A question in Banach space.

Question

Let $C$ be the Banach space of all complex continuous functions on $[0, 1]$, with the supremum norm. Let $B$ be the closed unit ball of $C$. Why there exist continuous linear functionals $\Gamma$ on $C$ for which $\Gamma (B)$ is an open subset of the complex plane? (In particular, $\left| \Gamma \right|$ attains no maximum on $B$.)

Answer 1

Continuous linear functionals on $C$ correspond to signed measures on $[0,1]$: $\Gamma(f) = \int_[0,1] f \; d\mu$. Consider an absolutely continuous measure $d\mu = \rho(x)\; dx$. Then $\|\Gamma\| = \int_0^1 |\rho(x)|\; dx$. In order to have $\Gamma(f) = \omega |\Gamma\|$ with $\|f\| \le 1$ and $|\omega|=1$ you would need $f \rho = \omega |\rho|$ a.e. If there is no continuous function that is equal a.e. to $|\rho|/\rho$ on $\{x: \rho(x) \ne 0\}$, this can't happen, and then $|\Gamma(f)| < \|\Gamma\|$ for all $f \in B$. This occurs e.g. for $$ \rho(x) = \cases{+1 & $x \le 1/2$\cr -1 & $x > 1/2$}$$