Show that for any finite dimensional subspace $E$ of a normed space $F$, there exists a finite set $A \subset F^{*}$ with... Show that for any finite dimensional subspace $E$ of a normed space $F$, there exists a finite set $A \subset F^{*}$ with $\|x \| \geq \max \{ |a(x)| : a \in A \} \geq {1 \over 2}\|x\|$ for all $x \in E$.
I know that $E$ is closed, since it is a finite-dimensional subspace of a normed space. I also know that I need to use compactness (weak? weak*?) to solve the problem.
 A: First, find a solution in $E$.
Note that ${1 \over 2} \|x\| \le \max_{a \in A} |a(x)| \le \|x\|$ for all $x$ iff
${1 \over 2} \le \max_{a \in A} |a(x)| \le 1$ for all $x$ of unit norm.
Note that $\|x\| = \max_{\|a\|_* \le 1} |a(x)|$, where $\|\cdot\|_*$ is the dual norm (on $E^*$). The $\max$ is attained because $E^*$ is finite dimensional. In particular, if
$\|x\|=1$ then $\max_{\|a\|_* \le 1} |a(x)| = 1$, and so
$\max_{\|a\|_* \le {3 \over 4}} |a(x)| = {3 \over 4}$.
In the following, I am assuming that $\|x\| = 1$.
Since $D^* = \overline{B}_*(0,{3 \over 4}) \subset E^*$ is compact and $\{B_*(a,{1 \over 8})\}_{a \in D^*}$ is an open cover, there is a finite subcover, that is for some $A= \{a_1,...,a_n\} \subset D^*$, $\{B_*(a,{1 \over 8})\}_{a \in A}$ is finite open cover of $D^*$.
Just from containment, we have $\max_{a \in A} |a(x)| \le \max_{\|a\|_* \le {3 \over 4}} |a(x)| = {3 \over 4} \le 1$.
For the other side:
For any $a \in D^*$, there is some $k$ such that $\|a-a_k\|_* < {1 \over 8}$ and we have
$|a(x)| < |a_k(x)| + {1 \over 8}$. In particular,
$|a(x)| < \max_k |a_k(x)| + {1 \over 8}$ for all $a \in D^*$ and so
$\max_{a \in D^*} |a(x)| \le \max_{a \in A} |a(x)|+ {1 \over 8}$.
Hence we have $\max_{a \in A} |a(x)| \ge \max_{a \in D^*} |a(x)| -{1 \over 8} \ge {3 \over 4} - {1 \over 8} \ge {1 \over 2}$.
To finish, use the Hahn Banach theorem to extend $a \in A$ to all of $F$.
