Functional giving open image for closed unit ball in C (Rudin Functional Analysis chapter 3, exercise 8) I am trying to solve the following problem:
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 in $C$. Show that there exist continuous linear functionals $\Lambda$ on $C$ for which $\Lambda(B)$ is an open subset of the complex plane; in particular, $|\Lambda|$ attains no maximum on $B$.
Here are some thoughts, edited somewhat after seeing the solution. Since $\Lambda$ is continuous it must be bounded, so we can assume that $|\Lambda f| \le M$ on $B$; we can take $M=1$ without loss of generality. Also, if $\Lambda f = \alpha$, then $\Lambda (r e^{i \theta} f) = re^{i\theta} \alpha,$ so the image of $B$ must be the open disk $\{z \in {\Bbb C}: |z|< 1\}$. To achieve this, $\Lambda$ should be designed so that the maximum can only be achieved by a discontinuous function.
 A: A beautiful solution was given by Ryszard Szwarc in the comments. I'm just fleshing out his solution with more details, to help myself understand it better, and in particular, to understand how to come up with it in the first place.
By Riesz, any bounded linear functional on $C$ is given by a unique complex regular Borel measure $\mu$, $$\Lambda(f) = \int_{[0,1]}f\,d\mu,$$ with $\lVert \Lambda \rVert = \lvert \mu\rvert([0,1]).$ We might reasonably start by considering real $\mu$. If $\mu = \mu^+ - \mu^-$ is the corresponding Jordan decomposition, then $$\Lambda f = \int f\,d\mu^+ - \int f \,d\mu^-,$$
which is maximized for unit vectors in $L^\infty\supset C$ by taking $f=1$ on the support of $\mu^+$ and $f=-1$ on the support of $\mu^-$. One sees immediately that by choosing $\mu^+$ and $\mu^-$ appropriately, one can force a discontinuity in the maximizing function at the transition points.
For example, taking $\mu^+(dx) = 1_{[0,1/2]}(dx)$ and $\mu^-(dx) = 1_{(1/2,1]}(dx)$ forces a discontinuity in the maximizing function at $x=1/2$. The maximizing function is unique up to equivalence a.e., so $|\Lambda f| < 1 = \lVert \Lambda \rVert$, for any nonequivalent function, which includes all $f\in C$. On the other hand, we can get arbitrarily close to 1 with functions in $C$:
\begin{equation*}
f_\epsilon(x) = \begin{cases}
   1 &0\le x\le \tfrac{1}{2} - \epsilon \\
\frac{1}{\epsilon}\left(\frac{1}{2} - x\right) &\frac{1}{2} - \epsilon < x < \frac{1}{2} + \epsilon \\
-1 & \frac{1}{2} + \epsilon \le x \le 1,
\end{cases}
\end{equation*}
for which $\lvert \Lambda f_\epsilon \rvert > 1 - 2\epsilon.$
Thus, $\{\Lambda f: f\in B\} = \{z: \lvert z \rvert < 1\}.$ The choice of $\mu$ is far from unique, although the given choice is the most elegant.
