Separation theorem on the space of all complex continuous functions Let $C$ be the Banach space of all complex continuous functions on $[0,1]$ endowed with the supremum norm. Let $B$ be the closed unit ball of $C$. Does there exist a $\Lambda\in C^*$ for which $\Lambda (B)$ is open?
Note that $C$ is a locally convex space and $B$ is convex, balanced and closed set, by the separation theorem of convex sets,we know there exists $\Lambda\in C^*$ such that $|\Lambda f|\leq 1$ for all $f\in B$ and $|\Lambda f|>1$ for $f\in B^c$. But how about this when we consider some concrete space above?

Take
$$\Lambda(f) = \int_0^{1/2} f(t)\,dt - \int_{1/2}^1 f(t)\, dt.$$
$\Lambda(B) = \{ z \in \mathbb{C} : \lvert z \rvert < 1\}$ can be verified with a little computation.

by the separation theorem of convex sets,we know there exists $\Lambda\in C^*$ such that $|\Lambda f|\leq 1$ for all $f\in B$ and $|\Lambda f|>1$ for $f\in B^c$

No, the complement $B^c$ of $B$ is not convex, so it cannot be separated from $B$ by a linear functional (every linear functional takes the value $0$ both in $B$ and $B^c$).
