Question related to Ham Brezis Functional Analysis Book Corollary 1.3

Question

I have a little confusion over Brezis book of "Functional Analysis, Sobolev Spaces, and Partial Differential Equations" regarding Corollary 1.3 in Chapater 1.
This is the statement of the Corollary:

Corollary 1.3
For every $x_0 \in E$, there exists $f_0 \in E^{*}$ such that $||f_0||_{E^{*}}=||x_0||$ and the duality pairing (inner product) $\langle f_0 , x_0 \rangle = ||x_0||^{2}$.

The proof says that I can use Corollary 1.2 and set $G=\mathbb{R}_{x_{0}}$ and $g(tx_0)=t ||x_{0}||^{2}$ given $g : G\to \mathbb{R}$ with $G \subset E$ a linear subspace of the vector space $E$ with its dual space $E^{*}$.

I have managed to prove Corollary 1.2 but I will also state it here.
Corollary 1.2
Let $G \subset E$ be a linear subspace. If $g : G \to \mathbb{R}$ is a continuous linear functional, then there exists $f \in E^{*}$ that extends $g$ and such that
$$||f||_{E^{*}}= \sup_{x\in G, ||x||\leq 1}{|g(x)|}= ||g||_{G^{*}}$$
Now, my question is, "What is the meaning of $\mathbb{R}_{x_{0}}$ in this context?". I cannot fully comprehend the proof because I don't know the precise definition of the set $\mathbb{R}_{x_{0}}$ described here.
Any help and clarification will be much appreciated! Thanks!

Answer 1

It seems that the notation $G = \mathbb{R}_{x_0}$ just means the span of the vector $x_0$ (this is more commonly denoted by $\mathbb{R}x_0, \operatorname{span} \{ x_0 \}, \left< x_0 \right>$). The idea of the proof is to define $g$ on the one-dimensional subspace $G$ such that $g(x_0) = \| x_0 \|^2$ and then extend it continuously using Corollary 1.2 to the whole space $E$.