9.1 Seminorms
Definition 9.1.1 (Convex). Let $E$ be a vector space over $K \in \RC$, then $A \subset E$ is convex if for any $x, y \in A$, $\bracs{\lambda x + (1 - \lambda) y| \lambda \in [0, 1]}\subset A$.
Lemma 9.1.2. Let $E$ be a TVS over $K \in \RC$, $A \subset E$ be convex, $x \in A^{o}$, and $y \in \ol{A}$, then
Proof. Fix $t \in (0, 1)$. Using translation, assume without loss of generality that $tx + (1 - t)y = 0$. In which case, $x = \alpha y$ where $\alpha = (1 - t)/t$. By (TVS2), $\alpha A^{o} \in \cn^{o}(y)$. Since $y \in \ol{A}$, $\alpha A^{o} \cap A \ne \emptyset$, so there exists $z \in A^{o}$ such that $\alpha z \in A$.
Let $\mu = \alpha/(\alpha - 1)$, then since $\alpha < 0$, $\mu \in (0, 1)$ and
By (TVS1) and (TVS2),
so $0 \in A^{o}$.$\square$
Lemma 9.1.3. Let $E$ be a TVS over $K \in \RC$, $A, B \subset E$ be convex, then the following sets are convex:
$A^{o}$.
$\ol{A}$.
$A + B$.
For any $\lambda \in K$, $\lambda A$.
Proof. (1): By Lemma 9.1.2.
(2): Let $x, y \in \ol{A}$. By Definition 4.5.2, there exists filters $\fF, \mathfrak{G}\subset 2^{A}$ such that $\fF$ converges to $x$ and $\mathfrak{G}$ converges to $y$. In which case,
converges to $tx + (1 - t)y$ by (TVS1) and (TVS2). Hence $tx + (1 - t)y \in \ol{A}$.$\square$
Proposition 9.1.4. Let $E$ be a TVS over $K \in \RC$ and $A \subset E$ be convex. If $A^{o} \ne \emptyset$, then $\ol{A}= \ol{A^o}$.
Proof. Since $A^{o} \subset A$, $\ol{A^o}\subset \ol{A}$. Let $x \in A^{o}$, then for any $y \in \ol{A}$,
by Lemma 9.1.2.$\square$
Definition 9.1.5 (Sublinear Functional). Let $E$ be a vector space over $K \in \RC$, then a sublinear functional is a mapping $\rho: E \to \real$ such that:
$\rho(0) = 0$.
For any $x \in E$ and $\lambda \ge 0$, $\rho(\lambda x) = \lambda \rho(x)$.
For any $x, y \in E$, $\rho(x + y) \le \rho(x) + \rho(y)$.
Definition 9.1.6 (Seminorm). Let $E$ be a vector space over $K \in \RC$, then a seminorm on $E$ is a mapping $\rho: E \to [0, \infty)$ such that:
$\rho(0) = 0$.
For any $x \in E$ and $\lambda \in K$, $\rho(\lambda x) = \abs{\lambda}\rho(x)$.
For any $x, y \in E$, $\rho(x + y) \le \rho(x) + \rho(y)$.
Lemma 9.1.7. Let $E$ be a TVS over $K \in \RC$ and $[\cdot]: E \times E \to [0, \infty)$ be a seminorm on $E$, then the following are equivalent:
$[\cdot]$ is uniformly continuous.
$[\cdot]$ is continuous.
$[\cdot]$ is continuous at $0$.
Proof. $(3) \Rightarrow (1)$: Let $\eps > 0$, then there exists $V \in \cn(0)$ such that $[x] < \eps$ for all $x \in V$. In which case, for any $x, y \in E$ with $x - y \in V$, $\abs{[x] - [y]}\le [x - y] < \eps$. By Proposition 8.1.6, $[\cdot]$ is uniformly continuous.$\square$
Definition 9.1.8 (Topology Induced by Seminorm). Let $E$ be a vector space over $K \in \RC$ and $\seqi{[\cdot]}$ be seminorms, then:
For each $i \in I$, $d_{i}: E \times E \to [0, \infty)$ defined by $(x, y) \mapsto [x - y]_{i}$ is a pseudo-metric.
The topology induced by $\seqi{d}$ makes $E$ a topological vector space.
For each $i \in I$, $[\cdot]_{i}: E \to [0, \infty)$ is continuous.
The topology induced by $\seqi{d}$ is the vector space topology induced by $\seqi{[\cdot]}$. In addition,
For any family $\seqj{[\cdot]}$ of seminorms continuous on $E$, the vector space topology induced by $\seqj{[\cdot]}$ is contained in the vector space topology induced by $\seqi{[\cdot]}$.
Definition 9.1.9 (Gauge/Minkowski Functional). Let $E$ be a vector space over $K \in \RC$ and $A \subset E$ be a radial set, then the mapping
is the gauge/Minkowski functional of $A$, and
For any $x \in E$ and $\lambda \ge 0$, $[\lambda x]_{A} = \lambda [x]_{A}$.
If $A$ is convex, then for any $x, y \in E$, $[x + y]_{A} \le [x]_{A} + [y]_{A}$.
If $A$ is circled, then for any $x \in E$ and $\lambda \in K$, $[\lambda x]_{A} = \abs{\lambda}[x]_{A}$.
In particular,
If $A$ is convex, then $[\cdot]_{A}$ is a sublinear functional.
If $A$ is convex and circled, then $[\cdot]_{A}$ is a seminorm.
Proof. (2): Let $\lambda, \mu > 0$ such that $\lambda^{-1}x, \mu^{-1}y \in A$. By convexity, $t\lambda^{-1}+ (1 - t)\mu^{-1}y \in A$ for all $t \in [0, 1]$. Let $t \in [0, 1]$ such that
then $(\lambda + \mu)^{-1}\in A$, and $\lambda + \mu \ge [x + y]_{A}$. Thus $[x + y]_{A} \le [x]_{A} + [y]_{A}$.$\square$
Definition 9.1.10 (Locally Convex Space). Let $E$ be a TVS over $\RC$, then the following are equivalent:
There exists a fundamental system of neighborhoods at $0$ consisting of convex sets.
There exists a fundamental system of neighbourhoods at $0$ consisting of convex, circled, and radial sets.
There exists a family of seminorms $\seqi{[\cdot]}$ that induces the topology on $E$.
If the above holds, then $E$ is a locally convex space.
Proof. $(1) \Rightarrow (2)$: Let $U \in \cn(0)$ be convex. By Proposition 8.1.11, there exists $V \in \cn(0)$ circled such that $V + V \subset U$. Let $W = \bracs{tx + (1 - t)y|x, y \in V}$ be the convex hull of $V$, then $W \subset U$ is convex and circled.
$(2) \Rightarrow (3)$: For each $V \in \cn(0)$ convex, circled, and radial, let $[\cdot]_{V}: E \to [0, \infty)$ be its gauge, then $[\cdot]_{V}$ is a seminorm. For each $x, y \in X$ and $r > 0$, $[x - y]_{V} < r$ if and only if $x - y \in rV$. Thus the uniformity induced by $[\cdot]_{V}$ corresponds to the uniformity generated by $\bracs{U_{rV}| r > 0}$, where $U_{V} = \bracs{(x, y) \in E|x - y \in V}$. Since this holds for all $U \in \cn(0)$, the topology of $E$ and the topology induced by $\bracs{[\cdot]_V| V \in \cn(0), \text{ convex, circled, radial}}$ coincide.
$(3) \Rightarrow (1)$: For each $i \in I$ and $r > 0$, $\bracs{x \in E| [x]_i < r}$ is convex.$\square$