pointwise convergence and convergence in $\ell_{1}$ norm

Question

Assume that we have a sequence $\{\mathbf{x}_{n}\}_{n=1}^{\infty}$ of non-negative vectors in $\ell_{1}$, i.e. $x_{n,i}\geq 0$ for all $n = 1,2 \dots $ and $i= 1,2 \dots $. Next, assume that for any $i$ we have $$ x_{n,i} \to a_{i} $$ and $||\textbf{a}||_{1} = b >0$.

The statement: $\mathbf{x}_{n} \to \textbf{a}$ in $\ell_{1}$.

I have the following proof, but I am note 100 % sure:

Let's fix some $\delta>0$. Then there exists $M$ such that $$ \sum_{i\leq M}a_{i} \geq b - \frac{\delta}{2}. $$

Since $\mathbf{x}_{n}$ converges to $\mathbf{a}$ pointwise, then there exists $n_{1}$ such that for all $n\geq n_{1}$ we have $$ \sup_{i\leq M}|x_{n,i} - a_{i}| \leq \frac{\delta}{2M}. $$ From those two inequalities above, it follows that $\sum_{i\leq M}x_{n,i} \geq b - \delta$, for all $n\geq n_{1}$.

Therefore, using inequalities above, we have $$ \sum_{i}|x_{n,i} - a_{i}| = \sum_{i\leq M}|x_{n,i} - a_{i}| + \sum_{i>M}|x_{n,i} - a_{i}| \leq \sum_{i\leq M}|x_{n,i} - a_{i}| + \sum_{i>M} x_{n,i} + \sum_{i>M}a_{i} \leq \frac{\delta}{2} + \delta + \frac{\delta}{2} = 2 \delta $$ holds for all $n\geq n_{1}$, which proves the oroginal statement.

edit: I see the problem now. In order to have the last inequality, we must also have that $||\textbf{x}_{n}||_{1} = b$, for all $n$.

hypothesis: I think, the statement is also true when we just have $\sup_{n}||\textbf{x}_{n}||_{1} < \infty$

Answer 1

As you mentioned in your edit, the proof will be correct if $\|\mathbf{x}_n\|_1 = b$ for all sufficiently large $n$. However, there is a more general condition under which it will hold:

Suppose that for every $\epsilon > 0$, there exists an $N \in \mathbb{N}$ such that

$$\sup_n \sum_{k=N}^\infty |\mathbf{x}_{n,k}| < \epsilon.$$

(I believe this is the $\ell_1$ analogue of uniform integrability). Then your proof works if we also assume $M$ is large enough that the above equation holds for $\epsilon = \delta$ and $N = M$.

Edit: This is actually necessary and sufficient. Suppose $\{\mathbb{x}_n\}$ does not satisfy the condition above. Then there must exist some $\epsilon > 0$ and sequences $n_i,N_i \to \infty$ such that $\sum_{k=N_i}^\infty |\mathbf{x}_{n_i}| > \epsilon$ for all $i$.

Then $x_{n_i}$ cannot converge to $a$. To see why, let $M$ be large enough that $\sum_{k=M}^\infty |a_k| < \frac{\epsilon}{2}$. Then,

$$\sum_{k=1}^\infty |\mathbf{x}_{n_i,k} - a_k| \geq \sum_{k=M}^\infty |\mathbf{x}_{n_i,k} - a_k| \geq \sum_{k=M}^\infty |\mathbf{x}_{n_i,k}| - \sum_{k=M}^\infty|a_k| > \frac{\epsilon}{2}.$$

Since this holds for infinitely many $i$ (all $i$ such that $N_i > M$), $\mathbf{x}_n$ does not converge to $a$ in $\ell^1$.

Answer 2

This is not true in general. Suppose $a_1=1$ and $x_{1,n}=1$ for all $n\in\mathbb N$. Let $a_i=0$ and $x_{i,n}=\frac{1}{n}$ for all $n\geq 1$ and $2\leq i\leq n$. For $i\geq n$ let $x_{i,n}=0$. Clearly $x_{i,n}\to a_i=0$, and $\|a\|_1=1>0$. However, for any $n\in \mathbb N$ $$\|x_n-a\|_1=\frac{1}{n}\sum_{i=2}^n1=\frac{n-1}{n}\to 1.$$