Let $I=[0,1]$, $Q=I\times I$ and $(u_n),(v_n)$ bounded sequences in $L^2(I)$. Assume $x\mapsto u_n(x), x\mapsto v(x)$ are continuous and monotone non decreasing on $I$ for all $n\in\mathbb{N}$; define\begin{equation}f_n(x,y)=u_n(x)v_n(y), (x,y)\in Q.\end{equation}
Prove that $f_n$ lies in $L^2(Q)$ for every $n$ and that the sequence is relatively compact.
My attempt
Using Holders inequality and boundedness of $u_n,v_n$ we have \begin{equation}\|f_n\|_{L^2}= \int |u_n(x)|^2|v_n(x)|^2<\infty.\end{equation}
My idea for relative compactness is to use Kolmogorov-Riesz compactness theorem to extract $(u_{n_k})$ and $(v_{n_{k_l}})$ which converge in $L^1$. So that $f_{n_{k_l}}=(u_{n_{k_l}})(v_{n_{k_l}})$ converges.
Since $I$ has finite measure, $L^2\subset L^1$. Let $\epsilon>0$ and $\delta>0$: $|y|<\delta$ implies $|u_n(x-y)-u_n(y)|<\epsilon$. Then $\int_I|u_n(x-y)-u_n(x)|<\epsilon \mathcal{m}(I)=\epsilon, \forall n.$
I want to show there is $R$: $\int_{|x|>R}|f(x)|dx<\epsilon, \forall n$. So that by the Kolmogorov-Riesz there is a subsequence of (u_n) converging in $L^1$. I guess I have to use monotonicity.