# Regarding Hahn Banach theorem and supporting hyperplane

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?

• What is $S(X)$? – Jihlbert Dec 10 '18 at 11:13

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 $$f \equiv 0$$.
• Here $I(f)$ is a subspace of hyperplane. What would you mean by $I(f)$ containing $B(x,r)$? Like in three dimensional space, $B(x,r)$ would be an open sphere with center $x$ and $I(f)$ would be a plane. – user510271 Jan 17 at 11:31
• Yes. if a set has an interior point $x$ then there would be some $r>0$ such that the set contains the open ball $B(x,r)$. – Kavi Rama Murthy Jan 17 at 11:44
• So going by your proof, if we consider $\mathbb{R}^3$, then $B(0,r)$ will be a open unit disc centered at $0$, so how will $f$ being zero at $B(0,r)$ imply that $f$ is the zero function? – user510271 Jan 17 at 12:55