In the proof of the second geometric form of the Hahn-Banach Theorem on page 7 of Brezis's Functional Analysis, we established that
$$ f(x-y) \le f(rz) \quad \forall x\in A, y\in B, |z|=1.$$
From this, Brezis deduces that $$f(x-z)\le -r||f||.$$
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Why can we do this? Is it because $f(rz)\le -r||f||$? But the statement is false. Rather, $-r||f||\le f(rz)\le r||f||$.
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Can we fix this argument by requiring that $$ f(x-y) \le -2\epsilon \quad \forall x\in A, y\in B?$$
Given that $H=[g=\alpha]$ separates sets $C=A-B$ and $\mathcal{B}(0,r)$, we can write down a new linear functional $f=g-\alpha-2\epsilon$, so that $H=[f=-2\epsilon]$.
Thank you!
Best Answer
In the proof of the theorem you have an open ball $B(0,r)$. Now you can considerer a closed ball $B\left[0,r'\right]\subseteq B(0,r)$ (for $r'$ small enough). Now note that $z\in B\left[0,1\right]$ iff $-z\in B\left[0,1\right]$, then $f(x-y)\leq f(-r'z)$ for every $z\in B\left[0,1\right]$ therefore $f(x-y)\leq -r'f(z)$ ($f$ is a linear functional) from which $r'f(z)\leq -f(x-y)$ for every $z\in B\left[0,1\right]$. Now recall that $||f||=sup_{x\in B\left[0,1\right]}f(x)$ which implies that $r'||f||\leq -f(x-y)$ in consequence $f(x-y)\leq -r'||f||$.