How to finish the proof of this corollary of Hahn-Banach theorem? I'm reading Corollary 0.1.3 of Hahn-Banach theorem:



Because $f \restriction G = g$, $\|g\| \le \|f\|$. I'm unable to deduce $\|f\| \le \|g\|$ from the fact that $f(x) \le p(x) = \|g\| \|x\|$ for all $x \in E$. All I can do is $\frac{f(x)}{\|x\|} \le \|g\|$ for all $x \in E \setminus \{0\}$.
Could you please shed me some light?
 A: The definitions are $\lVert f \rVert_* = \sup \{ \lvert f(x) \rvert \mid x \in E, \lVert x \rVert = 1 \}$ for $f \in E^*$, $\lVert g \rVert_{G^*} = \sup \{ \lvert g(x) \rvert \mid x \in G, \lVert x \rVert = 1 \}$ for $g \in G^*$. 
Given $g  \in G^*$, the Hahn-Banach theorem provides a linear extension $f : E \to \mathbb R$ of $f$ such that $f(x) \le p(x) = \lVert g \rVert_{G^*} \lVert x \rVert$ for all $x \in E$. We deduce $-f(x) = f(-x) \le \lVert g \rVert_{G^*} \lVert -x \rVert = \lVert g \rVert_{G^*} \lVert x \rVert$. Therefore $\lvert f(x) \rvert \le \lVert g \rVert_{G^*} \lVert x \rVert$ for all $x \in E$ which shows that $f$ is continuous, i.e. $f \in E^*$. We conclude
$$\lVert f \rVert_* = \sup \{ \lvert f(x) \rvert \mid x \in E, \lVert x \rVert = 1 \} \le \sup \{ \lVert g \rVert_{G^*} \lVert x \rVert \mid x \in E, \lVert x \rVert = 1 \} = \lVert g \rVert_{G^*} .$$
On the other hand
$$\lVert f \rVert_* = \sup \{ \lvert f(x) \rvert \mid x \in E, \lVert x \rVert = 1 \} \ge \sup \{ \lvert f(x) \rvert \mid x \in G, \lVert x \rVert = 1 \}  = \sup \{ \lvert g(x) \rvert \mid x \in G, \lVert x \rVert = 1 \}  = \lVert g \rVert_{G^*} .$$
