bounded double sequence

Question

Let $(a_{n,m})_{n,m\in\mathbb N}$ be a real double sequence. Suppose that $$ \forall m\in\mathbb N\quad \exists \lim_{n\to\infty} a_{n,m} = \lambda_m \in \mathbb R \,,$$ $$ \exists \lim_{m\to\infty} \lambda_{m} = \lambda \in \mathbb R \,,$$ Can we conclude that $(a_{n,m})_{n,m\in\mathbb N}$ is bounded?

Answer 1

Assuming that I’ve interpreted the question correctly, let

$$a_{n,m}=\begin{cases} m,&\text{if }n=0\\ 0,&\text{if }n>0\,; \end{cases}$$

then $\langle a_{n,m}:n,m\in\Bbb N\rangle$ is unbounded, but $\lim\limits_{n\to\infty}a_{n,m}=0$ for each $m\in\Bbb N$, so of course $\lim\limits_{m\to\infty}\lim\limits_{n\to\infty}a_{n,m}=0$.

You can’t even guarantee that the double sequence is bounded by further requiring that $\lim\limits_{m\to\infty}a_{n,m}$ exist for each $n\in\Bbb N$: let

$$a_{n,m}=\begin{cases} n,&\text{if }n=m\\ 0,&\text{otherwise.} \end{cases}$$

Then $\lim\limits_{n\to\infty}a_{n,m}=0$ for each $m\in\Bbb N$ and $\lim\limits_{m\to\infty}a_{n,m}=0$ for each $n\in\Bbb N$, but the double sequence is unbounded.

Answer 2

No, not in general. Consider, for example: $$a_{n,m} = \begin{cases} n & \text{if } 1 \le n \le m \\ 0 & \text{if }n > m.\end{cases}$$ Then for any $m$, the sequence $a_{n,m}$ is eventually constantly $0$, hence: $$\lim_{n \to \infty} a_{n,m} = 0.$$ The limit of these limits is clearly $0$ too. But, $a_{n,m}$ is not bounded, as $a_{n,n} = n \to \infty$ as $n \to \infty$.