Bounded linear functional separating a hyperplane and a point I want to find examples of normed linear space, a convex subset $C$ of $X$ and a point $x_0\in X\setminus C$ such that for all bounded linear functionals $f:X\to \mathbb{K}$, Re$f(x_0)<\sup\{\text{Re}f(x):x\in C\}$.
If I start with a proper dense subspace $C$ of $X$, then there exists a sequence $(x_n)$ in $C$ such that $x_n\to x_0$. Thus for all bounded linear functional $f$ on $X$, Re$f(x_n)\to \text{Re}f(x_0)$. Thus Re$f(x_0)\leq \sup\{\text{Re}f(x):x\in C\}$. But how to show the strict inequality? 
 A: First of all, as stated the problem fails trivially, since we have the zero functional (which is always bounded and it gives equality, not strict inequality). So suppose that we exclude the zero functional from the question. If the convex set is a subspace, the question has no interest:
If $Y$ is a proper, closed subspace of $X$ and $x_0\in X\setminus Y$, then there exists a functional $f\in X^*$ so that $f=0$ on $Y$ and $f(x_0)>0$, so $\text{Re}(f(x_0))=f(x_0)>0=\sup\{f(x): x\in Y\}$.
If $Y$ is a subspace of $X$ that is not closed, nor dense, then $\bar{Y}$ is a proper closed subspace of $X$ and the previous case yields a non-zero functional that fails to satisfy $\text{Re}(f(x_0))<\sup\text{Re}(f(x))$.
If $Y$ is a dense subspace of $X$, then any bounded functional $f$ that is zero on $Y$ is identically zero on $X$, bceause of continuity. Now suppose that $f$ is a non-zero functional on $Y$, so $f(y_0)\neq0$. Consider the element $y_1=\overline{f(y_0)}\cdot y_0\in Y$. Then $f(y_1)=|f(y_0)|^2>0$. But the elements $y_n=n\cdot y_1$ belong in $Y$ and $f(y_n)=n|f(y_0)|^2$, so $\sup\{\text{Re}(f(y)): y\in Y\}=\infty$ so for any $x\in X$ we have that $\text{Re}(f(x_0))<\sup\{\text{Re}(f(y)): y\in Y\}$.
