Vector Addition on Locally Convex Space is Continuous
Theorem
Let $\GF \in \set {\R, \C}$.
Let $\struct {X, \PP}$ be a locally convex space over $\GF$ equipped with the standard topolgoy.
Let $\struct {X \times X, \tau}$ be the Cartesian product $X \times X$ equipped with the product topology.
Let $+ : \struct {X \times X, \tau} \to \struct {X, \PP}$ be the vector addition defined on $X$.
Then $+ : \struct {X \times X, \tau} \to \struct {X, \PP}$ is a continuous mapping.
Proof
From the definition of the standard topolgoy, the topology on $\struct {X, \PP}$ has sub-basis:
- $\SS = \set {\map {B_p} {x, \epsilon} : p \in \PP, \, \epsilon > 0}$
where we define:
- $\map {B_p} {x, \epsilon} = \set {y \in X : \map p {y - x} < \epsilon}$
for each $p \in \PP$ and $\epsilon > 0$.
From Continuity Test using Sub-Basis, it suffices to show that:
- $+^{-1} \sqbrk {\map {B_p} {x_0, \epsilon} } \in \tau$ for each $p \in \PP$, $\epsilon > 0$ and $x_0 \in X$.
That is:
- $\set {\tuple {x, y} \in X \times X : \map p {x + y - x_0} < \epsilon} \in \tau$
For brevity, let:
- $U_{x_0} = \set {\tuple {x, y} \in X \times X : \map p {x + y - x_0} < \epsilon}$
Let $\tuple {x, y} \in U_{x_0}$ and $\tuple {x', y'} \in X \times X$.
We then have:
\(\ds \map p {x' + y' - x_0}\) | \(=\) | \(\ds \map p {x' + y' - x_0 + x + y - x - y}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \map p {\paren {x + y - x_0} + \paren {x' - x} + \paren {y' - y} }\) | ||||||||||||
\(\ds \) | \(\le\) | \(\ds \map p {x + y - x_0} + \map p {x' - x} + \map p {y' - y}\) | Seminorm Axiom $\text N 3$: Triangle Inequality |
Since $\tuple {x, y} \in U_{x_0}$, we have that:
- $\map p {x + y - x_0} < \epsilon$
Set:
- $\ds \map {\epsilon'} {x, y} = \frac {\epsilon - \map p {x + y - x_0} } 2$
Then, for $\tuple {x', y'} \in \map {B_p} {x, \map {\epsilon'} {x, y} } \times \map {B_p} {y, \map {\epsilon'} {x, y} }$, we have:
\(\ds \map p {x + y - x_0} + \map p {x' - x} + \map p {y' - y}\) | \(<\) | \(\ds \map p {x + y - x_0} + \frac {\epsilon - \map p {x + y - x_0} } 2 + \frac {\epsilon - \map p {x + y - x_0} } 2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \map p {x + y - x_0} + \epsilon - \map p {x + y - x_0}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \epsilon\) |
so that:
- $\map p {x + y - x_0} < \epsilon$
and so:
- $\map {B_p} {x, \map {\epsilon'} {x, y} } \times \map {B_p} {y, \map {\epsilon'} {x, y} } \subseteq U_{x_0}$
for each $\tuple {x, y} \in U_{x_0}$.
From Union of Subsets is Subset:
- $\ds \bigcup_{\tuple {x, y} \in U_{x_0} } \map {B_p} {x, \map {\epsilon'} {x, y} } \times \map {B_p} {y, \map {\epsilon'} {x, y} } \subseteq U_{x_0}$
On the other hand, for each $\tuple {x, y} \in U_{x_0}$ we clearly have:
- $\tuple {x, y} \in \map {B_p} {x, \map {\epsilon'} {x, y} } \times \map {B_p} {y, \map {\epsilon'} {x, y} }$
So, we have:
- $\ds U_{x_0} = \bigcup_{\tuple {x, y} \in U_{x_0} } \map {B_p} {x, \map {\epsilon'} {x, y} } \times \map {B_p} {y, \map {\epsilon'} {x, y} }$
From Natural Basis of Product Topology of Finite Product, we have that:
- $\set {\map {B_p} {x, \epsilon_1} \times \map {B_q} {y, \epsilon_2} : p, q \in \PP, \, x, y \in X, \, \epsilon_1, \epsilon_2 > 0}$
forms a basis for $\struct {X \times X, \tau}$.
So:
- $\map {B_p} {x, \map {\epsilon'} {x, y} } \times \map {B_p} {y, \map {\epsilon'} {x, y} } \in \tau$ for each $\tuple {x, y} \in U_{x_0}$.
Since topologies are closed under set union, we have $+^{-1} \sqbrk {\map {B_p} {x_0, \epsilon} } = U_{x_0} \in \tau$, which is what we wanted.
$\blacksquare$