# Limit of sequence of vectors in $\ell^2$ with coefficients approaching $0$

Let $$\{v_m\}_{m \in \mathbb{N}} \subset \ell^2$$ be a sequence in $$\ell^2$$ over the complex plane $$\mathbb{C}$$ such that: $$\{v_m\}_{m \in \mathbb{N}}$$ is linearly independend and $$v_m \to v$$

Let $$V= \operatorname{span} \{v_m\}_{m \in \mathbb{N}}$$

Let $$\{u_p\}_{p \in \mathbb{N}} \subset V$$ be a sequence in $$V$$ such that $$u_p \to u$$ so we have $$\forall p \in \mathbb{N}: u_p = \sum_{m=1}^\infty \left( a_{p,m} \cdot v_m \right)$$ with $$a_{m,p} \in \mathbb{C}$$ and for each fixed $$p \in \mathbb{N}$$ there are only finitely many $$m$$ with $$a_{p,m} \neq 0$$

Further, we have for each fixed $$m \in \mathbb{N}$$ $$\lim_{p \to \infty} a_{p,m} =0$$ My question is if it is true that: $$\lim_{p \to \infty} u_p = a \cdot v$$ with $$a \in \mathbb{C}$$

Thanks.

• Not clear: if $u_p$ is an element of $V$, what is the meaning of the given equality with the sum of a series, a complex number. – Pietro Majer Jan 3 at 16:20
• it's because $V$ is the span of $\{v_m\}_{m \in \mathbb{N}}$ so $u_p$ is a finite linear combination. I write that way to create the matrix $a_{p,m}$ – Matey Math Jan 3 at 16:35
• That's certainly not working, take something like $v_n=e_1/n + e_n/n^2$, $u_n=nv_n$. – Christian Remling Jan 3 at 17:22
• @ChristianRemling in your definition of $u_n$ the coefficient $n$ doesn't approach 0 – Matey Math Jan 3 at 18:03
• @MateyMath: It does: in your notation, I have $a_{nn}=n$ and $a_{mn}=0$ otherwise in my example, so $\lim_{p\to\infty} a_{pn}=0$, as desired. – Christian Remling Jan 3 at 19:15