I. Special case.

Suppose firstly that $E = V + \mathbf{R}a$. Since the proposition is trivial if $a \in V$, we confine ourselves to the case $a \notin V$. The conditions on V imply that the set $A''$ of $y \in V$ such that $a \leq y$ is not empty; similarly the set $A'$ of $z \in V$ such that $-z \geq -a$ (\emph{i.e.} $z \leq a$) is not empty. For $y \in A''$ and $z \in A'$, we have $z \leq a \leq y$, and thus by hypothesis $f(z) \leq f(y)$. Thus $\alpha', \alpha''$ are finite and $\alpha' \leq \alpha''$. Any linear form $f_1$ on E that extends $f$ is completely determined by $f_1(a)$ and for all $\lambda \in \mathbf{R}$ and all $x \in V$, we have

Thus $f_1$ is positive if and only if the relations

imply

As $f(\mu x) = \mu f(x)$ and the relations $x \geq 0$ and $\mu x \geq 0$ are equivalent for $\mu > 0$, it is sufficient to show that (2) implies (3) in the particular cases $\lambda = 0, \lambda = 1$ and $\lambda = -1$. For $\lambda = 0$, the fact that (2) implies (3) follows from the hypothesis that $f$ is positive. For $\lambda = 1$, to say that (2) implies (3) means that for $-x \in A'$, we have $f_1(a) \geq f(-x)$, \emph{i.e.} $f_1(a) \geq \alpha'$; for $\lambda = -1$, (2) implies (3), means that for $x \in A''$, we have $f(x) \geq f_1(a)$, \emph{i.e.} $f_1(a) \leq \alpha''$. The proposition is therefore proved in this case.

II. General case.

Let $\mathfrak{F}$ be the set of pairs $(W, g)$ where $W$ is a vector subspace of $E$ containing $V$ and $g$ is a positive linear form on $W$ which is an extension of $f$. We order $\mathfrak{F}$ putting $(W, g) \leq (W', g')$ if $W \subset W'$ and if $g'$ is an extension of $g$. Clearly $\mathfrak{F}$ is inductive and by th. 2 of S, III, § 2.4, there is a maximal element $(W_0, g_0)$. Suppose $W_0 \neq E$. Then there exists a vector $b \notin W_0$, and, if $W_1 = W_0 + \mathbf{R}b$, the special case above shows that there exists a positive linear form on $W_1$ which is an extension of $g_0$; this contradicts the hypothesis that $(W_0, g_0)$ is maximal. Thus $W_0 = E$, and the first part of the proposition is proved. When $a \in V$, the second assertion is obviously true with $\alpha' = \alpha'' = f(a)$; if, on the contrary, $a \notin V$ and one puts $V_1 = V + \mathbf{R}a$, the second assertion follows from the special case I of the proof.

By prop. 1 it is sufficient to show that for every $x \in E$, there exists $x' \in V$ such that $x' - x \in P$. Now let $U$ be a neighbourhood of $0$ in $E$ such that $x_0 + U \subset P$. Then $x + x_0 + U \subset x + P$, and, hence there exists $\varepsilon$ such that $0 < \varepsilon < 1$ and the point $y = x_0 + (1 - \varepsilon)x$ belongs to $x + P$; then every point of the form $x + \lambda(y - x)$ belongs to $x + P$ for $\lambda > 0$. If we take $\lambda = 1/\varepsilon$, then $x + \lambda(y - x) = \lambda x_0 \in V$, from which the conclusion follows.

The conclusion of the corollary is not necessarily valid if one does not assume that $V$ contains an interior point of $P$, even if $E$ is of finite dimension and if $P \cap V$ contains points interior in $V$ (II, p. 91, exerc. 25, b)).

The set of pairs $(x, a)$ such that $p(x) \leq a$ is a convex subset $P$ of the vector space $E_1 = E \times \mathbf{R}$ (II, p. 17, prop. 19), and it is clearly a pointed cone. Let $V_1$ be the subspace $V \times \mathbf{R}$ of $E_1$ and $g(y, a) = -f(y) + a$ for each point $(y, a) \in V_1$. Then $g$ is a positive linear form for the preorder structure on $V_1$ defined by $P \cap V_1$; for if $(y, a) \in P \cap V_1$, then $a \geq p(y) \geq f(y)$, therefore $g(y, a) \geq 0$. Next let $(x, a) \in E_1$; we show that $(x, a)$ is less than a point of $V_1$ for the preorder defined by $P$. If $(x', a') \in V_1$ then $(x, a) \leq (x', a')$ if, and only if, $p(x' - x) \leq a' - a$, taking $a' \geq p(-x) + a$, we see that $(0, a')$ of $V_1$ satisfies the requirements. Thus we can apply prop. I of II, p. 21; there is a linear form $u$ on $E_1$ extending $g$ and positive for the preorder defined by $P$. Therefore $u(0, 1) = g(0, 1) = 1$ and $u$ is of the form $u(x, a) = -h(x) + a$, where $h$ is a linear form on $E$ that extends $f$; further, for all $x \in E$ and all $a \geq p(x)$, we have $h(x) \leq a$, therefore $h(x) \leq p(x)$. Q.E.D.

For a semi-norm $q$ and a linear form $g$ on $E$, the relation $g \leq q$ is the same as $|g| \leq q$. The corollary follows from th. 1.

Apply cor. 1 to the vector subspace, $V$, generated by $x_0$ and to the linear form $\xi x_0 \mapsto \xi p(x_0)$ defined over $V$.

Apply cor. 1, taking $p(x) = \|f\| \cdot \|x\|$, which gives $\|h\| \leq \|f\|$; but clearly $\|h\| \geq \|f\|$, and the corollary follows.

The conclusion of cor. 3 is not necessarily valid for continuous linear mappings of a normed space into an arbitrary normed space (IV, p. 55, exerc. 16, c) and V, p. 65, exerc. 22).