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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 $a \in \mathbb{C}$

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  • $\begingroup$ Is it true when $V$ is $1$-dimensional? If so, can that be used to prove your case? $\endgroup$
    – Gerald Edgar
    Commented Dec 21, 2017 at 11:20
  • $\begingroup$ @GeraldEdgar $V$ is infinite dimensiona, now i edit $\endgroup$
    – Matey Math
    Commented Dec 21, 2017 at 11:31
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    $\begingroup$ @GeraldEdgar: If V is 1-dimensional and $u_0\neq 0$, isn't it tautologically true that $u=au_0$ for some a? Don't you mean 2-dimensional rather than 1-dimensional? $\endgroup$
    – Michael Renardy
    Commented Dec 21, 2017 at 11:47
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    $\begingroup$ I don't think it is true, if $\dim V>1$: given any $u$ and $u_0$ linearly independent, take $a_{n,m}:=\delta_{n,m}$, and $v_{n,n}:=u$ for all $n$ and $v_{m,n}:=u_0$ for $n\neq m$. $\endgroup$
    – Pietro Majer
    Commented Dec 21, 2017 at 16:17
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    $\begingroup$ To get the conclusion you wish, i would apply a dominated convergence for series, assuming the needed hypotheses. $\endgroup$
    – Pietro Majer
    Commented Dec 21, 2017 at 16:29

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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.

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  • $\begingroup$ This is very close to @PietroMajer's comment, I think. $\endgroup$
    – LSpice
    Commented Dec 24, 2017 at 14:40
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    $\begingroup$ @LSpice You are right (I just checked the Pietro's comment). $\endgroup$
    – Mikhail Ostrovskii
    Commented Dec 24, 2017 at 16:23

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