Convergence almost every where ( times) convergence weak star. Let $(v_n)$ be a sequence in $L^\infty (0,1, L^\infty(\mathbb{R}))$ such that there exists $v\in L^\infty (0,1, L^\infty(\mathbb{R})) $   with 
$$v_n \to v \quad \text{almost every where}.$$
Let $u_n$ and $u$ in $L^\infty (0,1, L^2(\mathbb{R}))$ such that 
$u_n$ converges to $u$ in $L^\infty (0,1, L^2(\mathbb{R}))$ weak star. 
Would it be possible to prove that $v_nu_n$ converges to $vu$ in $L^\infty (0,1, L^2(\mathbb{R}))$ weak star? 
I tried using the definition of such convergence: Let $\phi \in L^1(0,1, L^2(\mathbb{R})).$ We want to prove that 
$$\int_0^1\int_{\mathbb{R}} v_nu_n \phi \to \int_0^1\int_{\mathbb{R}} vu \phi.$$ 
First thing that I wanted to use is the dominated convergence theorem but not directly; Writing 
$$v_nu_n \phi - vu\phi = (v_n-v)u_n \phi + (u_n-u)v\phi,$$ 


*

*We might use from the one hand the dominated convergence theorem for $(v_n-v)u_n \phi.$ 

*On the other hand, $v\phi \in L^1(0,1, L^2(\mathbb{R}))$ and $u_n$ converges to $u$ in $L^\infty (0,1, L^2(\mathbb{R}))$ weak star, hence 
$$\int_0^1\int_{\mathbb{R}} u_n v\phi \to \int_0^1\int_{\mathbb{R}} uv \phi.$$
I'm not totally sure about the point 1. Do the assumptions imply the point wise convergence of $(v_n-v)u_n \phi$ to zero ? 


Thank you in advanced 
 A: Without assuming that the sequence $u_n$ is bounded in $L^\infty(0,1,L^\infty(\mathbb{R}))$, then the result is false.
Take $v_n(x)=n\chi_{(0,1)}$ if $0<x<\frac1n$ and $v_n(x)=0$ otherwise. Then $v_n\to 0$ a.e. Take $u_n(x)=\chi_{(0,1)}$ for every $x$ and $n$. 
Then for $\phi(x)=\chi_{(0,1)}$ for every $x$,
$$\int_0^1\int_{\mathbb{R}}u_nv_n\phi\,dtdtx=n\int_0^{1/n}\int_{0}^1 1\,dtdtx=1\not\to 0.$$
If $u_n$ is bounded in $L^\infty(0,1,L^\infty(\mathbb{R}))$, then you should be able to apply the Lebesgue dominated convergence theorem.
Assume first that there exist $L, M>0$ such that for every $x$, $\phi(x)$ is a function in $L^2(\mathbb{R})$ with compact support in $[-M,M]$ and $\Vert\phi(x)\Vert_\infty\le L$. 
Then by Holder's inequality
$$\int_0^1\int_{\mathbb{R}}|u_n(v_n-v)\phi|\,dtdtx=\int_0^1\int_{-M}^M|u_n(v_n-v)\phi|\,dtdtx\\\le \int_0^1\left(\int_{-M}^M|(v_n-v)\phi|^2dt\right)\left(\int_{-M}^M|u_n|^2dt\right)dx\le C \int_0^1\left(\int_{-M}^M|(v_n-v)\phi|^2dt\right)dx,$$ 
since the sequence $u_n$ is bounded in $L^\infty(0,1,L^2(\mathbb{R}))$.
Now you can just apply Lebesgue dominated convergence theorem on the right-hans side.
The general case of $\phi$ follows by density.
