If $\lim\limits_{i\to +\infty}\|x_i^1e_1+\dots+x_i^ne_n\|=0$ then $\lim\limits_{i\to +\infty}x_i^j=0,\forall j=1,\dots,n$?

Question

Trying to prove another thing, I realized that the following property would suffice to prove it:

Let $(E,\|\|)$ be a normed $\Bbb{K}$-vector space and $\{e_1,\dots,e_n\}\subset E$ be linearly independent. Consider $n$ scalar sequences $\{x^j_i\in \Bbb{K}\}_{i=1}^\infty$, $j=1,\dots,n$ such that $$\lim_{i\to +\infty}\|x^1_ie_1+\dots +x^n_ie_n\|=0.$$ Show that $\lim\limits_{i\to+\infty}x_i^j=0,\,\, \forall j=1,\dots,n$. Here $\Bbb{K}\in \{\Bbb{R},\Bbb{C}\}$.

It seems very plausible this fact to be true, but I have no ideia how to prove it:

• Triangle inequality seems to "fail" since $\|x^1_ie_1+\dots +x^n_ie_n\|\leq |x^1_i|\|e_i\|+\dots+|x^n_i|\|e_n\|$ does not help, because it is the left side which goes to zero;
• The fact is obviously true if $n=1$, but I could not see how to use induction over $n$ in this case...

Is this "fact" really true? Any hint?

Answer 1

Define for $(x_1, \ldots, x_n) \in \mathbb{R}^n$ the function $\|(x_1, \ldots, x_n)\|_E := \|x_1e_1 + \cdots + x_n e_n\|$. It is easy to see that $\| \cdot \|_E$ actually satisfies all the properties of a norm. Since $\mathbb{R}^n$ is a finite-dimensional space, all norms are equivalent. Therefore there exists a constant $C > $ so that for all $(x_1, \ldots, x_n) \in \mathbb{R}$ the inequality $$\max\limits_{i = 1}^n|x_i| \le C\|(x_1, \ldots, x_n)\|_E$$ holds. Now it is easy to see that the claim holds.