How to show $\text{Given any sequence} (x^{(n)}) _{n\in\mathbb{N}}\text{ converges to } x \text{ in } (X ,|| •||), \text {X :finite dimensional NLS?}$

Question

$(X, ||• ||) $ be a $d$ dimensional normed linear space over $K$.

$\beta = \{e_1,e_2,e_3,...,e_d\} \text {be a basis of } X $.

Given any, $x\in X$ has a unique representation of the form

$x= x_1 e_1 + x_2 e_2 +... +x_n e_n (x_j \in K,\forall j\in \mathbb{N}_d ) $

Then, $(x_1, x_2, x_3,..., x_d) $ is defined to be the coordinate of $x$ with respect to $\beta$.

Question : $$\text{ Given any sequence} (x^{(n)}) _{n\in \mathbb{N}}\text{ and } x \text{ in } X $$

$(x^{(n)})$$ \text{ converges to } x \text { iff it converges Co-ordinatewise. }$

My attempt :

$(x^{(n)}) _{n\in \mathbb{N}} \text{ is sequence in } X $

$\text{ where the n-th term }$

$x^{(n)}=(x_{1}^{(n)},x_{2}^{(n)},..., x_{d}^{(n)}) $

And, $x= (x_1, x_2,..., x_d) $

And, $(x_{j}^{(n)})$ converges to $x_j$

$|x_{j}^{(n)} - x_j | < \frac{\epsilon}{\sum_{n=1}^{d}{||e_{j}||}} $ $(\forall n\ge N_{j} \text { and } j \in \mathbb {N}_d) $

To show, $(x^{(n) })_{n \in \mathbb{N}}$ converges to $x$ in $(X, ||•||) $ \begin{align} ||x^{(n) }- x || &=||\sum_{n=1}^{d}{(x_{j}^{(n)} - x_j) e_j ||} \\ &\le \sum_{n=1}^{d}{|(x_{j}^{(n)} - x_j)| \text{ } {|| e_j||}}\\ &< \epsilon \text { }(\forall n \ge max{\mathbb\{N_j\}} \end{align}

Hence, $(x^{(n) })_{n \in \mathbb{N}}$ converges to $x$ in $(X, ||•||) $

Is my proof of first part is correct?

How can I prove the reverse implication?

Thanks.

Answer 1

Let $\lim_{n\to\infty}\|x(n)-x\|=0$ where $x(n)=\sum_{j=1}^de_jx_{n,j}$ and $x=\sum_{j=1}^de_jx_j.$ For simplicity of notation, let $z(n)=x(n)-x=\sum_{j=1}^de_j(x_{n,j}-x_j)=\sum_{j=1}^de_jz_{n,j}.$ Then $\lim_{n\to\infty}\|z(n)\|=0$ and we wish to show that $\lim_{n\to\infty}z_{n,j}=0$ for each $j\le d.$

Suppose by contradiction that $z(n)$ does not converge co-ordinate-wise to $0.$ Take $r>0$ such that tne set $A=\{n\in\Bbb N:\max_{j\le d}|z_{n,j}|>r\}$ is infinite. For each $n\in A$ take some $j(n)\le d$ such that $|x_{n,j(n)}|=\max_{j\le d} |x_{n,j}|.$ Now $d$ is finite so we may take some fixed $j'\le d$ such that the set $B=\{n\in A: j(n)=j'\}$ is infinite.

Consider the sequence $S_B=(\,(|z_{n,j'}|^{-1}\cdot z_{n,1},...,|z_{n,j'}|^{-1}\cdot z_{n,d})\,)_{n\in B}=((w_{n,1},...,w_{n,d}))_{n\in B}$ of members of $\Bbb R^d$ or $\Bbb C^d.$

We have $\forall n\in B\,\forall j\le d\,(|w_{n,j}|\le 1).$ Therefore there exists an infinite $C\subseteq B$ such that $\lim_{n\to\infty ;n\in C}w_{n,j}=w_j$ exists for each $j\le d$. This is a finite-dimensional generalization of the fact that a bounded sequence in $\Bbb R$ or $\Bbb C$ has a convergent subsequence.

We have $|z_{n,j'}|>r$ for $n\in C$ so we have $$(\bullet)\quad 0=r^{-1}\lim_{n\to\infty ;n\in C}\|z_n\|=$$ $$=\lim_{n\to\infty ;n\in C}r^{-1}\|z_n|\ge$$ $$\ge \lim_{n\to\infty ;n\in C}|w_{n,j'}|^{-1}\|z_n\|=$$ $$=\lim_{n\to\infty ;n\in C}\|\sum_{j=1}^de_jw_{n,j}\|.$$ And the sequence $T=(\sum_{j=1}^de_jw_{n,j})_{n\in C}$ converges co-ordinate-wise to $w=\sum_{j=1}^de_jw_j$. So by the first part (proved in your Q), $T$ converges in norm to $w$. And by $(\bullet)$ we must have $w=0.$

BUT $w_{n,j'}=|z_{n,j'}|^{-1}z_{n,j'}$ so we have $|w_{j'}|=\lim_{m\to\infty ;n\in C}|w_{n,j'}|=1.$ Therefore $$0=w=\sum_{j=1}^dw_je_j$$ but at least one co-ordinate of $w$ (namely $w_{j'}$) is non-zero, and this contradicts the linear independence of $\{e_1,...,e_d\}.$

Remark: The linear independence of $\{e_1,...,e_d\}$ must be used eventually. Here it is used only at the finish.