Let $V$ be a infinite dimensional Banach space over $\mathbb{C}$


Let $\{a\_{m,n} \cdot v\_{m,n}\}\_{m,n \in \mathbb{N}}$ be a double sequence with $a\_{m,n} \in \mathbb{C}$ and $v\_{m,n} \in V$ such that:
$$
\lim\_{m \to \infty}
\sum\_{n=1}^\infty
(a\_{m,n} \cdot v\_{m,n})
=
u
$$
and
$$
\forall n \in \mathbb{N}:
\lim\_{m \to \infty}
v\_{m,n}
=
u\_0
\neq
0
$$
I would like to know if is it true that $u= a \cdot u\_0$ with $a \in \mathbb{C}$



A double sequence in a Banach space
I think that the answer is "No", and the following
counterexample works in each space $V$ which is at least
two-dimensional. Let $u$ and $v$ be linearly independent vectors
in $V$.


We let $\{v\_{m,n}\}\_{n=1}^\infty$ be the sequence which starts
with $m$ vectors $u+v$, continues with $m$ vectors $u-v$, and all
further vectors are equal to $0$. Then the limits
$\lim\_{m\to\infty}v\_{m,n}$ are all equal to $u+v$.


Now let $\{a\_{m,n}\}\_{n=1}^\infty$ be a sequence which starts with
$2m$ coefficients $\frac1{2m}$ and continues arbitrarily.


It is clear that the sums $\sum\_{n=1}^\infty (a\_{m,n} \cdot
v\_{m,n})$ are all equal to $u$, and hence their limit as
$m\to\infty$ is equal to $u$. Since $u$ and $v$ are linearly
independent, we get the desired example.


