A well-known feature used in PDE's is the regularization by convolution with a mollifier sequence $\rho\_n$, i.e. $\rho\_n(x) := n^d \rho(nx)$ with $x \in \mathbb R^d$, $\rho \in C^\infty\_c(\mathbb R^d)$ and $\int \rho = 1$. If $f\in L^p(\mathbb R^d)$ for $p\in[1,\infty)$, then $f^n := f \* \rho\_n \in C^\infty(\mathbb R^d)$ and $f^n \to f$ in $L^p(\mathbb R^d)$.


My question is about what do we know about the regularization in $x$ as above when we have functions $f=f(t,x)$, in time $t\in \mathbb R^+$ and space $x\in \mathbb R^d$. More precisely, if $f \in L^\infty(0,T; L^1(\mathbb R^d)) \cap C([0,T]; \mathcal D'(\mathbb R^d))$ and we define, for all $0\leq t\leq T$, $f^n(t,\cdot) := f(t,\cdot) \*\_{x} \rho\_n$, what do we know about $f^n$ and the convergence $f^n \to f$ ? How can we prove it ?


I think I know how to prove that $f^n \in C([0,T]; L^1(\mathbb R^d))$.



Regularization by mollifier sequence
If you assume $f \in L^\infty(0,T; L^1(\mathbb R^d))$ then you get convergence 
$$
f\_n\to f \text{ in } L^r(0,T;L^1(\mathbb{R}^d))
$$
 for all $r$.


This follows by Lebesgue dominated convergence: for a.e $t\in(0,T)$ you have pointwise convergence $||f\_n(t,.)-f(t,.)||\_{L^1}\to 0$ when $n\to\infty$, and the $L^1-L^1$ Young inequality for convolution gives a uniform bound $||f\_n(t,.)||\_{L^1}=||\rho\_n\*f(t,.)||\_{L^1}\leq ||\rho\_n||\_{L^1} ||f(t,.)||
\_{L^1}\leq 1\cdot C$, thus $||f\_n(t,.)-f(t,.)||\_{L^1}\leq C$ for all $t,n$. You can conclude that, for any $r>0$,
$$
||f\_n-f||\_{L^r((0,T);L^1)}^r=\int\_0^T||f\_n(t)-f(t)||^r\_{L^1}dt\to 0.
$$


