Let $E$ and $F$ normed vector spaces of finite dimension. Prove that any linear function is Lipschitz I want an answer not only in terms of normed vector spaces over the fields $\Bbb R$ or $\Bbb C$, if not a general answer over any kind of normed vector spaces.

Let the linear function
$$A:E\to F$$
Im stuck with this problem. My strategy was:


*

*Trying to prove that $A$ is continuous (I tried composition of continuous functions but I dont found an easy strategy with this, or just tried direct proof for the $\epsilon-\delta$ definition of continuity but again I failed.)

*Trying to prove that the set $S=\{x\in E:\|x\|_E=1\}$ is compact (I cant conclude that is complete or something useful to prove compactness.)
If I prove these two things then I can conclude that $A$ is Lipschitz. But I failed to conclude anything about this.

Another basic work around was directly trying to prove that $A$ is Lipschitz:
$$\frac{\|Ax-Ay\|_F}{\|x-y\|_E}=\frac{\|Az\|_F}{\|z\|_E}=\frac{\|\sum_{k=1}^n z_k Ae_k\|_F}{\|\sum_{k=1}^n z_k e_k\|_E}\le \frac{|z_{\max}|C}{{\|\sum_{k=1}^n z_k e_k\|_E}}$$
but for the last equality I failed to conclude a bound. I used above a basis for $E$ of finite dimension $n$.

Some other work around was trying to relate the arbitrary norms to some known norm as the maximum norm and the taxicab norm, then it is easy to check that
$$\|x\|=\left\|\sum_{k=1}^n x_k e_k\right\|\le \sum_{k=1}^n |z_k| \|e_k\|\le \begin{cases}C\sum|z_k|=C\|x\|_1\\Cn|x_{\max}|=Cn\|x\|_{\infty}\end{cases}$$
Above I wasnt sure that I can consider bases with $\|e_k\|=1$ over a non necessarily inner product space, this is the reason for $C$.
But these relations dont help me to conclude anything useful. Some help will be appreciated, thank you.
 A: Hints:


*

*Let $\{v_1,\ldots,v_n\}$ be a basis of $E$ and prove that $(E,\|\cdot\|_1) \to (\Bbb R^n, |\cdot|_1)$, $x_1v_1 + \cdots + x_n v_n \mapsto x_1 e_1 + \cdots + x_n e_n$ is an isometry, where $\|\cdot\|_1$ is defined for $x =x_1v_1 + \cdots + x_nv_n$ by $\|x\|_1 = \sum_{i=1}^n |x_i|$. Use Heine-Borel to prove that the unit sphere in $\Bbb R^n$ is compact. Conclude that the unit sphere in $E$ for $\|\cdot\|_1$ is compact.

*Show that any norm on $E$ is equivalent to the norm $\|\cdot\|_1$ (one direction is easy; for the other one you'll need the compactness of the unit sphere for $\|\cdot\|_1$, the continuity of a norm and the fact that a continuous function on a compact set admits a minimum).  

*Prove that any linear map $L: E \to F$ satisfies: $\exists M > 0$ such that $\|Lx\|_F \le M \|x\|_1$ for all $x \in E$.
A: Note that I am using two norms below, $\|\cdot\|$ and a new norm $\|\cdot\|_1$.
Pick a basis $e_1,...,e_n$. Without loss of generality, assume that $\|e_k\| = 1$ 
 and define $\|x\|_1 = \sum_k |x_k|$ where
$x = \sum_k x_k e_k$. It is easy to see that
$\|x\| \le \sum_k |x_k| = \|x\|_1$. In particular, $B_1(0,1) \subset B(0,1)$.
Now I claim that there is some $r>0$ such that $B(0,r) \subset B_1(0,1)$.
If not, there there are $x_n$ such that $\|x_n\|_1 = 1$ but $\|x_n\| \to 0$.
Since $D=\{ x | \|x\|_1 = 1 \}$ is compact, we can presume (by renumbering the sequence) that there is some
$p \in D$ such that $\|x_n -p\|_1 \to 0$. Hence $\|x_n -p\| \to 0$, which
means that $p = 0$, a contradiction. Hence there is some $r$ such that
$B(0,r) \subset B_1(0,1)$. In particular, we have
$\|x\|_1 \le {1 \over r} \|x\|$.
You have $\|A x\| \le \sum_k \|A e_k\| |x_k| \le B \sum_k |x_k| = B \|x\|_1 \le {B \over r} \|x\|$, where
$B = \max_k \|A e_k\| $. Hence $A$ is bounded using the norm $\|\cdot\|$.
