Does This Inequality of Vector Norms Hold in General?

Question

Suppose $X$ is a vector space over $\mathbb{C}$ with finite dimension $n$, endowed with a norm $|| \cdot ||_X$. Let $x_1,\ldots,x_n$ be a basis.

Does there exist a real $C$ (maybe just $C=1$) such that for any $\alpha,\beta\in\mathbb{C}\setminus\{0\}$ and any $i,j\in\{1,\ldots,n\}$, we have $$ ||\alpha x_i + \beta x_j||_X \hspace{1mm}\geq\hspace{1mm} C||\alpha x_i||_X? $$ I know that in general, we certainly do not have $||a+b||_X \geq ||a||_X$ when $a,b \in X$. However, it's tempting to believe that when $a$ and $b$ are linearly independent (e.g., $a=\alpha x_i$, $b=\beta x_j$), that we may get something like this. Is this only naive temptation, or is there hope? Thanks.

Answer 1

On a finite dimensional normed linear space any linear function is continuous. Consider the map $L(\sum\limits_{j=1}^{n} a_jx_j)=a_ix_i$. Continuity of this map shows that $\|a_ix_i\| \leq M \|\sum\limits_{j=1}^{n} a_jx_j\|$ for some $M \in (0,\infty)$. Hence $\|\sum\limits_{j=1}^{n} a_jx_j\| \geq C \|a_ix_i\|$ where $C =\frac 1 M$.

Answer 2

Consider the function

$$ f_j(\beta) = \frac{\|\alpha x_i + \beta x_j \|}{\|\alpha x_i\|}. $$

As $|\beta| \to \infty$ so does $f_j(\beta)$ as we can see by applying the reverse triangle inequality:

$$ \|\alpha x_i + \beta x_j \| \ge \Big| \|\alpha x_i\| - \|\beta x_j\| \Big| \ge |\beta| \|x_j\|, $$ where the last inequality holds as long as $\|\beta x_j\| > \|\alpha x_i\|$.

Therefore $f_j$ has a global minimum $C_j$. If $C = \min_{i\ne j}\{C_j\}$ then we have

$$ C \le C_j \le \frac{\|\alpha x_i + \beta x_j \|}{\|\alpha x_i\|} $$

for all $\beta$ and all $j$.

We should also note that $C_j > 0$ since $x_i$ and $x_j$ are linearly independent.