In the above image from the book Linear Analysis by Bela Bollobas, corollary 7 gives the first consequence to the Hahn Banach Theorem. In the paragraph below corollary 8 they define a supporting functional and support plane. For a linear functional $f$ on a Banach space $X$ $$ I(f)=\{x\in X: f(x)=1\},$$ and $B(X)$ is the closed unit ball in $X$. The second last line of the bottomost paragraph states that $I(f)$ contains no interior point of $B(X)$. Can anyone tell why?
Best Answer
If $I(f)$ contains some open ball $B(x,r)$ then $y \in B(0,r)$ implies $f(y+x)=1$ so $f(y)=1-f(x)$. In particular this must hold for $y=0$ so $1-f(x)=0$ which in turn gives $f(y)=0$ for all $y \in B(0,r)$. But then $f \equiv 0$.