# Convergence of product of continuous functions and test functions

I suspect the following result is true but I"m not sure how to go about proving:

It is given that $\Omega \subset \mathbb{R}^{n}$ is an open bounded, connected domain.(Not sure if theses conditions on the domain are important)

If $\{v_{n}\}_{n} \subset C_{c}^{\infty}(\Omega)$ such that $v_{n} \rightarrow v$ in $L^{1}(\Omega)$. Does it follow that for $f \in C(\Omega)$($f$ is continuous) that $fv_{n} \rightarrow fv$ in $L^{1}(\Omega)$?

thanks for any assistance

• Look for a counterexample. Say with $v \equiv 1$. – Daniel Fischer Jun 1 '14 at 19:35
• Since $f$ is only continuous on the open domain it can grow arbitrarily fast at the boundary, even for $\Omega=(0,1)$. If $\Omega$ is compact this is true even for $f\in L^\infty$. – Conifold Jun 1 '14 at 19:56
• Exhaust $\Omega$ by compact sets $K_n$ and let $v_n$ be a cutoff function for $K_n$ ($v_n\lvert_{K_n} \equiv 1$, $0 \leqslant v_n(x) \leqslant 1$, and $v_n$ has compact support in $\Omega$). – Daniel Fischer Jun 1 '14 at 20:07
• This would be (rather trivially) true if $f\in L^\infty(\Omega)\)$, without any other assumption. – Giuseppe Negro Jun 1 '14 at 20:19
• Yes, that is a standard choice for the $K_n$. To construct the $v_n$, convolve the characteristic function $\chi_{K_{n+1}}$ of $K_{n+1}$ with a mollifier with small enough support. – Daniel Fischer Jun 1 '14 at 20:36