \begin{cases} \partial_t u + Lu =0 & \hbox{ in } \Omega\times[0,T] \\ u = f & \hbox{ on } \Omega\times \{0\} \\ u = g & \hbox{ on } \partial\Omega\times [0,T] \end{cases} $$ where $\Omega\subset\mathbb{R}^n$ is a smooth bounded domain, and $$ L = -\sum_{i,j=1}^n \partial_{x_i}( a_{ij}(x)\partial_{x_j}) + \sum_{k=1}^n b_k(x)\partial_{x_k} + c(x) $$ where $a_{ij},b_k,c\in C^\infty(\overline\Omega)$ and $$ \sum_{i,j=1}^n a_{ij}(x)\xi_i\xi_j\ge \lambda |\xi|^2 $$ for some $\lambda>0$, for all $\xi\in\mathbb{R}^n$ and $x\in\Omega$.
What conditions must we impose on $f,g$ to have $u\in C^\infty(\Omega\times (0,T])$? A modification of arguments from Evans PDE gives that $f,g\in C^\infty$ suffices, for instance by taking $h$ on $\Omega\times[0,T]$ such that $L h=0$ and $h=g$ on $\partial\Omega\times[0,T]$ and using that $$ \partial_t (u-h) + L(u-h) = -\partial_th \quad\hbox{ on }\Omega\times[0,T] $$ and that $u-h$ has homogeneous Dirichlet boundary conditions. Looking at the heat equation, I would expect $u\in C^\infty(\Omega\times(0,T])$ even when we only have $f,g\in L^1$, but is there a good way to prove this?