Proof of A Dominated Linear Functional on a Line

First, $\mathbb{R}x_{0}$ is a linear subspace. It contains $0_{X}=0\,x_{0}$ by Elementary Identities in a Vector Space (ii), and for $t,s,\lambda\in\mathbb{R}$,

$$t\,x_{0}+s\,x_{0}=(t+s)\,x_{0},\qquad \lambda(t\,x_{0})=(\lambda t)\,x_{0}$$

by Vector Space over a Field (viii) and Vector Space over a Field (v); so $\mathbb{R}x_{0}$ satisfies Linear Subspace of a Real Vector Space (i), Linear Subspace of a Real Vector Space (ii) and Linear Subspace of a Real Vector Space (iii). The coefficient of a point of $\mathbb{R}x_{0}$ is unique: if $t\,x_{0}=t'\,x_{0}$, then

$$(t-t')\,x_{0}=t\,x_{0}+(-t')\,x_{0}=t\,x_{0}-t'\,x_{0}=t\,x_{0}-t\,x_{0}=0_{X}$$

by Vector Space over a Field (viii), Elementary Identities in a Vector Space (iv) with the definition of differences, and Elementary Identities in a Vector Space (vii). Since $x_{0}\neq0_{X}$, Elementary Identities in a Vector Space (xii) gives $t-t'=0$, that is, $t=t'$. This proves (i).

By (i), each point of $\mathbb{R}x_{0}$ is $t\,x_{0}$ for exactly one $t$, so $g$ is well defined. It is linear: by Vector Space over a Field (viii) and Vector Space over a Field (v),

$$g(t\,x_{0}+s\,x_{0})=g\bigl((t+s)\,x_{0}\bigr)=(t+s)\,c,\qquad g\bigl(\lambda(t\,x_{0})\bigr)=g\bigl((\lambda t)\,x_{0}\bigr)=\lambda t\,c,$$

which are Linear Functional on a Linear Subspace (i) and Linear Functional on a Linear Subspace (ii).

Finally, $g$ is dominated by $p$. Let $t\in\mathbb{R}$. If $t\ge0$, the hypothesis $c\le p(x_{0})$ and Sublinear Functional (ii) give

$$g(t\,x_{0})=t\,c\le t\,p(x_{0})=p(t\,x_{0}).$$

If $t<0$, put $s=-t>0$. Then $s(-x_{0})=(-s)\,x_{0}=t\,x_{0}$ by Elementary Identities in a Vector Space (iv), and the hypothesis $-c\le p(-x_{0})$ together with Sublinear Functional (ii) gives

$$g(t\,x_{0})=s\,(-c)\le s\,p(-x_{0})=p\bigl(s(-x_{0})\bigr)=p(t\,x_{0}).$$

This proves (ii).