Bound on finite dimensional space Let $V$ be a finite dimensional normed space over $\mathbb{R}$, with norm $||.||$
Show that there exists $C>0$ such that for all $x\in V$,  $\sum_{i=1}^n|x_i|\leq C||x||$.
My attempt:
Suppose $dimV=n$. Let $\{$ $e_1,e_2...,e_n$ $\}$ be a basis for $V$. Consider the unit ball,
$K=\{$ $x\in \mathbb{R}^n$ $:$ $||x||_1=1$ $\}$ , where $||.||_1$ is simply the maximum of the coordinates. Observe, by Heine Borel, and equivalence of norms, $K$ is compact.
Define $f:\mathbb{R}^n \rightarrow \mathbb{R}$ by $f(x_1,x_2....,x_n) =||x_1e_1+....+x_ne_n||$
$f$ is indeed Lipschitz, and thus continuous. Restricting the domain to $K$ yields a continuous map defined on a compact space. Hence $f|_K$ attains its minimum.
So, there exists $k \in K$ such that $\forall y\in K$, $f(k)\leq f(y)$.
i.e. $||k_1e_1+...+k_ne_n||\leq ||y_1e_1+...+y_ne_n ||$ for all $(y_1,y_2...,y_n)$ satisfying $\sum_{i=1}^n|y_i|=1$.
What should I do next?

In the following by $x$, I mean $\sum_k x_k e_k$.
Note that $\|x\| \le \sum_k |x_k| \|e_k\| \le K_1 \|x\|_1$ where $K_1 = \max_k \|e_k\|$.
This is the 'easy' direction.
In particular, $\|\cdot\|$ is continuous with respect to $\|\cdot\|_1$.
Let $K_2 = \min_{\|x\|_1 = 1} \|x\|$. Since the $\| \cdot\|_1$ sphere is compact the $\min$ is
attained and since $\|\cdot\|$ is a norm, $K_2 >0$. Since $\| {x \over \|x\|_1} \|\ge K_2$, it follows that $\|x\| \ge K_2 \|x\|_1$.
