> [!definition] > > Let $E$ be a [[Topological Vector Space|topological vector space]], then the following are equivalent: > - There exists a family $\seqi{[\cdot]}$ of [[Seminorm|seminorms]] such that the [[Topological Space|topology]] on $E$ is generated by the family $\bracs{B_i(x, \eps): i \in I, x \in E, \eps > 0}$, where for each $i \in I$, $x \in E$, and $\eps > 0$, $B_i(x, \eps) = \bracs{y \in E: [x - y]_i < \eps}$. > - There exists a [[Neighbourhood Base|neighbourhood base]] at $0$, consisting of [[Convexity|convex]], [[Balanced Set|balanced]], and [[Absorbing|absorbing]] sets. > > If the above holds, then $E$ is a **locally convex topological vector space** defined by $\seqi{[\cdot]}$, where > > 1. For each $x \in E$, finite intersections of $\bracs{B_i(x, \eps): i \in I, \eps > 0}$ is a neighbourhood base at $x$. > 2. For each $i \in I$, the seminorm $[\cdot]_i: E \to \real^+$ is [[Continuity|continuous]]. > 3. For any [[Net|net]] $\net{x} \subset E$, $x_\alpha \to x$ if and only if $[x_\alpha - x]_i \to 0$ for all $i \in I$. > > Moreover, if $\{[\cdot]_j\}_{j \in J}$ is any family of continuous seminorms on $E$, then $\{[\cdot]_j\}_{j \in J}$ combined with $\seqi{[\cdot]}$ induces the same topology on $E$. The set of *all* continuous seminorms on $E$ is known as *the* system of seminorms defining the topology on $E$. > > *Proof*. Let $\seqi{[\cdot]}$ be a family of seminorms defining the topology on $E$, then finite intersections of $\bracs{B_i(0, \eps): i \in I, \eps > 0}$ forms a neighbourhood base at $0$. Since finite intersections of convex, balanced, and absorbing sets are also convex, balanced, and absorbing, $E$ admits a neighbourhood base at $0$ consisting of the desired sets. > > Now suppose that there is a neighbourhood base $\seqi{U}$ at $0$ consisting of convex, balanced, and absorbing sets. For each $i \in I$, let $[\cdot]_i$ be the [[Gauge|Minkowski functional]] of $U_i$, then $U_i = B_i(0, 1) = \bracs{x \in E: [x]_i < 1}$, so the topology defined by $\seqi{[\cdot]}$ contains the topology on $E$. On the other hand, for any $\eps > 0$, $B_i(0, \eps) = \eps B_i(0, 1)$. Since the topology on $E$ is invariant under translation and scaling, the topology on $E$ contains the topology defined by $\seqi{[\cdot]}$. > > Lastly, if $[\cdot]_j: E \to \real^+$ is a seminorm continuous on $E$, then the topology on $E$ contains $B_j(x, \eps)$ for all $x \in E$ and $\eps > 0$. Therefore adding $[\cdot]_j$ to the family does not change the topology. > [!theorem] > > Let $E, F$ be topological vector spaces, with topologies induced by $\seqi{[\cdot]}$ and $\{[\cdot]_j\}_{j \in J}$, respectively. Let $T: E \to F$ be a [[Linear Transformation|linear map]], then $T$ is continuous if and only for each $j \in J$, there exists $\seqf{i_k} \subset I$ and $C \ge 0$ such that > $ > [Tx]_j \le C\sum_{k = 1}^n [x]_{i_k} \quad \forall x \in E > $ > [!theorem] > > Let $E$ be a locally convex topological vector space, $M \subset E$ be a subspace, and $V \subset M$ be a convex, balanced, and absorbing neighbourhood of $0$, then there exists a convex, balanced, and absorbing neighbourhood $W$ of $0$ in $E$ such that $W \cap M = V$. > > Moreover, if $M$ is closed and $x \not\in M$, then $W$ can be chosen to not contain $x$. > > *Proof*. Let $U \subset E$ be a convex neighbourhood of $0$ such that $U \cap M \subset V$. Let $W$ be the convex hull containing $U$ and $V$, then $W$ is also convex, balanced, and absorbing neighbourhood of $0$ satisfying the desired criteria. > > If $M$ is closed, then $E/M$ is Hausdorff and locally convex. Choosing a neighbourhood here yields the desired neighbourhood.