From Norms to Seminorm Families
Every space we have studied so far carried a single
norm, and that one
function controlled everything: the open sets, the convergent sequences, the continuous maps. Yet the most
important topologies in functional analysis are not generated by any norm. When we weakened
convergence to study existence of minimizers — declaring \(x_n \to x\) whenever
\(\varphi(x_n) \to \varphi(x)\) for every continuous functional \(\varphi\) — we built a topology in which
the closed unit ball becomes compact, something no norm topology on an infinite-dimensional space can achieve.
That topology is not induced by a metric, and the tools calibrated to a single norm no longer reach it.
The right level of generality replaces one norm by a whole family of weaker measurements. We work throughout
over a fixed scalar field \(\mathbb{F}\), either \(\mathbb{R}\) or \(\mathbb{C}\); the constructions are
identical for both unless stated otherwise. The measurements are
seminorms: maps
\(p : \mathcal{X} \to [0, \infty)\) that are subadditive and absolutely homogeneous but may vanish on nonzero
vectors. A single seminorm is too coarse to separate points, but a sufficiently rich family of them
recovers a usable geometry. Before isolating which families qualify, we record the structure that any sensible
topology on a vector space must respect.
Topological Vector Spaces
A topology on a vector space is useful only when it is compatible with the algebra. The two vector-space
operations — addition and scalar multiplication — should be continuous, so that limits interact with
the linear structure exactly as they do in a normed space.
Definition: Topological Vector Space
A topological vector space (TVS) over \(\mathbb{F}\) is a vector space \(\mathcal{X}\)
equipped with a topology
for which the two structure maps are continuous:
(a) addition \(\mathcal{X} \times \mathcal{X} \to \mathcal{X}\), \((x, y) \mapsto x + y\),
is continuous;
(b) scalar multiplication \(\mathbb{F} \times \mathcal{X} \to \mathcal{X}\),
\((\alpha, x) \mapsto \alpha x\), is continuous.
Here \(\mathcal{X} \times \mathcal{X}\) and \(\mathbb{F} \times \mathcal{X}\) carry the
product topology,
with \(\mathbb{F}\) given its usual metric topology.
Continuity of these two maps has an immediate structural consequence that we will use repeatedly. For any fixed
\(x_0 \in \mathcal{X}\), the translation \(x \mapsto x + x_0\) is a homeomorphism of \(\mathcal{X}\): it is
continuous by (a), and its inverse \(x \mapsto x - x_0\) is continuous for the same reason. Likewise, for any
fixed nonzero scalar \(\alpha\), the dilation \(x \mapsto \alpha x\) is a homeomorphism, with inverse
\(x \mapsto \alpha^{-1} x\). Translations and dilations therefore move the topology around rigidly: the open sets
near any point are exact translates of the open sets near the origin. The local structure of a TVS is completely
determined by the neighborhoods of \(0\).
Every normed space is a topological vector space. If \(\mathcal{X}\) carries a norm, then the reverse triangle
inequality \(\bigl|\,\|x\| - \|x'\|\,\bigr| \leq \|x - x'\|\) shows the norm is continuous, and the estimates
\[
\|(x + y) - (x_0 + y_0)\| \;\leq\; \|x - x_0\| + \|y - y_0\|,
\]
\[
\|\alpha x - \alpha_0 x_0\| \;\leq\; |\alpha|\,\|x - x_0\| + |\alpha - \alpha_0|\,\|x_0\|,
\]
show addition and scalar multiplication are continuous at every point. The second estimate follows by adding and
subtracting \(\alpha x_0\) inside the norm and applying the triangle inequality together with absolute homogeneity
\(\|\alpha x\| = |\alpha|\,\|x\|\). Norms thus furnish topological vector spaces; we turn now to the examples that
no norm produces.
Convergence Without a Metric: Nets
In a metric space, sequences are enough: a point lies in the closure of a set exactly when some sequence from the
set converges to it, and a map is continuous exactly when it preserves convergent sequences. Both equivalences
rest silently on the existence of a countable neighborhood basis at each point — the shrinking
balls of radius \(1/n\). The topologies we are heading toward have no such countable basis. In the weak topology
on an infinite-dimensional space, the neighborhoods of a point are cut out by finitely many functionals at a
time, and there are uncountably many functionals to choose from; no single sequence of neighborhoods can be
cofinal among them. Sequences become blind, and statements about closure and continuity that we take for granted
can fail if phrased through them.
The repair is to allow index sets richer than \(\mathbb{N}\). What made \(\mathbb{N}\) work was not that it was
countable but that it was directed: any two indices have a common successor, so “eventually”
makes sense. Abstracting exactly that property yields the notion of a net, which restores both equivalences in
full generality.
Directed Sets and Nets
Definition: Directed Set
A directed set is a set \(I\) equipped with a partial order \(\leq\) (a reflexive,
antisymmetric, transitive relation) such that any two elements have a common upper bound: for all
\(i_1, i_2 \in I\) there exists \(i_3 \in I\) with \(i_1 \leq i_3\) and \(i_2 \leq i_3\).
The natural numbers under their usual order form a directed set, so every construction below specializes to the
familiar one for sequences. The example that matters for topology is different. Fix a point \(x_0\) in a
topological space and let \(\mathcal{U}\) be the collection of all open sets containing \(x_0\). Order
\(\mathcal{U}\) by reverse inclusion: declare \(U \leq V\) to mean \(U \supseteq V\), so that
“larger” means “smaller neighborhood.” Given two neighborhoods \(U_1, U_2\) of \(x_0\),
their intersection \(U_1 \cap U_2\) is again an open set containing \(x_0\), and it is contained in both; under
reverse inclusion it is an upper bound for the pair. Thus the neighborhood system at a point is a directed set,
and the smaller a neighborhood is, the further along the direction it sits.
Definition: Net
A net in a set \(X\) is a function \(x : I \to X\) from a directed set \(I\) into \(X\). We
write \(x_i\) for \(x(i)\) and denote the net by \(\{x_i\}_{i \in I}\), or simply \(\{x_i\}\) when the index
set is understood. A sequence is precisely a net indexed by \(\mathbb{N}\).
Definition: Convergence of a Net
Let \(\{x_i\}_{i \in I}\) be a net in a topological space \(X\) and let \(x_0 \in X\). The net
converges to \(x_0\), written \(x_i \to x_0\), if for every open set \(U\) containing
\(x_0\) there is an index \(i_0 \in I\) such that
\[
x_i \in U \qquad \text{for all } i \geq i_0.
\]
We then call \(x_0\) a limit of the net. As with sequences, it suffices to check this
condition for \(U\) ranging over any neighborhood basis at \(x_0\), since every open set containing \(x_0\)
then contains a basic one.
Closure and Continuity Through Nets
On a metric space, a point belongs to the closure of a set precisely when it is the limit of a sequence drawn
from the set. The
neighborhood
characterization of closure — that \(x \in \overline{A}\) iff every neighborhood of \(x\)
meets \(A\) — holds on any topological space, and nets convert it into a convergence statement verbatim.
Theorem: Net Characterization of Closure
Let \(X\) be a topological space and \(A \subseteq X\). Then \(x \in \overline{A}\) if and only if there is a
net \(\{a_i\}\) in \(A\) with \(a_i \to x\).
Proof
From a net to the closure.
Suppose \(\{a_i\}\) is a net in \(A\) with \(a_i \to x\). Let \(U\) be any open set containing \(x\). By
convergence there is an index \(i_0\) with \(a_i \in U\) for all \(i \geq i_0\); in particular
\(a_{i_0} \in U \cap A\), so the neighborhood \(U\) meets \(A\). Since every neighborhood of \(x\) intersects
\(A\), the neighborhood characterization of closure gives \(x \in \overline{A}\).
From the closure to a net.
Suppose \(x \in \overline{A}\). Let \(I\) be the collection of all open sets containing \(x\), directed by
reverse inclusion as above. For each \(U \in I\), the point \(x\) lies in \(\overline{A}\), so \(U\) meets
\(A\); choose a point \(a_U \in U \cap A\). This defines a net \(\{a_U\}_{U \in I}\) in \(A\). To see that it
converges to \(x\), let \(W\) be any open set containing \(x\). Then \(W \in I\), and for every
\(U \geq W\) — that is, every \(U \subseteq W\) — we have \(a_U \in U \subseteq W\). Thus
\(a_U \in W\) for all \(U \geq W\), which is exactly \(a_U \to x\).
Continuity admits the same upgrade. The preimage-of-open-sets definition is the official one, but the working
criterion in analysis is that continuous maps preserve limits. For sequences this is only necessary in general
spaces; for nets it is both necessary and sufficient at every point.
Theorem: Net Characterization of Continuity
Let \(X\) and \(Y\) be topological spaces, \(f : X \to Y\), and \(x_0 \in X\). Then \(f\) is continuous at
\(x_0\) if and only if \(f(x_i) \to f(x_0)\) for every net \(\{x_i\}\) in \(X\) with \(x_i \to x_0\).
Proof
Continuity implies preservation of limits.
Assume \(f\) is continuous at \(x_0\) and let \(x_i \to x_0\). Let \(V\) be any open set containing
\(f(x_0)\). By continuity at \(x_0\) there is an open set \(U\) containing \(x_0\) with \(f(U) \subseteq V\).
Since \(x_i \to x_0\), there is an index \(i_0\) with \(x_i \in U\) for all \(i \geq i_0\); hence
\(f(x_i) \in f(U) \subseteq V\) for all \(i \geq i_0\). As \(V\) was an arbitrary open set containing
\(f(x_0)\), this is \(f(x_i) \to f(x_0)\).
Preservation of limits implies continuity.
We prove the contrapositive. Suppose \(f\) is not continuous at \(x_0\). Then there is an open set \(V\)
containing \(f(x_0)\) such that no open neighborhood of \(x_0\) maps into \(V\); that is, for every open set
\(U\) containing \(x_0\) there is a point \(x_U \in U\) with \(f(x_U) \notin V\). Index these points by the
neighborhood system \(I\) of \(x_0\), directed by reverse inclusion. The resulting net \(\{x_U\}_{U \in I}\)
converges to \(x_0\): given any open \(W \ni x_0\), every \(U \geq W\) satisfies \(x_U \in U \subseteq W\).
Yet \(f(x_U) \notin V\) for all \(U\), so the net \(\{f(x_U)\}\) never enters the neighborhood \(V\) of
\(f(x_0)\) and therefore cannot converge to \(f(x_0)\). Thus limit preservation fails, completing the
contrapositive.
Weak convergence was defined precisely so that a net converges weakly exactly when it converges against every
continuous functional; the net language is what makes that definition interact correctly with closures and
continuous maps. With convergence now available on any topological vector space, we return to the central
question: which topologies on a vector space arise from a family of seminorms.
Locally Convex Spaces and Seminorm Topologies
We now make precise the construction sketched at the outset: a family of seminorms generates a topology, and the
topological vector spaces that arise this way are the locally convex ones. Let \(\mathcal{X}\) be a vector space
over \(\mathbb{F}\) and let \(\mathcal{P}\) be a family of
seminorms on \(\mathcal{X}\). Take as
a subbase the collection of
all sets
\[
\{\, x \in \mathcal{X} : p(x - x_0) \lt \varepsilon \,\}, \qquad p \in \mathcal{P}, \; x_0 \in \mathcal{X}, \;
\varepsilon > 0.
\]
A set \(U\) is then open precisely when, for each \(x_0 \in U\), there are finitely many seminorms
\(p_1, \dots, p_n \in \mathcal{P}\) and \(\varepsilon_1, \dots, \varepsilon_n > 0\) with
\[
\bigcap_{j=1}^{n} \{\, x : p_j(x - x_0) \lt \varepsilon_j \,\} \;\subseteq\; U.
\]
This is the smallest topology making every \(p \in \mathcal{P}\) continuous and every translate of a
seminorm-ball open. Under it, addition and scalar multiplication are continuous, so \(\mathcal{X}\) becomes a
topological vector space; the verification is the same triangle-inequality estimate used for a single norm,
applied one seminorm at a time.
One condition is still needed to make the topology Hausdorff: the seminorms must collectively detect every
vector, so that distinct points can be separated.
Definition: Locally Convex Space
A locally convex space (LCS) is a topological vector space whose topology is generated, in
the manner above, by a family \(\mathcal{P}\) of seminorms that separates points:
\[
\bigcap_{p \in \mathcal{P}} \{\, x \in \mathcal{X} : p(x) = 0 \,\} \;=\; \{0\}.
\]
The separation condition is imposed precisely so that the topology is Hausdorff. Indeed, if \(x \neq y\),
then \(x - y \neq 0\), so some \(p \in \mathcal{P}\) has \(p(x - y) > 0\); writing
\(\delta := \tfrac{1}{2} p(x - y)\), the basic neighborhoods \(\{z : p(z - x) \lt \delta\}\) and
\(\{z : p(z - y) \lt \delta\}\) are disjoint, since a common point \(z\) would give
\(p(x - y) \leq p(x - z) + p(z - y) \lt 2\delta = p(x - y)\), a contradiction.
Every normed space is locally convex: the single norm forms a separating family by itself. The name records the
geometry we will expose in the final section — each basic neighborhood of \(0\) is a convex set — but
the working definition is the seminorm one, and it is the form in which the weak topologies appear.
Continuity of a Seminorm
Before cataloguing examples, we settle when an individual seminorm is continuous for a given TVS topology. The
answer is a chain of equivalences that reduces global continuity to a single condition at the origin — the
payoff of translation-invariance — and connects continuity to domination by an already-continuous seminorm.
The proof is the first place the net machinery of the previous section earns its keep.
Theorem: Characterizations of a Continuous Seminorm
Let \(\mathcal{X}\) be a topological vector space and let \(p\) be a seminorm on \(\mathcal{X}\). The
following are equivalent.
(a) \(p\) is continuous.
(b) \(\{x : p(x) \lt 1\}\) is open.
(c) \(0\) belongs to the interior of \(\{x : p(x) \lt 1\}\).
(d) \(0\) belongs to the interior of \(\{x : p(x) \leq 1\}\).
(e) \(p\) is continuous at \(0\).
(f) there is a continuous seminorm \(q\) on \(\mathcal{X}\) with \(p \leq q\).
Proof
The implications (a) \(\Rightarrow\) (b) \(\Rightarrow\) (c) \(\Rightarrow\) (d) are immediate: if \(p\) is
continuous then \(\{p \lt 1\}\) is the preimage of the open ray \([0,1)\) and hence open, giving (b); an open
set containing \(0\) witnesses (c) because \(p(0) = 0 \lt 1\); and \(\{p \lt 1\} \subseteq \{p \leq 1\}\) makes
any interior point of the former an interior point of the latter, giving (d).
(d) \(\Rightarrow\) (e).
Assume \(0\) is interior to \(\{x : p(x) \leq 1\}\). Fix \(\varepsilon > 0\). By absolute homogeneity,
\(\varepsilon\{x : p(x) \leq 1\} = \{x : p(x) \leq \varepsilon\}\), and scalar multiplication by
\(\varepsilon\) is a homeomorphism, so \(0\) is interior to \(\{x : p(x) \leq \varepsilon\}\) as well; that
is, this set is a neighborhood of \(0\). Now let \(x_i \to 0\) be any net. Since the set is a neighborhood of
\(0\), there is an index \(i_0\) with \(x_i \in \{x : p(x) \leq \varepsilon\}\) for all \(i \geq i_0\), i.e.
\(p(x_i) \leq \varepsilon\) for \(i \geq i_0\). As \(\varepsilon > 0\) was arbitrary, \(p(x_i) \to 0\). By the
net characterization of continuity, \(p\)
is continuous at \(0\).
(e) \(\Rightarrow\) (a).
The reverse triangle inequality for seminorms, \(|p(x) - p(y)| \leq p(x - y)\), holds because
\(p(x) \leq p(x - y) + p(y)\) and symmetrically \(p(y) \leq p(y - x) + p(x) = p(x - y) + p(x)\). Let
\(x_i \to x\) be a net. Then \(x_i - x \to 0\): subtracting the constant net \(x\) is continuous in a TVS, so
the net characterization applied to \(z \mapsto z - x\) gives \(x_i - x \to 0\). By (e),
\(p(x_i - x) \to 0\), and the reverse triangle inequality forces \(|p(x_i) - p(x)| \to 0\), i.e.
\(p(x_i) \to p(x)\). Hence \(p\) is continuous at every \(x\) by the net characterization, which is (a).
(a) \(\Rightarrow\) (f).
Take \(q = p\), which is a continuous seminorm dominating \(p\).
(f) \(\Rightarrow\) (e).
Let \(q\) be continuous with \(p \leq q\). If \(x_i \to 0\), then \(q(x_i) \to 0\) by continuity of \(q\) at
\(0\) (which (a) \(\Rightarrow\) (e) supplies for \(q\)), and \(0 \leq p(x_i) \leq q(x_i)\) forces
\(p(x_i) \to 0\). Thus \(p\) is continuous at \(0\).
The catalogue of continuous seminorms is closed under the operations one would hope for. This is what lets us
enlarge a generating family without changing the topology: passing to finite sums and suprema produces no new
open sets.
Theorem: Operations on Continuous Seminorms
Let \(\mathcal{X}\) be a topological vector space.
(a) If \(p_1, \dots, p_n\) are continuous seminorms, then \(p_1 + \dots + p_n\) and
\(\max(p_1, \dots, p_n)\) are continuous seminorms.
(b) If \(\{p_i\}_{i \in I}\) is a family of continuous seminorms for which there exists a
single continuous seminorm \(q\) with \(p_i \leq q\) for all \(i\), then \(x \mapsto \sup_i p_i(x)\) is a
continuous seminorm.
Proof
(a).
A finite sum and a finite maximum of seminorms are again seminorms: each satisfies absolute homogeneity
termwise, and subadditivity passes through because \((p_1 + p_2)(x + y) \leq (p_1 + p_2)(x) + (p_1 + p_2)(y)\)
and \(\max_k p_k(x + y) \leq \max_k\bigl(p_k(x) + p_k(y)\bigr) \leq \max_k p_k(x) + \max_k p_k(y)\). For
continuity, let \(x_i \to x\). By the equivalence above each \(p_k\) is continuous, so
\(p_k(x_i) \to p_k(x)\) for each \(k\). For finitely many real nets, both the sum and the maximum are
continuous functions of the entries, so a finite sum of convergent nets converges to the sum of the limits,
giving \((p_1 + \dots + p_n)(x_i) \to (p_1 + \dots + p_n)(x)\); and a finite maximum converges to the maximum
of the limits — concretely, \(|\max_k a_k - \max_k b_k| \leq \max_k |a_k - b_k|\) drives the differences
to zero — giving \(\max_k p_k(x_i) \to \max_k p_k(x)\). Both are therefore continuous.
(b).
Write \(p(x) := \sup_i p_i(x)\). The bound \(p_i \leq q\) gives \(p(x) \leq q(x) \lt \infty\), so \(p\) is
finite-valued. It is a seminorm: absolute homogeneity is
\(p(\alpha x) = \sup_i |\alpha|\, p_i(x) = |\alpha|\, p(x)\), and for subadditivity,
\[
p_i(x + y) \;\leq\; p_i(x) + p_i(y) \;\leq\; p(x) + p(y) \qquad \text{for every } i,
\]
so taking the supremum over \(i\) on the left yields \(p(x + y) \leq p(x) + p(y)\). Finally \(p \leq q\) with
\(q\) continuous, so \(p\) is continuous by part (f) of the preceding theorem.
The Weak and Weak* Topologies as Seminorm Topologies
The motivating examples now fall into place as instances of the general construction. Let \(\mathcal{X}\) be a
normed space with continuous dual \(\mathcal{X}^*\). For each functional \(x^* \in \mathcal{X}^*\), define
\[
p_{x^*}(x) \;=\; |x^*(x)|.
\]
Each \(p_{x^*}\) is a seminorm: it is absolutely homogeneous because \(x^*\) is linear, and subadditive because
the modulus is. The family \(\mathcal{P} = \{\, p_{x^*} : x^* \in \mathcal{X}^* \,\}\) separates points exactly
when the dual separates points — if \(x \neq 0\) there is an \(x^*\) with \(x^*(x) \neq 0\), so
\(p_{x^*}(x) > 0\) — and that separation is the guarantee the extension theorem was proved to supply. The
topology this family generates is the weak topology: a net converges weakly,
\(x_i \to x\), if and only if \(p_{x^*}(x_i - x) = |x^*(x_i) - x^*(x)| \to 0\) for every \(x^*\), which is
convergence tested against every continuous functional. The weak topology is thus a locally convex topology, and
the description of weak convergence used earlier — one net at a time, against each functional — is
precisely net convergence in this seminorm topology.
Dually, fix the space \(\mathcal{X}^*\) and let each vector \(x \in \mathcal{X}\) act on it by
\[
p_x(x^*) \;=\; |x^*(x)|.
\]
The family \(\{\, p_x : x \in \mathcal{X} \,\}\) of seminorms on \(\mathcal{X}^*\) separates points of
\(\mathcal{X}^*\), since a functional vanishing on every \(x\) is the zero functional. The locally convex
topology it generates is the weak* topology on \(\mathcal{X}^*\), under which a net of
functionals converges exactly when it converges pointwise on \(\mathcal{X}\). These two topologies — weak
on \(\mathcal{X}\), weak* on \(\mathcal{X}^*\) — are the reason locally convex spaces enter functional
analysis at all, and both are now seen to be ordinary seminorm topologies with no metric behind them.
Convex Geometry and the Minkowski Functional
The name “locally convex” promises geometry, and we now deliver it. The seminorm description and a
purely geometric description of these spaces — in terms of convex sets — are two faces of one object,
and the dictionary translating between them is the Minkowski functional. It manufactures a
seminorm out of a convex set and, conversely, realizes every seminorm as the gauge of its own unit ball. This is
the device that the geometric, separation-type consequences of the extension theorem will run on, so we develop
the set-theoretic vocabulary it requires before constructing it.
Convex Sets, Hulls, and Symmetry
Recall the notion of a
convex set: a set \(C\) is convex
when the line segment joining any two of its points lies in \(C\). The segment \(\{(1-t)a + tb : t \in [0,1]\}\)
is defined using only addition and scalar multiplication, so although the definition was first introduced for a
normed space, it applies verbatim in any vector space, and we use it at that generality here. Convexity, stated
for two points, automatically governs all finite convex combinations.
Theorem: Convex Combinations and Intersections
Let \(\mathcal{X}\) be a vector space.
(a) A set \(C \subseteq \mathcal{X}\) is convex if and only if, whenever
\(x_1, \dots, x_n \in C\) and \(t_1, \dots, t_n \in [0,1]\) with \(\sum_{k=1}^n t_k = 1\), the combination
\(\sum_{k=1}^n t_k x_k\) lies in \(C\).
(b) If \(\{C_\alpha\}_{\alpha \in A}\) is any collection of convex sets, then
\(\bigcap_{\alpha \in A} C_\alpha\) is convex.
Proof
(a).
If the convex-combination condition holds, then taking \(n = 2\) recovers the definition of convexity, so
\(C\) is convex. Conversely, suppose \(C\) is convex; we induct on \(n\). The cases \(n = 1\) (trivial) and
\(n = 2\) (the definition) hold. Assume the claim for combinations of \(n - 1\) points, and let
\(x_1, \dots, x_n \in C\) with weights \(t_k \in [0,1]\) summing to \(1\). If \(t_n = 1\), then all other
weights vanish and the combination is \(x_n \in C\). Otherwise \(s := 1 - t_n = \sum_{k=1}^{n-1} t_k > 0\),
and we write
\[
\sum_{k=1}^{n} t_k x_k \;=\; s \underbrace{\left( \sum_{k=1}^{n-1} \frac{t_k}{s}\, x_k \right)}_{=: \, y}
\;+\; t_n x_n.
\]
The weights \(t_k / s\) for \(1 \leq k \leq n-1\) are non-negative and sum to \(1\), so by the inductive
hypothesis \(y \in C\). Then \(s + t_n = 1\) with \(s, t_n \in [0,1]\), so \(s y + t_n x_n\) is a convex
combination of the two points \(y, x_n \in C\), hence lies in \(C\) by convexity.
(b).
Let \(a, b \in \bigcap_\alpha C_\alpha\) and \(t \in [0,1]\). For each \(\alpha\), both \(a\) and \(b\) lie in
the convex set \(C_\alpha\), so \((1-t)a + tb \in C_\alpha\). As this holds for every \(\alpha\), the point
lies in the intersection. Thus the intersection is convex.
Part (b) guarantees that the following constructions are well posed: an arbitrary set has a smallest convex
superset, obtained by intersecting all convex sets that contain it.
Definition: Convex Hull
Let \(A \subseteq \mathcal{X}\). The convex hull of \(A\), denoted
\(\operatorname{co}(A)\), is the intersection of all convex sets containing \(A\). It is the smallest convex
set containing \(A\): it is convex by the intersection property, contains \(A\), and is contained in any
convex set that contains \(A\).
The convex hull is intrinsic to the linear structure. In a topological vector space one frequently needs the
smallest closed convex set containing \(A\); since an intersection of closed convex sets is both closed
and convex, this too is well defined.
Definition: Closed Convex Hull
Let \(\mathcal{X}\) be a topological vector space and \(A \subseteq \mathcal{X}\). The closed convex
hull of \(A\), denoted \(\overline{\operatorname{co}}(A)\), is the intersection of all closed convex
sets containing \(A\). It is the smallest closed convex set containing \(A\).
The two hulls are related in the simplest possible way: closing the convex hull gives the closed convex hull. The
proof uses that the closure of a convex set is convex, which we record first.
Theorem: The Closed Convex Hull Is the Closure of the Convex Hull
Let \(\mathcal{X}\) be a topological vector space and \(A \subseteq \mathcal{X}\). Then the closure of a
convex set is convex, and consequently
\[
\overline{\operatorname{co}}(A) \;=\; \overline{\operatorname{co}(A)}.
\]
Proof
The closure of a convex set is convex.
Let \(C\) be convex and fix \(t \in [0,1]\). Consider the map
\(m : \mathcal{X} \times \mathcal{X} \to \mathcal{X}\), \(m(x, y) = (1-t)x + t y\), which is continuous
because addition and scalar multiplication are continuous in a topological vector space. Convexity of \(C\)
says \(m(C \times C) \subseteq C\). Taking closures and using that \(m\) is continuous, together with the
product-topology identity \(\overline{C \times C} = \overline{C} \times \overline{C}\), we get
\[
m\bigl(\overline{C} \times \overline{C}\bigr) \;=\; m\bigl(\overline{C \times C}\bigr) \;\subseteq\;
\overline{m(C \times C)} \;\subseteq\; \overline{C}.
\]
Hence \((1-t)x + ty \in \overline{C}\) for all \(x, y \in \overline{C}\), so \(\overline{C}\) is convex.
Equality of the two hulls.
Since \(\operatorname{co}(A)\) is convex, its closure \(\overline{\operatorname{co}(A)}\) is convex by the
first part, and it is closed and contains \(A\); being a closed convex set containing \(A\), it contains the
smallest such set, so \(\overline{\operatorname{co}}(A) \subseteq \overline{\operatorname{co}(A)}\).
Conversely, \(\overline{\operatorname{co}}(A)\) is a closed convex set containing \(A\), hence contains
\(\operatorname{co}(A)\); being closed, it contains the closure \(\overline{\operatorname{co}(A)}\). The two
inclusions give equality.
The interior of a convex set interacts with its closure through a segment principle: starting at an interior point
and moving toward any point of the closure, every point short of the destination is again interior.
Theorem: Segment-Interior Property of Convex Sets
Let \(\mathcal{X}\) be a topological vector space and \(A \subseteq \mathcal{X}\) a convex set. If
\(a \in \operatorname{int} A\) and \(b \in \operatorname{cl} A\), then
\[
(1 - t)\, a + t\, b \in \operatorname{int} A \qquad \text{for every } 0 \leq t \lt 1 .
\]
In particular the half-open segment \(\{(1 - t) a + t b : 0 \leq t \lt 1\}\) lies in \(\operatorname{int} A\).
Proof
For \(t = 0\) the point is \(a \in \operatorname{int} A\). Fix \(0 \lt t \lt 1\) and set
\(c = (1 - t) a + t b\). Because \(a \in \operatorname{int} A\), there is an open neighborhood \(W\) of \(0\)
with \(a + W \subseteq A\). The set \(b - \tfrac{1 - t}{t}\, W\) is a neighborhood of \(b\), since translation
and multiplication by the nonzero scalar \(-\tfrac{1 - t}{t}\) are homeomorphisms of \(\mathcal{X}\); as
\(b \in \operatorname{cl} A\), it meets \(A\). Choose \(b' \in A\) with
\(b - b' \in \tfrac{1 - t}{t}\, W\), so that \(t(b - b') \in (1 - t) W\). Then
\[
c = (1 - t) a + t b = (1 - t) a + t b' + t(b - b') \in (1 - t)(a + W) + t b' .
\]
Every point of \((1 - t)(a + W) + t b'\) has the form \((1 - t) a'' + t b'\) with \(a'' \in a + W \subseteq A\)
and \(b' \in A\); by convexity it lies in \(A\). Hence \((1 - t)(a + W) + t b' \subseteq A\). This set equals
\((1 - t) a + t b' + (1 - t) W\), a translate of the open set \((1 - t) W\) (open because
\(x \mapsto (1 - t) x\) is a homeomorphism for \(1 - t \neq 0\)), so it is open; and it contains \(c\).
Therefore \(c\) lies in an open subset of \(A\), giving \(c \in \operatorname{int} A\).
Two further symmetry properties of sets complete the vocabulary. They describe how a set sits relative to scalar
multiplication and how thoroughly it fills space around the origin.
Definition: Balanced and Absorbing Sets
Let \(A \subseteq \mathcal{X}\) be a subset of a vector space over \(\mathbb{F}\).
\(A\) is balanced if \(\alpha x \in A\) whenever \(x \in A\) and \(|\alpha| \leq 1\).
\(A\) is absorbing if for each \(x \in \mathcal{X}\) there is an \(\varepsilon > 0\) such
that \(t x \in A\) for all \(0 \leq t \lt \varepsilon\).
A balanced set contains \(0\) (take \(\alpha = 0\), provided \(A\) is nonempty), and an absorbing set contains
\(0\) (take \(x = 0\)). We say \(A\) is absorbing at a point \(a \in A\) if \(A - a\) is
absorbing — equivalently, if for each \(x\) there is an \(\varepsilon > 0\) with
\(a + tx \in A\) for \(0 \leq t \lt \varepsilon\). In a topological vector space, every open set containing
\(0\) is absorbing, and is absorbing at each of its points, because scalar multiplication
\(t \mapsto a + tx\) is continuous and sends \(t = 0\) into the open set.
The Minkowski Functional
We can now state the central correspondence. A seminorm \(p\) gives rise to the set \(\{x : p(x) \lt 1\}\), which
is convex (a sublevel set of a seminorm), balanced (by absolute homogeneity), and absorbing at each of its points
(given \(x\), the value \(p(a + tx)\) varies continuously and stays below \(1\) for small \(t\) when
\(p(a) \lt 1\)). The Minkowski functional reverses this passage: from such a set it reconstructs the seminorm.
Theorem: The Minkowski Functional (Gauge)
Let \(\mathcal{X}\) be a vector space over \(\mathbb{F}\) and let \(V\) be a nonempty convex balanced set that
is absorbing at each of its points. Then the Minkowski functional (or gauge)
of \(V\),
\[
p(x) \;=\; \inf\{\, t \geq 0 : x \in tV \,\},
\]
is the unique seminorm on \(\mathcal{X}\) for which \(V = \{\, x : p(x) \lt 1 \,\}\).
Proof
The infimum is well defined and \(p(0) = 0\).
Since \(V\) is absorbing (it is absorbing at the point \(0 \in V\)), for each \(x\) there is
\(\varepsilon > 0\) with \(tx \in V\) for small \(t > 0\); equivalently \(x \in s V\) for large
\(s = 1/t\), so the set \(\{t \geq 0 : x \in tV\}\) is nonempty and \(p(x) \lt \infty\). Taking \(x = 0\),
every \(t > 0\) has \(0 \in tV\), so \(p(0) = 0\).
Absolute homogeneity.
Fix \(x\) and a scalar \(\alpha\). If \(\alpha = 0\), both sides of \(p(\alpha x) = |\alpha| p(x)\) vanish.
Suppose \(\alpha \neq 0\). Because \(V\) is balanced, \(\alpha V = |\alpha| V\): writing
\(\alpha = |\alpha|\,\omega\) with \(|\omega| = 1\), the map \(v \mapsto \omega v\) carries \(V\) onto itself,
so \(\alpha V = |\alpha|\,\omega V = |\alpha| V\). Therefore
\[
\begin{align*}
p(\alpha x) &= \inf\{\, t \geq 0 : \alpha x \in tV \,\} \\\\
&= \inf\{\, t \geq 0 : x \in t\,\alpha^{-1} V \,\} \\\\
&= \inf\{\, t \geq 0 : x \in (t/|\alpha|)\, V \,\}.
\end{align*}
\]
Substituting \(s = t/|\alpha|\) gives \(p(\alpha x) = |\alpha| \inf\{\, s \geq 0 : x \in sV \,\} =
|\alpha|\, p(x)\).
Subadditivity.
Let \(x, y \in \mathcal{X}\) and let \(\delta > 0\). Choose \(s, u \geq 0\) with \(p(x) \leq s \lt p(x) + \delta\)
and \(x \in sV\), and likewise \(u\) with \(p(y) \leq u \lt p(y) + \delta\) and \(y \in uV\); such \(s, u\)
exist by definition of the infimum (and may be taken positive, enlarging slightly if a value is \(0\)). Then
\(x/s, y/u \in V\), and by convexity of \(V\),
\[
\frac{x + y}{s + u} \;=\; \frac{s}{s+u}\cdot\frac{x}{s} \;+\; \frac{u}{s+u}\cdot\frac{y}{u} \;\in\; V,
\]
since the coefficients \(\tfrac{s}{s+u}\) and \(\tfrac{u}{s+u}\) are non-negative and sum to \(1\). Hence
\(x + y \in (s + u) V\), so \(p(x+y) \leq s + u \lt p(x) + p(y) + 2\delta\). Letting \(\delta \to 0\) gives
\(p(x + y) \leq p(x) + p(y)\). With absolute homogeneity, \(p\) is a seminorm.
The unit set is \(V\).
Suppose \(p(x) \lt 1\). Then there is \(t\) with \(p(x) \leq t \lt 1\) and \(x \in tV\), say \(x = t v\) with
\(v \in V\). Since \(V\) is balanced and \(|t| \leq 1\), we have \(tv \in V\), so \(x \in V\). (If \(t = 0\)
then \(x = 0 \in V\) because \(V\) is nonempty and balanced.) Thus \(\{p \lt 1\} \subseteq V\). Conversely, let \(x \in V\). Because \(V\) is absorbing at the point \(x\), there
is \(\varepsilon > 0\) with \(x + t x \in V\) for \(0 \leq t \lt \varepsilon\); fixing one such \(t > 0\),
\((1 + t) x \in V\), so \(x \in (1+t)^{-1} V\) and therefore \(p(x) \leq (1 + t)^{-1} \lt 1\). Thus
\(V \subseteq \{p \lt 1\}\), and the two sets coincide.
Uniqueness.
If \(q\) is any seminorm with \(V = \{x : q(x) \lt 1\}\), then for every \(x\) and every \(t > 0\),
\[
x \in tV \;\iff\; q(x/t) \lt 1 \;\iff\; q(x) \lt t.
\]
Thus the positive numbers \(t\) with \(x \in tV\) are exactly those with \(t > q(x)\), so
\(p(x) = \inf\{t \geq 0 : x \in tV\} = \inf\{t > 0 : t > q(x)\} = q(x)\). Therefore \(q = p\).
The correspondence is now a closed loop. Every seminorm is the gauge of its open unit set, and every nonempty
convex balanced set absorbing at each of its points is the open unit set of a unique seminorm. Geometry and
seminorms carry exactly the same information.
The Geometric Characterization of Local Convexity
With the gauge in hand, the defining feature of a locally convex space can be stated without mentioning seminorms
at all — in purely geometric terms, as a condition on the neighborhoods of the origin. One hypothesis must be
retained explicitly. A locally convex space is Hausdorff, since its generating seminorms separate points; conversely,
a convex balanced neighborhood basis alone does not force separation — under the indiscrete topology the single
set \(\mathcal{X}\) is an open convex balanced neighborhood basis of \(0\), yet the space is not Hausdorff and is not
locally convex. We therefore assume the space Hausdorff, which is exactly the missing separation.
Theorem: Geometric Characterization of Local Convexity
Let \(\mathcal{X}\) be a Hausdorff topological vector space and let \(\mathcal{U}\) be the
collection of all open convex balanced subsets of \(\mathcal{X}\). Then \(\mathcal{X}\) is locally convex if
and only if \(\mathcal{U}\) is a neighborhood basis at \(0\) — that is, every open set containing \(0\)
contains a member of \(\mathcal{U}\).
Proof
Seminorm topology gives a convex balanced basis.
Suppose \(\mathcal{X}\) is locally convex, with topology generated by a separating family \(\mathcal{P}\) of
seminorms. A basic neighborhood of \(0\) is a finite intersection
\(W = \bigcap_{j=1}^n \{x : p_j(x) \lt \varepsilon_j\}\). Each set \(\{x : p_j(x) \lt \varepsilon_j\}\) is convex
(sublevel set of a seminorm) and balanced (by absolute homogeneity), and it is open. A finite intersection of
open convex balanced sets is open, convex (by the intersection property), and balanced (if \(|\alpha| \leq 1\)
and \(x\) lies in each set, so does \(\alpha x\)). Thus \(W \in \mathcal{U}\). Since the sets \(W\) form a
neighborhood basis at \(0\), so does the larger collection \(\mathcal{U}\).
A convex balanced basis gives a seminorm topology.
Conversely, suppose \(\mathcal{U}\) is a neighborhood basis at \(0\). For each \(V \in \mathcal{U}\), the set
\(V\) is open, hence absorbing at each of its points, and it is convex and balanced; by the gauge theorem its
Minkowski functional \(p_V\) is a seminorm with \(V = \{x : p_V(x) \lt 1\}\). Each \(p_V\) is continuous,
because \(\{p_V \lt 1\} = V\) is open and part (b) of the continuity characterization applies. Let
\(\mathcal{P} = \{\, p_V : V \in \mathcal{U} \,\}\). We claim \(\mathcal{P}\) generates the topology of
\(\mathcal{X}\) and separates points.
For the topology: each \(p_V\) is continuous, so every seminorm-ball \(\{x : p_V(x - x_0) \lt \varepsilon\}\) is
open in \(\mathcal{X}\), meaning the seminorm topology is coarser than the given one. For the reverse, let
\(U\) be open with \(x_0 \in U\). Then \(U - x_0\) is an open neighborhood of \(0\), so it contains some
\(V \in \mathcal{U}\); scaling, \(\varepsilon V \subseteq U - x_0\) for a suitable \(\varepsilon\), and
\(\varepsilon V = \{x : p_V(x) \lt \varepsilon\}\). Hence \(\{x : p_V(x - x_0) \lt \varepsilon\} \subseteq U\),
exhibiting \(U\) as a union of seminorm-balls. The two topologies coincide.
For separation: by hypothesis \(\mathcal{X}\) is Hausdorff, so given \(x \neq 0\) there is an open
neighborhood of \(0\) excluding \(x\), hence a \(V \in \mathcal{U}\) with \(x \notin V\). Then
\(p_V(x) \geq 1 > 0\), so the family \(\mathcal{P}\) separates \(x\) from \(0\). As this holds for every
nonzero \(x\), the intersection \(\bigcap_V \{p_V = 0\}\) is \(\{0\}\), and \(\mathcal{X}\) is locally convex
in the sense of the definition.
Local convexity is therefore not a statement about any particular family of seminorms but a property of the space
itself: that its origin has a neighborhood basis of convex balanced sets. The two presentations — a
separating family of seminorms, or a convex balanced neighborhood basis — determine each other through the
gauge. This is the geometric foundation on which the separation of convex sets by continuous functionals, and the
extreme-point structure of compact convex sets, will be built.