Properties of $\{x\in X\mid f(x)=||f||\}$ Let $X$ be a normed space, $f\in X^*\setminus\{0\}$ (the continuous dual), $E:=\{x\in X\mid f(x)=\|f\|\}$. Prove that $E$ is a nonempty closed set and that $\inf \{\|x\|\mid x\in E\}=1$.
I have no idea how to prove that $E$ is non empty and that $\inf \{\|x\|\mid x\in E\}=1$.
$E$ is closed because it is the inverse image of a point ($\|f\|$) and $f$ is continuos, so inverse images of closed sets are closed.

Since $f \neq 0$ we have $f(x^*) \neq 0$ for some $x^*$, so then $f({\|f\| \over f(x^*)} x^*) = \|f\|$ so $E$ is not empty.
Since $f$ is continuous, $f^{-1} ( \{ \|f\| \})$ is closed.
Note that $f(x) = \|f\|$ for all $x \in E$. Since $\|f\| = f(x) \le \|f\| \|x\|$ we see that $\|x\| \ge 1$.
