In the limited sequence $v_m \in \ell^2$ is $\sup_{n \in \mathbb{N}} \sup_{m \in \mathbb{N}} |v_{m,n}|<\infty$

Question

Let $v_m \in \ell^2$ be a sequence of $\ell^2$ over the field $\mathbb{C}$ such that $$\lim_{m \to \infty} v_m = v$$ Let $v_{m,n}$ be the $n$-th scalar of the sequence $v_m$

Is it true that $$ \sup_{n \in \mathbb{N}} \sup_{m \in \mathbb{N}} |v_{m,n}|<\infty $$

thanks.

Answer 1

Fix $\epsilon = 1$. Since $v_m \rightarrow v$ in $\ell^2$, then it follows there exists $N$ such that for all $m\geq N$ we have \begin{align} |v_{m, n}-v_n|\leq \sqrt{\sum^\infty_{k=1}|v_{m, k}-v_k|^2}=\|v_m - v\|_{\ell^2}<1 \end{align} which means \begin{align} |v_{m, n}| \leq 1+|v_n| \leq 1+\|v\|_{\ell^2}=:M_1. \end{align}

For all $m< N$, we see that \begin{align} |v_{m, n}| \leq \max_{1 \leq j < N}\|v_j\|_{\ell^2}=:M_2. \end{align} Thus, we have for all $n , m$ \begin{align} |v_{m, n}| \leq \max\left(M_1, M_2 \right). \end{align}