Continuity, Hyperplanes, and the Gauge
The extension form of the Hahn-Banach theorem produced functionals: given a bounded functional on a subspace,
it manufactured a norm-preserving extension to the whole space. That is an analytic statement about
domination by a sublinear bound. There is a second face of the same theorem, entirely geometric, which
speaks not of extending functionals but of separating sets. A nonzero continuous linear functional
\(f\) cuts a real
topological vector space
into two pieces, \(\{f \lt \alpha\}\) and \(\{f \gt \alpha\}\), meeting along the hyperplane \(\{f = \alpha\}\).
To say that two convex sets can be separated is to say that some such functional places them on opposite
sides of a hyperplane. The whole development rests on converting geometry — an open convex set — into
analysis — a sublinear functional — and then feeding that functional to the extension theorem.
We work throughout over a scalar field \(\mathbb{F}\), either \(\mathbb{R}\) or \(\mathbb{C}\), and specialize to
\(\mathbb{R}\) where the geometry demands it. Two preliminary results carry the entire section: a criterion for
when a linear functional on a topological vector space is continuous, and the construction that extracts a
sublinear functional from an open convex set. Both adapt their normed-space ancestors with only the topology of
the space available.
Continuity of Linear Functionals
On a normed space, a linear functional is continuous exactly when it is bounded. Without a norm the criterion must
be phrased topologically, but the list of equivalent conditions is longer and sharper: continuity is detected at a
single point, or by the closedness of the kernel alone.
Theorem: Continuity of a Linear Functional
Let \(\mathcal{X}\) be a topological vector space over \(\mathbb{F}\) and let \(f : \mathcal{X} \to \mathbb{F}\)
be a nonzero linear functional. The following are equivalent.
(a) \(f\) is continuous;
(b) \(f\) is continuous at \(0\);
(c) \(f\) is continuous at some point;
(d) \(\ker f\) is closed;
(e) \(x \mapsto |f(x)|\) is a continuous seminorm.
Proof
(a) \(\Rightarrow\) (b) \(\Rightarrow\) (c).
Continuity everywhere implies continuity at \(0\), which is one particular point; both implications are
immediate.
(c) \(\Rightarrow\) (a).
Suppose \(f\) is continuous at a point \(x_0\). Given any \(x_1\) and \(\varepsilon \gt 0\), continuity at
\(x_0\) furnishes an open neighborhood \(W\) of \(0\) with \(|f(x_0 + w) - f(x_0)| \lt \varepsilon\) for all
\(w \in W\); by linearity this says \(|f(w)| \lt \varepsilon\) for \(w \in W\). Since translation
\(x \mapsto x + (x_1 - x_0)\) is a homeomorphism, \(x_1 + W\) is a neighborhood of \(x_1\), and for
\(x = x_1 + w\) in it,
\[
|f(x) - f(x_1)| = |f(w)| \lt \varepsilon .
\]
Thus \(f\) is continuous at the arbitrary point \(x_1\), hence continuous.
(a) \(\Rightarrow\) (d).
If \(f\) is continuous then \(\ker f = f^{-1}(\{0\})\) is the preimage of the closed set \(\{0\} \subseteq
\mathbb{F}\), hence closed.
(d) \(\Rightarrow\) (b).
Assume \(\ker f\) is closed. Because \(f \neq 0\), pick \(x_1\) with \(f(x_1) = 1\). Since \(x_1 \notin \ker f\)
and \(\ker f\) is closed, its complement is an open neighborhood of \(x_1\); translating, there is an open
neighborhood \(N\) of \(0\) with \((x_1 + N) \cap \ker f = \varnothing\). By continuity of scalar
multiplication at \((0,0)\), the neighborhood \(N\) contains a
balanced
open neighborhood \(U\) of \(0\): there are \(\delta \gt 0\) and a neighborhood \(W\) of \(0\) with
\(\alpha W \subseteq N\) for \(|\alpha| \leq \delta\), and \(U = \bigcup_{|\alpha| \leq \delta} \alpha W\) is
balanced, open, and contained in \(N\). Then \((x_1 + U) \cap \ker f = \varnothing\) as well, since
\(U \subseteq N\). We claim \(|f(u)| \lt 1\) for all \(u \in U\). If instead some \(u \in U\) had
\(|f(u)| \geq 1\), then \(\lambda := f(u)\) satisfies \(|\lambda| \geq 1\), so \(\lambda^{-1} u \in U\) by
balancedness, and
\[
f\bigl(x_1 - \lambda^{-1} u\bigr) = 1 - \lambda^{-1} f(u) = 1 - 1 = 0,
\]
placing \(x_1 - \lambda^{-1} u \in (x_1 + U) \cap \ker f\), a contradiction. Hence \(|f(u)| \lt 1\) on \(U\).
For any \(\varepsilon \gt 0\) the balanced neighborhood \(\varepsilon U\) then satisfies \(|f| \lt \varepsilon\),
so \(f\) is continuous at \(0\).
(b) \(\Leftrightarrow\) (e).
The map \(p(x) = |f(x)|\) is absolutely homogeneous, \(p(\alpha x) = |\alpha|\,|f(x)|\), and subadditive by the
triangle inequality in \(\mathbb{F}\); it is therefore a
seminorm. A seminorm is continuous
exactly when it is continuous at \(0\), because \(|p(x) - p(x_0)| \leq p(x - x_0)\). Continuity of \(p\) at
\(0\) is the statement that for every \(\varepsilon \gt 0\) there is a neighborhood of \(0\) on which
\(|f| \lt \varepsilon\), which is precisely continuity of \(f\) at \(0\). The equivalence follows.
The crux is the implication from a closed kernel to continuity. It rests on a structural dichotomy peculiar to
hyperplanes: the kernel of a nonzero functional is a maximal proper subspace, so its closure has nowhere to grow
except to the whole space.
Theorem: A Hyperplane Is Closed or Dense
Let \(\mathcal{X}\) be a topological vector space over \(\mathbb{F}\) and let \(f : \mathcal{X} \to \mathbb{F}\)
be a nonzero linear functional. Then \(\ker f\) is either closed or dense in \(\mathcal{X}\). It is closed if
and only if \(f\) is continuous.
Proof
Write \(\mathcal{M} = \ker f\). Because \(f\) is linear and surjective onto \(\mathbb{F}\), the subspace
\(\mathcal{M}\) has codimension one: fixing \(x_1\) with \(f(x_1) = 1\), every \(x\) decomposes uniquely as
\(x = \bigl(x - f(x) x_1\bigr) + f(x) x_1\) with the first summand in \(\mathcal{M}\), so
\(\mathcal{X} = \mathcal{M} \oplus \mathbb{F} x_1\). Consequently \(\mathcal{M}\) is a maximal proper
subspace: if a subspace \(\mathcal{N}\) strictly contains \(\mathcal{M}\), it contains some \(x\) with
\(f(x) \neq 0\); writing \(x = m + f(x) x_1\) with \(m = x - f(x) x_1 \in \mathcal{M} \subseteq \mathcal{N}\),
the difference \(f(x) x_1 = x - m\) lies in \(\mathcal{N}\), so \(x_1 \in \mathcal{N}\) and hence
\(\mathcal{N} = \mathcal{M} \oplus \mathbb{F} x_1 = \mathcal{X}\).
Now the closure \(\overline{\mathcal{M}}\) is again a subspace, because the closure of a linear subspace in a
topological vector space is a subspace: continuity of addition and scalar multiplication carries the subspace
relations \(\overline{\mathcal{M}} + \overline{\mathcal{M}} \subseteq \overline{\mathcal{M}}\) and
\(\alpha \overline{\mathcal{M}} \subseteq \overline{\mathcal{M}}\) over from \(\mathcal{M}\) by taking closures
of continuous images, exactly as in the closed-hull argument for convex sets. Since
\(\mathcal{M} \subseteq \overline{\mathcal{M}} \subseteq \mathcal{X}\) and \(\mathcal{M}\) is maximal proper,
either \(\overline{\mathcal{M}} = \mathcal{M}\) (so \(\mathcal{M}\) is closed) or
\(\overline{\mathcal{M}} = \mathcal{X}\) (so \(\mathcal{M}\) is dense). No third possibility exists.
The final equivalence is the content of conditions (a) and (d) of the preceding theorem: \(\ker f\) closed is
equivalent to continuity of \(f\).
The Gauge of an Open Convex Set
We now perform the conversion of geometry into analysis. From an open convex set containing the origin we read off
a functional that measures, in each direction, how far one must scale the set to swallow a given point. When the
set is also balanced this functional is a seminorm — the
Minkowski functional
already constructed in that symmetric case. Dropping balancedness costs absolute homogeneity but keeps positive
homogeneity and subadditivity: the result is a
sublinear functional,
exactly the kind of bound the Hahn-Banach extension theorem accepts.
Theorem: The Gauge of an Open Convex Neighborhood
Let \(\mathcal{X}\) be a topological vector space and let \(G\) be an open convex subset containing the origin.
Then the gauge of \(G\),
\[
q(x) \;=\; \inf\{\, t \geq 0 : x \in tG \,\},
\]
is a non-negative continuous sublinear functional on \(\mathcal{X}\), and
\[
G \;=\; \{\, x : q(x) \lt 1 \,\}.
\]
Proof
Finiteness and non-negativity.
Every open set containing \(0\) is absorbing: scalar multiplication \(t \mapsto tx\) is continuous and sends
\(t = 0\) into the open set \(G\), so for each \(x\) there is \(\varepsilon \gt 0\) with \(tx \in G\) for
\(0 \leq t \lt \varepsilon\). Hence \(x \in s G\) for \(s = 1/t\) large, the defining set
\(\{t \geq 0 : x \in tG\}\) is nonempty, and \(0 \leq q(x) \lt \infty\).
Positive homogeneity.
Fix \(x\) and a scalar \(\alpha \gt 0\). Substituting \(s = t/\alpha\),
\[
\begin{align*}
q(\alpha x) &= \inf\{\, t \geq 0 : \alpha x \in tG \,\} \\\\
&= \inf\{\, t \geq 0 : x \in (t/\alpha) G \,\} \\\\
&= \alpha \,\inf\{\, s \geq 0 : x \in sG \,\} \;=\; \alpha\, q(x).
\end{align*}
\]
For \(\alpha = 0\) both sides vanish, using \(0 \in G\) so that \(q(0) = 0\). Absolute homogeneity is
not claimed: \(G\) need not be balanced, so \(q(-x)\) and \(q(x)\) may differ.
Subadditivity.
Let \(x, y \in \mathcal{X}\) and \(\delta \gt 0\). Choose \(s, u \gt 0\) with
\(q(x) \leq s \lt q(x) + \delta\), \(x \in sG\), and \(q(y) \leq u \lt q(y) + \delta\), \(y \in uG\); such
positive \(s, u\) exist by the definition of the infimum. Then \(x/s, y/u \in G\), and convexity of \(G\) gives
\[
\frac{x + y}{s + u} \;=\; \frac{s}{s+u}\cdot\frac{x}{s} \;+\; \frac{u}{s+u}\cdot\frac{y}{u} \;\in\; G ,
\]
since the coefficients are non-negative and sum to \(1\). Hence \(x + y \in (s+u) G\), so
\(q(x+y) \leq s + u \lt q(x) + q(y) + 2\delta\). Letting \(\delta \to 0\) yields
\(q(x+y) \leq q(x) + q(y)\). With positive homogeneity, \(q\) is sublinear.
The unit set is \(G\).
Suppose \(q(x) \lt 1\). Then there is \(t\) with \(q(x) \leq t \lt 1\) and \(x \in tG\), say \(x = t g\) with
\(g \in G\). Since \(0 \in G\) and \(G\) is convex, the segment from \(0\) to \(g\) lies in \(G\), and
\(x = tg = (1-t)\cdot 0 + t\cdot g \in G\). Thus \(\{q \lt 1\} \subseteq G\). Conversely let \(x \in G\). Because
\(G\) is open, it is absorbing at the point \(x\): there is \(\varepsilon \gt 0\) with \(x + tx \in G\) for
\(0 \leq t \lt \varepsilon\); fixing one such \(t \gt 0\), \((1+t) x \in G\), so \(x \in (1+t)^{-1} G\) and
\(q(x) \leq (1+t)^{-1} \lt 1\). Hence \(G \subseteq \{q \lt 1\}\), and the two sets coincide.
Continuity.
Subadditivity and positive homogeneity give, for any \(x, h\),
\(q(x + h) \leq q(x) + q(h)\) and \(q(x) \leq q(x + h) + q(-h)\), so
\[
-q(-h) \;\leq\; q(x + h) - q(x) \;\leq\; q(h).
\]
Thus it suffices to control \(q(h)\) and \(q(-h)\) for \(h\) near \(0\). Given \(\varepsilon \gt 0\), the set
\(\varepsilon G\) is an open neighborhood of \(0\), and for \(h \in \varepsilon G\) we have
\(q(h) \lt \varepsilon\) by the unit-set identity applied to \(\varepsilon^{-1} h \in G\). Likewise
\(-\varepsilon G\) is an open neighborhood of \(0\) on which \(q(-h) \lt \varepsilon\). On the intersection
\(\varepsilon G \cap (-\varepsilon G)\), both bounds hold, so \(|q(x+h) - q(x)| \lt \varepsilon\). Hence \(q\) is
continuous.
The gauge is the precise generalization of the Minkowski functional to sets that need not be symmetric about the
origin. When \(G\) is in addition balanced, \(q(-x) = q(x)\) is restored and \(q\) becomes a seminorm, recovering
the symmetric construction; the asymmetry permitted here is exactly what lets a single open convex set, sitting
anywhere, generate the dominating functional that the separation arguments will require.
Separating a Point from an Open Convex Set
The first geometric consequence is the prototype for everything that follows: a point lying outside an open convex
set can be peeled away from it by a closed hyperplane. The proof is the conversion principle in action. The gauge
turns the open convex set into a sublinear functional; a one-dimensional functional is defined on the line through
the excluded point and dominated by that gauge; the Hahn-Banach extension theorem spreads it to the whole space
without losing domination; and the resulting hyperplane misses the set. We carry out the real case in full, then
reduce the complex case to it.
Theorem: A Point Outside an Open Convex Set Lies on a Closed Hyperplane Missing It
Let \(\mathcal{X}\) be a topological vector space over \(\mathbb{F}\) and let \(G\) be a nonempty open convex
subset with \(0 \notin G\). Then there is a closed hyperplane \(\mathcal{M} = \ker f\), with
\(f : \mathcal{X} \to \mathbb{F}\) a nonzero continuous linear functional, such that
\(\mathcal{M} \cap G = \varnothing\).
Proof
Case 1: \(\mathbb{F} = \mathbb{R}\).
Pick any \(x_0 \in G\) and set \(H = x_0 - G\). Then \(H\) is open (translation and reflection are
homeomorphisms), convex (the image of the convex \(G\) under the affine map \(x \mapsto x_0 - x\)), and
contains \(0 = x_0 - x_0\). Let \(q\) be its
gauge, a non-negative continuous
sublinear functional with \(H = \{x : q(x) \lt 1\}\). Since \(0 \notin G\), we have
\(x_0 = x_0 - 0 \notin x_0 - G = H\); equivalently \(q(x_0) \geq 1\).
On the one-dimensional subspace \(\mathcal{Y} = \mathbb{R} x_0\) define \(f_0(\alpha x_0) = \alpha\, q(x_0)\).
We verify \(f_0 \leq q\) on \(\mathcal{Y}\). For \(\alpha \geq 0\), positive homogeneity gives
\(f_0(\alpha x_0) = \alpha\, q(x_0) = q(\alpha x_0)\). For \(\alpha \lt 0\), the left side
\(f_0(\alpha x_0) = \alpha\, q(x_0) \leq \alpha \lt 0\) (using \(q(x_0) \geq 1\)) is negative, while the right
side \(q(\alpha x_0) \geq 0\); so \(f_0(\alpha x_0) \leq q(\alpha x_0)\) holds in this case too. Thus
\(f_0 \leq q\) on \(\mathcal{Y}\).
By the Hahn-Banach theorem
there is a linear functional \(f : \mathcal{X} \to \mathbb{R}\) with \(f|_{\mathcal{Y}} = f_0\) and
\(f \leq q\) on all of \(\mathcal{X}\). It is nonzero, since \(f(x_0) = q(x_0) \geq 1\). It is continuous:
\(f \leq q\) and \(f(-x) \leq q(-x)\) give \(-q(-x) \leq f(x) \leq q(x)\), and as \(q\) is continuous with
\(q(0) = 0\), the functional \(f\) is squeezed to continuity at \(0\), hence continuous by the
continuity criterion.
Set \(\mathcal{M} = \ker f\), a closed hyperplane. To see \(\mathcal{M} \cap G = \varnothing\), it is cleaner to
show \(f \lt f(x_0)\) on all of \(G\). Take \(x \in G\). Then \(x_0 - x \in x_0 - G = H\), so
\(q(x_0 - x) \lt 1\), and therefore
\[
f(x_0) - f(x) = f(x_0 - x) \leq q(x_0 - x) \lt 1 .
\]
Hence \(f(x) \gt f(x_0) - 1 = q(x_0) - 1 \geq 0\) for every \(x \in G\). In particular \(f(x) \gt 0\) on \(G\),
so no point of \(G\) lies in \(\mathcal{M} = \{f = 0\}\), giving \(\mathcal{M} \cap G = \varnothing\).
Case 2: \(\mathbb{F} = \mathbb{C}\).
View \(\mathcal{X}\) as a real topological vector space by restricting scalars to \(\mathbb{R}\); the set \(G\)
is still open, convex, and avoids \(0\). Case 1 produces a continuous \(\mathbb{R}\)-linear functional
\(f : \mathcal{X} \to \mathbb{R}\) with \(G \cap \ker f = \varnothing\). Define
\(F(x) = f(x) - i\,f(ix)\). By the
recovery lemma,
\(F\) is \(\mathbb{C}\)-linear with \(f = \operatorname{Re} F\), and \(F\) is continuous because \(f\) and
\(x \mapsto f(ix)\) are. Now \(F(x) = 0\) forces \(\operatorname{Re} F(x) = f(x) = 0\), so
\(\ker F \subseteq \ker f\), whence \(\ker F \cap G = \varnothing\). Taking \(\mathcal{M} = \ker F\) completes
the complex case.
The same engine separates a point not from a set through the origin but from an arbitrary
affine object — a translate of a subspace. The reduction is to quotient out the subspace and apply
the theorem in the quotient.
Affine Subspaces and the Quotient Reduction
Recall that a linear subspace \(\mathcal{Y} \subseteq \mathcal{X}\) carried along by a single translate forms an
affine subspace: a set of the form \(x_1 + \mathcal{Y}\). When \(\mathcal{X}\) is a topological
vector space and \(\mathcal{Y}\) is closed, the
quotient
\(\mathcal{X}/\mathcal{Y}\), with the natural map \(Q : \mathcal{X} \to \mathcal{X}/\mathcal{Y}\), is itself a
topological vector space, and \(Q\) is continuous and open. Continuity is the defining property of the quotient
topology; openness follows because for an open \(U \subseteq \mathcal{X}\) the saturation
\(Q^{-1}(Q(U)) = U + \mathcal{Y} = \bigcup_{y \in \mathcal{Y}} (U + y)\) is a union of translates of \(U\), hence
open, so \(Q(U)\) is open by definition of the quotient topology. We use only these two facts.
Corollary: Separating an Affine Subspace from an Open Convex Set
Let \(\mathcal{X}\) be a topological vector space and let \(G\) be a nonempty open convex subset. If
\(\mathcal{A}\) is an affine subspace of \(\mathcal{X}\) with \(\mathcal{A} \cap G = \varnothing\), then there
is a closed affine hyperplane \(\mathcal{M}\) with \(\mathcal{A} \subseteq \mathcal{M}\) and
\(\mathcal{M} \cap G = \varnothing\).
Proof
Since \(G\) is open and \(\mathcal{A} \cap G = \varnothing\), the closure also satisfies
\(\overline{\mathcal{A}} \cap G = \varnothing\): any \(x \in \overline{\mathcal{A}} \cap G\) would have the
open neighborhood \(G\) meeting \(\mathcal{A}\), contradicting disjointness. As the separating hyperplane we
build is closed, separating \(\overline{\mathcal{A}}\) yields the same conclusion for
\(\mathcal{A} \subseteq \overline{\mathcal{A}}\), so we may replace \(\mathcal{A}\) by \(\overline{\mathcal{A}}\)
and assume it closed. Fix \(x_1 \in \mathcal{A}\) and replace \((\mathcal{A}, G)\) by
\((\mathcal{A} - x_1, G - x_1)\); this translation preserves openness, convexity, disjointness, and the
conclusion, so we may assume \(\mathcal{A}\) is a closed linear subspace \(\mathcal{Y}\) containing \(0\). Let
\(Q : \mathcal{X} \to \mathcal{X}/\mathcal{Y}\) be the quotient map. Then
\(Q(G)\) is open (since \(Q\) is open) and convex (the image of a convex set under the linear map \(Q\)), and
\(0 \notin Q(G)\): if \(0 = Q(g)\) for some \(g \in G\) then \(g \in \mathcal{Y} = \mathcal{A}\), contradicting
\(\mathcal{A} \cap G = \varnothing\).
By the preceding theorem applied in \(\mathcal{X}/\mathcal{Y}\), there is a nonzero continuous linear
functional \(g\) on \(\mathcal{X}/\mathcal{Y}\) with \(\ker g \cap Q(G) = \varnothing\). Set \(f = g \circ Q\),
a nonzero continuous linear functional on \(\mathcal{X}\) (continuous as a composition of continuous maps),
and \(\mathcal{M} = \ker f = Q^{-1}(\ker g)\). Since \(\mathcal{Y} = \ker Q \subseteq \mathcal{M}\), we have
\(\mathcal{A} = \mathcal{Y} \subseteq \mathcal{M}\). If some \(x \in \mathcal{M} \cap G\), then
\(Q(x) \in \ker g \cap Q(G) = \varnothing\), impossible. Hence \(\mathcal{M} \cap G = \varnothing\), and
\(\mathcal{M}\), being the kernel of a continuous functional, is a closed (affine) hyperplane.
Half-Spaces and Separated Sets
A nonzero continuous real-linear functional splits the space into the two open regions on either side of a level
hyperplane. Naming these regions fixes the vocabulary in which every separation statement is phrased.
Definition: Open and Closed Half-Spaces
Let \(\mathcal{X}\) be a real topological vector space. A subset \(S \subseteq \mathcal{X}\) is an
open half-space if there is a continuous linear functional \(f : \mathcal{X} \to \mathbb{R}\)
and a scalar \(\alpha\) with
\[
S = \{\, x \in \mathcal{X} : f(x) \gt \alpha \,\} ,
\]
and a closed half-space if \(S = \{\, x : f(x) \geq \alpha \,\}\) for some such \(f\) and
\(\alpha\). The hyperplane \(\{f = \alpha\}\) is the boundary of the half-space.
Definition: Separated and Strictly Separated Sets
Two subsets \(A\) and \(B\) of a real topological vector space are strictly separated if they
are contained in disjoint open half-spaces, and separated if they are contained in two
closed half-spaces whose intersection is a closed affine hyperplane. Equivalently, \(A\) and \(B\) are
separated by a continuous linear functional \(f\) and scalar \(\alpha\) when
\[
f(a) \leq \alpha \leq f(b) \qquad \text{for all } a \in A,\ b \in B,
\]
and strictly separated when these inequalities can be made strict on both sides with a gap.
The geometric force of the definition is visible already: when \(f\) is a nonzero continuous real-linear
functional, its kernel hyperplane disconnects the space into the two
connected components
\(\{f \gt 0\}\) and \(\{f \lt 0\}\). To separate two sets is to certify that one such functional sees them on
opposite sides. The theorems of the next section establish when this certificate exists.
The Separation Theorems
We now reach the results that justify the name. The point-separation theorem is upgraded to separate two convex
sets, then sharpened so that disjoint closed convex sets — one of them compact — are pried fully
apart with a gap between them. The progression is governed by how much openness or compactness is available: an
open set lets the gauge construction run, while a compact set can be fattened into an open one without colliding
with its closed companion.
Functionals and Half-Spaces
Two technical facts make the half-space language interchangeable with the functional language: a half-space and
its boundary behave as expected under closure and interior, and a continuous functional maps an open convex set to
an open interval. The second is what later forces separating inequalities to be strict.
Proposition: Half-Spaces and the Functional Criterion
Let \(\mathcal{X}\) be a real topological vector space.
(a) The closure of an open half-space \(\{f \gt \alpha\}\) is the closed half-space
\(\{f \geq \alpha\}\), and the interior of \(\{f \geq \alpha\}\) is \(\{f \gt \alpha\}\), provided \(f\) is a
nonzero continuous linear functional.
(b) If \(f : \mathcal{X} \to \mathbb{R}\) is a nonzero continuous linear functional and \(A\)
is open and convex, then \(f(A)\) is an open interval; in particular \(f(A)\) contains none of its endpoints.
Proof
(a).
Because \(f\) is nonzero, pick \(e\) with \(f(e) = 1\). The map \(t \mapsto x + t e\) is continuous, and
\(f(x + te) = f(x) + t\), so along this line \(f\) takes every real value near \(f(x)\). If \(f(x) = \alpha\),
then for \(t \gt 0\) the point \(x + te\) has \(f \gt \alpha\), and \(x + te \to x\) as \(t \to 0^+\); hence
every boundary point of \(\{f \gt \alpha\}\) is a limit of interior points, giving
\(\overline{\{f \gt \alpha\}} \supseteq \{f \geq \alpha\}\). The reverse inclusion holds because
\(\{f \geq \alpha\} = f^{-1}([\alpha, \infty))\) is closed and contains \(\{f \gt \alpha\}\). The interior
statement is dual: \(\{f \gt \alpha\} = f^{-1}((\alpha, \infty))\) is open and contained in
\(\{f \geq \alpha\}\), and any point with \(f(x) = \alpha\) has points \(x - te\) (\(t \gt 0\)) arbitrarily
close with \(f \lt \alpha\), so it is not interior.
(b).
Since \(A\) is convex and \(f\) is linear, \(f(A)\) is convex in \(\mathbb{R}\), hence an interval. It remains
to show \(f(A)\) is open. Let \(\beta = f(a) \in f(A)\) with \(a \in A\), and choose \(e\) with \(f(e) = 1\).
As \(A\) is open, there is \(\varepsilon \gt 0\) with \(a \pm te \in A\) for \(0 \leq t \lt \varepsilon\)
(openness gives a neighborhood of \(a\), and \(t \mapsto a + te\) is continuous through \(t = 0\)). Then
\(f(a \pm te) = \beta \pm t\) ranges over \((\beta - \varepsilon, \beta + \varepsilon)\), so this whole
interval lies in \(f(A)\). Thus every point of \(f(A)\) is interior, and \(f(A)\) is open.
Part (b) is the engine behind strict separation: as soon as a separating functional is applied to an open convex
set, the resulting interval cannot touch the separating value, which forces a strict inequality. With this in
hand the main separation theorem follows by reducing two sets to one through their difference.
Separation of Disjoint Convex Sets
Theorem: Separation of Disjoint Convex Sets
Let \(\mathcal{X}\) be a real topological vector space and let \(A\) and \(B\) be disjoint convex subsets with
\(A\) open. Then there is a continuous linear functional \(f : \mathcal{X} \to \mathbb{R}\) and a real scalar
\(\alpha\) with
\[
f(a) \lt \alpha \leq f(b) \qquad \text{for all } a \in A,\ b \in B .
\]
If \(B\) is also open, then \(A\) and \(B\) are strictly separated.
Proof
Form the difference set \(G = A - B = \{\, a - b : a \in A,\ b \in B \,\}\). It is convex: a convex
combination \((1-t)(a_1 - b_1) + t(a_2 - b_2)\) equals \(\bigl((1-t)a_1 + t a_2\bigr) - \bigl((1-t)b_1 + t
b_2\bigr)\), a difference of points of \(A\) and \(B\). It is open, being the union
\(G = \bigcup_{b \in B}(A - b)\) of translates of the open set \(A\). Moreover \(0 \notin G\): if
\(0 = a - b\) then \(a = b \in A \cap B\), contradicting disjointness.
By the point-separation theorem,
applied to the open convex set \(G\) avoiding \(0\), there is a nonzero continuous linear functional \(f_1\)
with \(\ker f_1 \cap G = \varnothing\); the proof of that theorem produced a fixed sign on \(G\), so we may
take \(f_1 \gt 0\) on \(G\). Set \(f = -f_1\), matching the stated orientation, so \(f \lt 0\) on \(G\). Then
for all \(a \in A\) and \(b \in B\),
\[
f(a) - f(b) = f(a - b) \lt 0, \qquad \text{that is} \qquad f(a) \lt f(b) .
\]
Hence \(\sup_{a \in A} f(a) \leq \inf_{b \in B} f(b)\); pick \(\alpha\) between them, so that
\(f(a) \leq \alpha\) for all \(a \in A\) and \(f(b) \geq \alpha\) for all \(b \in B\).
Since \(A\) is open and \(f \neq 0\), part (b) of the preceding proposition shows \(f(A)\) is an open interval,
so it cannot contain its supremum; therefore \(f(a) \lt \alpha\) strictly for every \(a \in A\), giving
\(f(a) \lt \alpha \leq f(b)\). If \(B\) is also open, the same argument applied to \(f(B)\) — whose
infimum cannot be attained — shows \(f(b) \gt \alpha\) strictly for every \(b \in B\), so \(A\) and
\(B\) lie in the disjoint open half-spaces \(\{f \lt \alpha\}\) and \(\{f \gt \alpha\}\) and are strictly
separated.
The hypothesis that one set be open is essential to the construction: it is what makes \(G = A - B\) open and what
sharpens the inequality on its side. When neither set is open we cannot run the gauge, and a different lever is
needed — compactness. The next lemma supplies it, fattening a compact set into an open neighborhood that
still avoids a given closed companion.
Fattening a Compact Set
In a topological vector space, a
compact
set sitting inside an open set can be enlarged by a fixed neighborhood of
the origin and still remain inside. This is the topological-vector-space refinement of the tube lemma, and it is
where the additive structure interacts with compactness.
Lemma: A Compact Set Has a Fattening Inside Any Open Superset
Let \(\mathcal{X}\) be a topological vector space, let \(K \subseteq \mathcal{X}\) be
compact,
and let \(V \subseteq \mathcal{X}\) be open with \(K \subseteq V\). Then there is an open neighborhood \(U\) of
\(0\) such that
\[
K + U \;\subseteq\; V .
\]
Proof
Fix \(x \in K\). Then \(x \in V\), and \(V - x\) is an open neighborhood of \(0\). Continuity of addition at
\((0,0)\) furnishes an open neighborhood \(W_x\) of \(0\) with \(W_x + W_x \subseteq V - x\), i.e.
\(x + W_x + W_x \subseteq V\). The translates \(\{x + W_x : x \in K\}\) form an open cover of the compact set
\(K\), so finitely many suffice: there are \(x_1, \ldots, x_n \in K\) with
\(K \subseteq \bigcup_{j=1}^n (x_j + W_{x_j})\). Put
\[
U \;=\; \bigcap_{j=1}^n W_{x_j} ,
\]
an open neighborhood of \(0\) as a finite intersection of such.
Let \(y \in K\) and \(u \in U\). Then \(y \in x_j + W_{x_j}\) for some \(j\), so \(y = x_j + w\) with
\(w \in W_{x_j}\), and \(u \in U \subseteq W_{x_j}\). Therefore
\[
y + u = x_j + w + u \in x_j + W_{x_j} + W_{x_j} \subseteq V .
\]
Hence \(K + U \subseteq V\).
The compactness of \(K\) is indispensable: the cover by translated neighborhoods admits a finite subcover only
because \(K\) is compact, and it is the finite intersection that produces a single \(U\) working uniformly across
\(K\). Phrased through
nets, the failure for
a merely closed \(K\) is exactly that a net escaping to infinity along \(K\) has no
convergent
behavior to anchor the uniform neighborhood; compactness restores the cluster point that the finite subcover
encodes.
Strict Separation with a Compact Set
Combining the fattening lemma with the convex-separation theorem yields the result used most often in practice:
in a locally convex
space, two disjoint closed convex sets can be strictly separated as soon as one of them is
compact. Local convexity is what guarantees the fattening can be taken convex, so that the separation theorem
applies to the enlarged sets.
Theorem: Strict Separation of a Compact and a Closed Convex Set
Let \(\mathcal{X}\) be a real locally convex space
and let \(A\) and \(B\) be disjoint closed convex subsets of \(\mathcal{X}\). If \(B\) is compact, then \(A\)
and \(B\) are strictly separated: there is a continuous linear functional \(f\) and scalars
\(\alpha_1 \lt \alpha_2\) with \(f(a) \leq \alpha_1 \lt \alpha_2 \leq f(b)\) for all \(a \in A\), \(b \in B\).
Proof
Since \(A\) and \(B\) are disjoint and \(A\) is closed, \(B\) is a compact subset of the open set
\(\mathcal{X} \setminus A\). By the
fattening lemma, there is an open
neighborhood \(U_1\) of \(0\) with \(B + U_1 \subseteq \mathcal{X} \setminus A\), that is,
\((B + U_1) \cap A = \varnothing\). Because \(\mathcal{X}\) is locally convex, there is a continuous
seminorm \(p\) with
\(\{x : p(x) \lt 1\} \subseteq U_1\); set \(U = \{x : p(x) \lt \tfrac{1}{2}\}\), an open convex balanced
neighborhood of \(0\) (a sublevel set of a seminorm) with \(U + U \subseteq U_1\).
Consider \(A + U\) and \(B + U\). Each is open (a union of translates of \(U\)) and convex (a sum of convex
sets is convex), and they contain \(A\) and \(B\) respectively. They are disjoint: if \(a + u = b + u'\) with
\(a \in A\), \(b \in B\), \(u, u' \in U\), then \(a = b + (u' - u) \in B + (U - U) \subseteq B + U_1\), since
\(U\) is balanced so \(U - U \subseteq U + U \subseteq U_1\); this places \(a \in (B + U_1) \cap A\), which is
empty. Hence \((A + U) \cap (B + U) = \varnothing\).
Now \(A + U\) and \(B + U\) are disjoint open convex sets, so by the
convex-separation theorem they
are strictly separated by some continuous linear functional \(f\): there is \(\alpha\) with \(f \lt \alpha\)
on \(A + U\) and \(f \gt \alpha\) on \(B + U\). Restricting to \(A \subseteq A + U\) and \(B \subseteq B + U\)
gives \(f \lt \alpha\) on \(A\) and \(f \gt \alpha\) on \(B\). Since \(f(B)\) is the continuous image of the
compact set \(B\), it is compact in \(\mathbb{R}\) and attains its minimum, so
\(\alpha_2 := \min_{b \in B} f(b) \gt \alpha\). Setting \(\alpha_1 := \alpha\) gives
\(f(a) \lt \alpha_1 \lt \alpha_2 \leq f(b)\) for all \(a \in A\), \(b \in B\), the asserted strict separation
with a gap.
Compactness of one set cannot be dropped. In the plane, the closed convex region below the \(x\)-axis and the
closed convex region on or above a branch of the hyperbola \(y = 1/x\) for \(x \gt 0\) are disjoint and closed,
yet they approach each other arbitrarily closely as \(x \to \infty\); no horizontal gap separates them, and no
line strictly separates the two. The gap in the conclusion is precisely what compactness buys.
Convex Sets as Intersections of Half-Spaces
The separation theorems have a structural payoff that closes the circle opened in the first section. There we
said a functional cuts the space into half-spaces; we can now run that statement backwards. A closed convex set
is exactly the intersection of all the closed half-spaces that contain it — it is carved out of the space by
the continuous linear functionals that stay on one side of it. This is the geometric counterpart of the analytic
fact that a closed subspace is the common kernel of the functionals vanishing on it, and it is the precise sense
in which a weakly convergent sequence cannot escape the convex hull of its terms.
Separating a Point from a Closed Convex Set
Specializing the compact-versus-closed theorem to a single point gives the workhorse corollary: any point outside
a closed convex set is strictly separated from it. A single point is compact, so the hypotheses are met with no
further effort.
Corollary: A Point Outside a Closed Convex Set Is Strictly Separated From It
Let \(\mathcal{X}\) be a real locally convex space, let \(A\) be a closed convex subset, and let
\(x \notin A\). Then \(x\) is strictly separated from \(A\): there is a continuous linear functional \(f\) and
scalars \(\alpha_1 \lt \alpha_2\) with \(f(a) \leq \alpha_1 \lt \alpha_2 \leq f(x)\) for all \(a \in A\).
Proof
The singleton \(B = \{x\}\) is convex, closed, and compact, and \(A \cap B = \varnothing\) because
\(x \notin A\). The strict-separation
theorem applied to \(A\) and \(B\) yields a continuous linear functional \(f\) and scalars
\(\alpha_1 \lt \alpha_2\) with \(f(a) \leq \alpha_1\) for all \(a \in A\) and \(\alpha_2 \leq f(x)\).
The Closed Convex Hull as an Intersection of Half-Spaces
A closed half-space is itself closed and convex, so any intersection of closed half-spaces is closed and convex.
The separation corollary supplies the converse: every closed convex set arises this way, and applied to the
smallest closed convex set containing a given set, it identifies the
closed convex hull
as the intersection of all closed half-spaces over the set.
Theorem: The Closed Convex Hull Is an Intersection of Closed Half-Spaces
Let \(\mathcal{X}\) be a real locally convex space and let \(A \subseteq \mathcal{X}\). Then the closed convex
hull \(\overline{\operatorname{co}}(A)\) equals the intersection of all closed half-spaces containing \(A\):
\[
\overline{\operatorname{co}}(A) \;=\; \bigcap\{\, H : H \text{ a closed half-space},\ A \subseteq H \,\} .
\]
In particular, every closed convex set is the intersection of the closed half-spaces containing it.
Proof
Let \(\mathcal{H}\) be the collection of all closed half-spaces containing \(A\), and write
\(P = \bigcap\{H : H \in \mathcal{H}\}\).
\(\overline{\operatorname{co}}(A) \subseteq P\).
Each \(H \in \mathcal{H}\) is closed and convex and contains \(A\), so it contains the smallest closed convex
set containing \(A\), namely \(\overline{\operatorname{co}}(A)\). Intersecting over all \(H \in \mathcal{H}\)
gives \(\overline{\operatorname{co}}(A) \subseteq P\).
\(P \subseteq \overline{\operatorname{co}}(A)\).
We prove the contrapositive: if \(x \notin \overline{\operatorname{co}}(A)\), then \(x \notin P\). The set
\(\overline{\operatorname{co}}(A)\) is closed and convex, and \(x\) lies outside it, so by the
preceding corollary there is a
continuous linear functional \(f\) and scalars \(\alpha_1 \lt \alpha_2\) with \(f(y) \leq \alpha_1\) for all
\(y \in \overline{\operatorname{co}}(A)\) and \(f(x) \geq \alpha_2\). The closed half-space
\(H = \{\, y : f(y) \leq \alpha_1 \,\}\) then contains \(A \subseteq \overline{\operatorname{co}}(A)\), so
\(H \in \mathcal{H}\); but \(f(x) \geq \alpha_2 \gt \alpha_1\) means \(x \notin H\), hence \(x \notin P\).
The two inclusions give \(\overline{\operatorname{co}}(A) = P\). Taking \(A\) itself closed and convex makes
\(\overline{\operatorname{co}}(A) = A\), so \(A\) is the intersection of the closed half-spaces containing it.
This is the geometric form of Mazur's principle. A point in the closure of the convex hull of a set cannot be
told apart from that set by any continuous linear functional kept on one side; equivalently, if some functional
strictly separates a point from the set, that point lies outside the closed convex hull. The same dichotomy,
read on a subspace rather than a general convex set, recovers the analytic characterization of the closed span.
Corollary: The Closed Span Is an Intersection of Closed Hyperplanes
Let \(\mathcal{X}\) be a real locally convex space and let \(A \subseteq \mathcal{X}\). Then the closed linear
span of \(A\) equals the intersection of all closed hyperplanes \(\{f = 0\}\) containing \(A\), as \(f\)
ranges over the continuous linear functionals vanishing on \(A\).
Proof
Let \(\mathcal{Y} = \overline{\operatorname{span}}(A)\), a closed linear subspace, and let \(P\) be the
intersection of all closed hyperplanes \(\ker f\) containing \(A\) with \(f\) a continuous linear functional.
If \(f\) is continuous and vanishes on \(A\), it vanishes on the span of \(A\) by linearity and on its closure
\(\mathcal{Y}\) by continuity; hence \(\mathcal{Y} \subseteq \ker f\) for each such \(f\), giving
\(\mathcal{Y} \subseteq P\).
Conversely, suppose \(x \notin \mathcal{Y}\). Then \(\mathcal{Y}\) is a closed convex set not containing \(x\),
so by the separation corollary there
is a continuous linear functional \(g\) with \(g(y) \leq \alpha_1 \lt \alpha_2 \leq g(x)\) for all
\(y \in \mathcal{Y}\). A linear functional bounded above on the subspace \(\mathcal{Y}\) must vanish on it: if
\(g(y_0) \neq 0\) for some \(y_0 \in \mathcal{Y}\), then \(g(t y_0) = t\, g(y_0)\) is unbounded above as
\(t \to \pm\infty\) within \(\mathcal{Y}\), contradicting the bound \(g \leq \alpha_1\). Hence \(g\) vanishes on
\(\mathcal{Y} \supseteq A\), so \(\ker g\) is a closed hyperplane containing \(A\); but \(g(x) \geq \alpha_2
\gt \alpha_1 \geq 0\) gives \(g(x) \neq 0\), so \(x \notin \ker g\) and therefore \(x \notin P\). Thus
\(P \subseteq \mathcal{Y}\), and the two inclusions give \(P = \mathcal{Y}\).
The Complex Case
Every complex locally convex space is also a real locally convex space, obtained by restricting scalar
multiplication to real scalars; the topology and convexity are untouched. The separation theorems therefore apply
to the underlying real space, and the real functionals they produce are promoted to complex functionals by the
recovery lemma. The separating inequalities are then phrased through the real part, since that is what the real
theory controls.
Theorem: Strict Separation in a Complex Locally Convex Space
Let \(\mathcal{X}\) be a complex locally convex space and let \(A\) and \(B\) be disjoint closed convex
subsets with \(B\) compact. Then there are a continuous linear functional \(f : \mathcal{X} \to \mathbb{C}\),
a real scalar \(\alpha\), and an \(\varepsilon \gt 0\) such that
\[
\operatorname{Re} f(a) \;\leq\; \alpha \;\lt\; \alpha + \varepsilon \;\leq\; \operatorname{Re} f(b)
\qquad \text{for all } a \in A,\ b \in B .
\]
Proof
Regard \(\mathcal{X}\) as a real locally convex space. The sets \(A\) and \(B\) remain disjoint, closed, and
convex, and \(B\) remains compact. By the
real strict-separation theorem
there is a continuous \(\mathbb{R}\)-linear functional \(u : \mathcal{X} \to \mathbb{R}\) and scalars
\(\alpha \lt \alpha + \varepsilon\) with \(u(a) \leq \alpha\) for all \(a \in A\) and
\(u(b) \geq \alpha + \varepsilon\) for all \(b \in B\).
Define \(f(x) = u(x) - i\, u(ix)\). By the
recovery lemma,
\(f\) is \(\mathbb{C}\)-linear with \(u = \operatorname{Re} f\), and \(f\) is continuous because \(u\) and
\(x \mapsto u(ix)\) are continuous. Substituting \(u = \operatorname{Re} f\) into the inequalities above gives
\[
\operatorname{Re} f(a) \;\leq\; \alpha \;\lt\; \alpha + \varepsilon \;\leq\; \operatorname{Re} f(b)
\]
for all \(a \in A\) and \(b \in B\), as claimed.
With this the geometric theory is complete in both scalar fields. A continuous linear functional is the basic
instrument of separation; over \(\mathbb{R}\) it acts directly, and over \(\mathbb{C}\) it acts through its real
part, which is the real functional the separation theorems actually build. The closed convex sets of a locally
convex space are thus exactly the sets cut out by these functionals — the intersections of the closed
half-spaces that contain them — and the existence of the functionals, the deep input that makes the whole
geometry possible, traces back through the gauge construction to the single extension principle from which this
chapter began.