Let $X$ be a normed space, $K$ and $L$ are two non-empty convex subsets such that $\inf\{\|k-l\|:k\in K, l \in L\}>0$. Prove that there exists a non-zero linear continuous functional $\varphi:X \rightarrow \mathbb{K}$ s.t.
$$\sup\{\mathrm{Re}\varphi(k): k \in K\} \leq \inf\{\mathrm{Re}\varphi(l): l \in L\}$$.
My attempt at solution:
Assume that $\mathbb{K} = \mathbb{R}$. Denote $Z:=K-L=\{k-l:k\in K, l \in L\}$. Notice that $Z$ is a non-empty, convex subset of $X$. $\inf\{\|k-l\|:k\in K, l \in L\}>0 \Rightarrow \exists \ x_0 \in X \text{ s.t. } x_0\notin Z$. Let $V := \mathrm{Lin}(x_0) = \{\lambda x_0:\lambda \in \mathbb{R}\}$. Consider a functional $L:V\rightarrow \mathbb{R}$, $L(\lambda x_0) = \lambda$. $L$ is linear and continuous. Since $Z$ is convex the Minkowski functional $\mu_Z$ is well-defined. Moreover, for any $\lambda x_0 \in V$:
$$ \left|L(\lambda x_0)\right|= |\lambda| \leq \mu_Z (\lambda x_0)$$
because $x_0 \notin Z$. We now use Hahn-Banach theorem and state that there exists $\varphi:X \rightarrow \mathbb{R}$ such that:
- $\varphi|_V = L$.
- $\left|\varphi(x)\right|\leq \mu_Z (x)$ for all $x \in X$.
It is true that:
$$\varphi(z) \leq \varphi(x_0)\ \forall z \in Z$$.
Why? $\varphi(z) \leq \mu_Z (z) \leq 1 = L(x_0) = \varphi(x_0)$.
$$z = k - l, k\in K, l \in L \Rightarrow \varphi(z) = \varphi(k)-\varphi(l)$$
I thought that the second part of the proof will arise if we consider $x_0 = 0$, but apparently it does not work, so now I wonder how to generalize this for $\mathbb{C}$ and how to prove the second part. I will appreciate any insight!