Convergence of a sequence of linearly independent vectors in normed space

Question

In an infinite dimensional normed vector space is it possible to find a sequence ${v_n}$ of linearly independent vector (so the sequence is a set of linearly independent vectors) each has norm 1 such that the sequence converges i.e. $\mbox{lim}_{n \rightarrow \infty} v_n$ exists?

This is possible if we don't have the unit norm condition since $v_n = (0,..,0,1/n,0,...)$ converges to $0$ in $l^p$ yet they are linearly independent but I can't really reason why sequence of unit vectors would or would not converge.

Answer 1

This is not a proof of the general statement you ask. However, the answer is no if you're in an inner-product space and the sequence is orthonormal:

If $u,v$ are orthonormal, then $$ |u-v|^2 = \langle u-v, u-v \rangle = |u|^2 + |v|^2 = 2 $$ as the cross terms cancel. Therefore, if $\{v_n\}_{n=1}^{\infty}$ is an orthonormal sequence, it will not be Cauchy.

Answer 2

It is possible.

Consider $\ell^\infty$ and the sequence of vectors $(x_n)_{n=1}^\infty$ defined as:

$$x_n = e_1 + \frac{1}{n+1}e_{n+1} = \left(\underbrace{1, 0, \ldots, 0}_{n \text{ places}}, \frac{1}{n+1}, 0, \ldots\right), \quad\text{ for every } n \in \mathbb{N}$$

where $e_n$ is the $n$-th canonical vector.

$(x_n)_{n=1}^\infty$ is obviously a linearly independent sequence, and $\|x_n\|_\infty = 1$ for all $n \in \mathbb{N}$.

Furthermore, $x_n \xrightarrow{n\to\infty} e_1$:

$$\|x_n - e_1\|_\infty = \left\|\frac{1}{n+1}e_{n+1}\right\|_\infty = \frac{1}{n+1} \xrightarrow{n\to\infty} 0$$