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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?

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  • $\begingroup$ What is $S(X)$? $\endgroup$ – Jihlbert Dec 10 '18 at 11:13
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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 $f \equiv 0$.

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  • $\begingroup$ 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. $\endgroup$ – user510271 Jan 17 at 11:31
  • $\begingroup$ Yes. if a set has an interior point $x$ then there would be some $r>0$ such that the set contains the open ball $B(x,r)$. $\endgroup$ – Kavi Rama Murthy Jan 17 at 11:44
  • $\begingroup$ 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? $\endgroup$ – user510271 Jan 17 at 12:55

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