Proof of One-Step Extension of a Dominated Linear Functional

Throughout, sums and scalar multiples in $V$ are rearranged by the vector space axioms Vector Space over a Field (i), Vector Space over a Field (ii), Vector Space over a Field (vii), Vector Space over a Field (viii), Vector Space over a Field (v) and Vector Space over a Field (vi) and the identities Elementary Identities in a Vector Space (ii), Elementary Identities in a Vector Space (iv), Elementary Identities in a Vector Space (i), Elementary Identities in a Vector Space (viii) and Elementary Identities in a Vector Space (ix).

First, $M+\mathbb{R}x_{0}$ satisfies Linear Subspace of a Real Vector Space (iv). It contains $0_{V}=0_{V}+0\,x_{0}$, as $0_{V}\in M$ by Linear Subspace of a Real Vector Space (i). For $m,m'\in M$ and $t,t',\lambda\in\mathbb{R}$,

$$(m+t\,x_{0})+(m'+t'\,x_{0})=(m+m')+(t+t')\,x_{0},\qquad \lambda(m+t\,x_{0})=\lambda m+(\lambda t)\,x_{0},$$

with $m+m'\in M$ and $\lambda m\in M$ by Linear Subspace of a Real Vector Space (ii) and Linear Subspace of a Real Vector Space (iii); so $M+\mathbb{R}x_{0}$ is closed under sums and scalar multiples. Finally $m=m+0\,x_{0}$ for $m\in M$, and $x_{0}=0_{V}+1\,x_{0}$, so $M+\mathbb{R}x_{0}$ contains $M$ and $x_{0}$. This proves (i).

Next, suppose $m+t\,x_{0}=m'+t'\,x_{0}$ with $m,m'\in M$ and $t,t'\in\mathbb{R}$. Cancelling gives $(t-t')\,x_{0}=m'-m$, and $m'-m=m'+(-1)m$ lies in $M$ by Linear Subspace of a Real Vector Space (iii) and Linear Subspace of a Real Vector Space (ii). If $t\neq t'$, then $x_{0}=(t-t')^{-1}(m'-m)\in M$ by Linear Subspace of a Real Vector Space (iii), contradicting $x_{0}\notin M$. So $t=t'$, and then $m=m'$ by cancelling again. This proves (ii).

The construction rests on one inequality. Let $m,m'\in M$. Since $(m-x_{0})+(m'+x_{0})=m'+\bigl((m-x_{0})+x_{0}\bigr)=m'+m$, Linear Functional on a Linear Subspace (i), the hypothesis $f\le p$ on $M$ and Sublinear Functional (i) give

$$f(m)+f(m')=f(m+m')\le p(m+m')=p\bigl((m-x_{0})+(m'+x_{0})\bigr)\le p(m-x_{0})+p(m'+x_{0}),$$

that is,

$$f(m)-p(m-x_{0})\le p(m'+x_{0})-f(m')\qquad\text{for all }m,m'\in M.$$

Now the set $\{f(m)-p(m-x_{0}):m\in M\}$ is nonempty, since $0_{V}\in M$, and by the last inequality every number $p(m'+x_{0})-f(m')$ with $m'\in M$ is an upper bound of it. Its supremum $c$ therefore exists by The Real Numbers, with the Natural Numbers, Integers and Rationals Identified with Subsets of the Reals, and Completeness (vii). As $c$ is an upper bound, $f(m)-p(m-x_{0})\le c$ for every $m\in M$; as $c$ is the least upper bound (Bounds, Least and Greatest Elements, Suprema and Infima for a Partial Order (iv)), $c\le p(m'+x_{0})-f(m')$ for every $m'\in M$. This proves (iii).

It remains to examine $g$. By (ii) every $y\in M+\mathbb{R}x_{0}$ has exactly one representation $y=m+t\,x_{0}$, so $g$ is a well-defined function. By the formulas for sums and scalar multiples above and by Linear Functional on a Linear Subspace (i) and Linear Functional on a Linear Subspace (ii) for $f$,

$$g\bigl((m+t\,x_{0})+(m'+t'\,x_{0})\bigr)=f(m+m')+(t+t')\,c=g(m+t\,x_{0})+g(m'+t'\,x_{0}),\qquad g\bigl(\lambda(m+t\,x_{0})\bigr)=f(\lambda m)+\lambda t\,c=\lambda\,g(m+t\,x_{0}),$$

so $g$ is a linear functional, and $g(m)=g(m+0\,x_{0})=f(m)$ for $m\in M$.

For domination let $y=m+t\,x_{0}$. If $t=0$, then $g(y)=f(m)\le p(m)=p(y)$ by the hypothesis. If $t>0$, then $t^{-1}m\in M$, and the right-hand inequality of (iii) at $t^{-1}m$ reads $c\le p(t^{-1}m+x_{0})-f(t^{-1}m)$. Multiplying by $t$, and using $t(t^{-1}m+x_{0})=m+t\,x_{0}$, Sublinear Functional (ii) and Linear Functional on a Linear Subspace (ii), we get

$$t\,c\le p(m+t\,x_{0})-f(m),\qquad\text{that is,}\qquad g(y)=f(m)+t\,c\le p(y).$$

If $t<0$, put $s=-t>0$. The left-hand inequality of (iii) at $s^{-1}m\in M$ reads $f(s^{-1}m)-p(s^{-1}m-x_{0})\le c$. Since $s(s^{-1}m-x_{0})=m-s\,x_{0}$, multiplying by $s$ gives

$$f(m)-p(m-s\,x_{0})\le s\,c .$$

As $m-s\,x_{0}=m+t\,x_{0}$, this says $f(m)+t\,c\le p(m+t\,x_{0})$, that is, $g(y)\le p(y)$. This proves (iv).