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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.

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  • $\begingroup$ 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. $\endgroup$ – Pietro Majer Jan 3 at 16:20
  • $\begingroup$ 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 $a_{p,m}$ $\endgroup$ – Matey Math Jan 3 at 16:35
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    $\begingroup$ That's certainly not working, take something like $v_n=e_1/n + e_n/n^2$, $u_n=nv_n$. $\endgroup$ – Christian Remling Jan 3 at 17:22
  • $\begingroup$ @ChristianRemling in your definition of $u_n$ the coefficient $n$ doesn't approach 0 $\endgroup$ – Matey Math Jan 3 at 18:03
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    $\begingroup$ @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. $\endgroup$ – Christian Remling Jan 3 at 19:15

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