TITLE: Inequality in normed linear space implies independence
QUESTION [5 upvotes]: Suppose $X$ is a normed linear space over $\mathbb{R}$, and $v_1,
 \cdots, v_n \in X$ are unit vectors. Furthermore, assume that there
   exists $\epsilon \in (0, 1/2)$ such that for any constants $c_i \in
 \mathbb{R}$, we have $$\left \| \sum_{i=1}^{n} c_i v_i \right \| \le
 (1+\epsilon) \max_{1\le i \le n} |c_i|.$$ Prove that the collection
   $\{v_i\}$ is linearly independent.

My first thought is to try bounding $\left \|\sum_{i=1}^{n} c_i v_i \right \|$ below in terms of $\epsilon$ and $\max_{1\le i \le n} |c_i|$ but I'm unsure how to go about doing this. If this can be done such that the bound is always positive, then linear independence would of course follow. Any help would be much appreciated!

REPLY [2 votes]: The requirement $\epsilon \in (0, 1/2)$ can be replaced by an weaker one of $\epsilon \in [0, 1)$. Arguing by contradiction, suppose $\{v_i\}$ are linearly dependent. Hence, there exist $a_1, a_2, \ldots, a_n$, not all of them $0$, such that$$\sum_{i = 1}^n a_iv_i = 0.$$Without loss of generality, suppose$$|a_1| = \max\{|a_1|, |a_2|, \ldots, |a_n|\}.$$We have$$2a_1v_1 = a_1v_1 - \sum_{i = 2}^n a_iv_i.$$Therefore,$$2|a_1| = \left\|a_1v_1 - \sum_{i = 2}^n a_iv_i\right\| \le (1 + \epsilon) \max_{1 \le i \le n} |a_i| = (1 + \epsilon)|a_1|,$$a contradiction, since $a_1 \neq 0$.