By translation the problem can be reduced to the case $0 \in A$, so that $A$ is absorbent. Let $p$ be the gauge of $A$ (II, p. 20) so that $A$ is the set of points $x \in E$ such that $p(x) < 1$. On the other hand, let $V$ be the vector subspace of $E$ generated by $M$; thus $M$ is a hyperplane in $V$ that does not contain $0$, and hence there is a unique linear form $f$ on $V$ such that $M$ is the set of points $y \in V$ for which $f(y) = 1$. The hypothesis $M \cap A = \varnothing$ implies therefore that for all $y \in V$ for which $f(y) = 1$, we have $p(y) \geqslant 1$; as $f$ and $p$ are positively homogeneous we have $f(y) \leqslant p(y)$ for all $y \in V$ such that $f(y) > 0$; finally as $p(y) \geqslant 0$ for all $y \in V$, we see that $f(y) \leqslant p(y)$ for all $y \in V$. By the analytical form of the Hahn-Banach theorem (II, p. 22, th. 1) there exists a linear form $h$ on $E$ which extends $f$ and is such that, for all $x \in E$, $h(x) \leqslant p(x)$. Let $H$ be the hyperplane in $E$ with the equation $h(x) = 1$. Clearly $H \cap V = M$ and $H \cap A = \varnothing$. On the other hand the complement of $H$ in $E$ contains the open non-empty set $A$, therefore $H$ is closed in $E$ (I, p. 11, corollary).

Q.E.D.

For the set $C = A - B$ is open, convex (II, p. 9, prop. 7) and non-empty, also $0 \notin C$. By theorem 1 of II, p. 36, there exists a continuous linear form $f \neq 0$ on $E$ such that $f(z) > 0$ in $C$. Then, for all $x \in A$, and $y \in B$, we have $f(x) > f(y)$. Write $\alpha = \inf_{x \in A} f(x)$; $\alpha$ is finite and we have $f(x) \geq \alpha$ for all $x \in A$ and $f(y) \leq \alpha$ for all $y \in B$; the closed hyperplane $H$ with the equation $f(z) = \alpha$ separates $A$ from $B$.

For, if $f(x) = \gamma$ is an equation of $H$, where $f$ is a continuous linear form in $E$, the restriction of $f$ to $A$ is continuous, therefore bounded and attains its bounds in $A$ (GT, IV, § 6.1, th. 1).

This demonstrates that there exist one or two support hyperplanes of $A$ parallel to $H$; the first case can only arise when $A$ is completely contained in a hyperplane parallel to $H$.

Every support hyperplane of $A$ is closed, since all the points of $A$ are on the same side of the hyperplane (II, p. 15, prop. 17). Also if $x_0$ is a frontier point of $A$, then $x_0$ does not belong to the open non-empty convex set $\overset{\circ}{A}$; after th. 1 of II, p. 36 there exists a hyperplane $H$ that contains $x_0$ and does not meet $\overset{\circ}{A}$. As $A$ is the closure of $\overset{\circ}{A}$ (II, p. 14, cor. 1 to prop. 16), it follows from prop. 17 of II, p. 15 that $H$ is a support hyperplane of $A$.

For there exists an open convex neighbourhood $V$ of 0 in $E$ such that $A + V$ and $K + V$ do not meet (GT, II, § 4.3, prop. 4). As $A + V$ and $K + V$ are convex and open in $E$, prop. 1 of II, p. 37 shows that there exists a closed hyperplane $H$ that strictly separates $A + V$ from $K + V$, and $a$ fortiori $A$ from $K$.

In fact, for every point $x \notin A$, there exists a closed hyperplane that separates $x$ strictly from $A$ (using prop. 4).

For, let $x_0 \notin A$; $\{ x_0 \}$ is closed, therefore there exists a closed hyperplane $H$ which separates $x_0$ strictly from $A$ (prop. 4); let $f(x) = \alpha$ be an equation of $H$ ($f$ a continuous linear form) and suppose that $f(x) > \alpha$ for all $x \in A$. If we put $\gamma = \inf_{x \in A} f(x)$, the half-space defined by $f(x) \geq \gamma$ contains $A$, is determined by the support hyperplane of equation $f(x) = \gamma$, and does not contain $x_0$; whence the corollary.

It is possible that a closed convex set that is not compact and has no interior point, in a locally convex space, does not have any closed support hyperplane (II, p. 86, exerc. 18 : cf. also V, p. 71, exerc. 11).

For all $x \notin \overline{M}$, let $H$ be a closed hyperplane that separates $x$ strictly from $\overline{M}$;

thus $\overline{M}$ is parallel to $H$; the closed hyperplane $H_1$, containing $\overline{M}$ and parallel to $H$ does not contain $x$. The corollary follows.

The condition is obviously necessary. Conversely we show that it is sufficient; if a point $x \in A$ is not contained in $C$, there exists a closed hyperplane of equation $f(z) = \alpha$ separating $x$ strictly from $C$; if we suppose for example that $f(x) < \alpha$, then the continuous affine function $u = f - \alpha$ contradicts the hypotheses.

For $\overline{C}$ is a convex cone of vertex $0$ (II, p. 13, prop. 14). For $x \notin \overline{C}$, there exists a closed hyperplane $H$ that separates $x$ strictly from $\overline{C}$(prop. 4). It is now just necessary to apply the following lemma :

Let $f(z) = \alpha$ with $\alpha < 0$ be an equation of $H$, so that $f(z) = 0$ is the equation of $H_0$. If there exists $z \in A$ such that $f(z) < 0$, then there would exist $\lambda > 0$, such that $f(\lambda z) = \alpha$, and as $\lambda z \in A$, this would contradict the hypothesis.

For, the set $A \subset E \times \mathbf{R}$ of points $(x, t)$ such that $x \in X$ and $t \geqslant f(x)$ is convex (II, p. 17, prop. 19) and closed, since the function $(x, t) \mapsto f(x) - t$ is lower semi-continuous. Then let $x$ be any point of $X$ and let $a \in \mathbf{R}$ be such that $a < f(x)$. By cor. 1 of II, p. 38, there exists a closed hyperplane $H$ in $E \times \mathbf{R}$, that contains $(x, a)$ and does not meet $A$. Every linear continuous form on $E \times \mathbf{R}$ being of the form

where $\lambda \in \mathbf{R}$ and $u$ is a continuous linear form on $E$, it follows that $H$ has an equation of the form $u(z) + \lambda t = \alpha$, and as $H$ contains $(x, a)$ we have $\alpha = u(x) + \lambda a$. Now the point $(x, f(x)) \in A$ does not belong to $H$ and therefore $\lambda \neq 0$. Dividing by $-\lambda$, if necessary, we can write the equation of $H$ as $t - a = u(z - x)$. As $f(x) - a > 0$, we have, therefore, $f(z) > u(z - x) + a$ for all $z \in X$ and this proves the proposition.

We may suppose that $X$ is non-empty. Let $u, v$ be two elements of $L$, such that $u(x) > f(x)$ and $v(x) > f(x)$ for all $x \in X$, and let $b$ be a constant that is an upper bound of $u$ and $v$. Let $U$ (resp. $V$) be the compact convex set of points $(x, t)$ of $X \times \mathbf{R}$ such that $u(x) \leq t \leq b$ (resp. $v(x) \leq t \leq b$), and let $F$ be the set of $(x, t) \in X \times \mathbf{R}$ such that $t \leq f(x)$; $F$ is convex and closed in $X \times \mathbf{R}$. The convex envelope $K$ of $U \cup V$ does not meet $F$, since $U \cup V$ is contained in the set of $(x, t) \in X \times \mathbf{R}$ such that $f(x) < t$, a set which is convex and does not meet $F$. As $K$ is compact (II, p. 14, prop. 15), we can separate $F$ strictly from $K$ by a closed hyperplane $H$ in $E \times \mathbf{R}$. For every $x \in X$, the hyperplane $H$ separates $(x, f(x))$ strictly from $(x, b)$, and therefore meets the line $\{ x \} \times \mathbf{R}$ in a single point $w(x)$; thus $H$ is the graph of a continuous affine function whose restriction $w$ to $X$ is a member of $L$, that is a lower bound for $u$ and $v$ and that satisfies the inequality $w(x) > f(x)$ for all $x \in X$. This proves that the set $L'$ is decreasing directed. Prop. 5 of II, p. 39, applied to $-f$ shows that $f$ is the lower envelope of $L'$.

For, prop. 6 and Dini’s theorem (GT, X, § 4.1, th. 1) show that for all $n$ there exists $h_n \in L$ such that $f \leq h_n \leq f + 1/n$.