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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 $f^n \in C([0,T]; L^1(\mathbb R^d))$.

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  • $\begingroup$ Concerning this topic I have my own question :-P: How do you obtain $f_n\rightarrow f$ in the $L^\infty([0,T],L^1)$ norm without using the convergence $L^p([0,T],L^1)\rightarrow L^\infty([0,T],L^1)$? Is there an obvious way to do that? $\endgroup$
    – guacho
    Commented May 15, 2013 at 11:48

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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 $r>0$, $$ ||f_n-f||_{L^r((0,T);L^1)}^r=\int_0^T||f_n(t)-f(t)||^r_{L^1}dt\to 0. $$

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  • $\begingroup$ In the same spirit, if you assume that $f\in L^{s}(0,T;L^p)$ then you get convergence in the same space. $\endgroup$ Commented May 9, 2013 at 21:25

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