Let $x_n$ be a $n$-dimensional vector and $A_n$ be an $n\times n$ matrix. We can assume that both $x_n$ and $A_n$ are real-valued vector and matrix. I know that each component of $x_n$ is $o(1)$ uniformly, i.e. $\max_i \lvert x_{n,i}\rvert = o(1)$, where $x_{n,i}$ is the $i$th component of $x_{n}$. I also know that $A_n$ is uniformly bounded in column sum. I want to show that $x_n^{\prime}A_n = o(1)$.
I use the column sum norm $L_1$. We can show that $\lVert x_n^{\prime} A_n\rVert_1 \leq \lVert x_n^{\prime}\rVert_1 \lVert A_n\rVert_1$. As $\max_i\lvert x_{n,i}\rvert = o(1)$, we have $\lVert x_n^{\prime}\rVert_1 = o(1)$ and $\lVert A_n\rVert_1 = O(1)$ because $A_n$ is uniformly bounded in column sum. As a result $x_n^{\prime}A_n = o(1)$.
Is there a problem with my proof?
Can I do the same thing if the space is endowed with a probability measure? That is $\max_i x_{n,i} = o_p(1)$ and I will show that $x_n^{\prime}A_n = o_p(1)$.