I'm reading a proposition related to the geometric form of the Hahn-Banach Theorem:
My lecture note deduces $f\left(x_{0}+r x\right)<\alpha$ for all $x \in E$ such that $\|x\| \leq 1$. I think this not not correct. This is because $\|(x_{0}+r x) - x_0 \| =r\|x\| \le r$ for all $\|x\| \leq 1$. As such, $\|x\| \leq 1$ does not imply $x$ belongs to the open ball $\mathbb B (x_0,r)$.
To fix this bug, we reduce $r$ by half and get $\|(x_{0}+(r/2) x) - x_0 \| =(r/2)\|x\| \le r/2 <r$ for all $\|x\| \leq 1$. As such, $f (x_{0}+(r/2) x) < \alpha$. By the linearity of $f$, we get $$f(x) \leq \frac{2}{r}\left(\alpha-f\left(x_{0}\right)\right), \quad \forall x\in E: \|x\|\le 1 \tag{1}$$
For all $x \in E$ such that $\|x\| \leq 1$, we have $\|-x\| \le 1$. As such $$f(-x) \leq \frac{2}{r}\left(\alpha-f\left(x_{0}\right)\right), \quad \forall x\in E: \|x\|\le 1$$ or equivalently $$-f(x) \leq \frac{2}{r}\left(\alpha-f\left(x_{0}\right)\right), \quad \forall x\in E: \|x\|\le 1 \tag{2}$$
Combining $(1),(2)$, we get $$|f(x)| \leq \frac{2}{r}\left(\alpha-f\left(x_{0}\right)\right), \quad \forall x\in E: \|x\|\le 1$$
As such, $$\|f\| = \sup_{\|x\|\le 1} |f(x)| \le \frac{2}{r}\left(\alpha-f\left(x_{0}\right)\right)$$
Hence $f$ is bounded and continuous.
My question:
Could you please confirm if my understanding is correct? Thank you so much!