Vector Addition is Continuous in Weak Topology

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Theorem

Let $K$ be a topological field.

Let $X$ be a topological vector space over $K$ with weak topology $w$.

Define $s : \struct {X, w} \times \struct {X, w} \to \struct {X, w}$ by:

$\map s {x, y} = x + y$

for each $x, y \in X$.


Then $s$ is continuous.

That is, vector addition remains continuous when restricting to the weak topology.


Proof

Let $X^\ast$ be the topological dual space of $X$.

From Continuity in Initial Topology, it suffices to show that for each $f \in X^\ast$ we have:

$f \circ s : \struct {X, w} \times \struct {X, w} \to K$ is continuous.

Define the projections $\pr_1 : \struct {X, w} \times {X, w} \to \struct {X, w}$ and $\pr_2 : \struct {X, w} \times {X, w} \to \struct {X, w}$ as the projection onto the first and second factors.

Then for each $x, y \in X$ we have:

\(\ds \map {\paren {f \circ s} } {x, y}\) \(=\) \(\ds \map f {x + y}\)
\(\ds \) \(=\) \(\ds \map f x + \map f y\) Definition of Linear Functional
\(\ds \) \(=\) \(\ds \map f {\map {\pr_1} {\tuple {x, y} } } + \map f {\map {\pr_2} {\tuple {x, y} } }\)

That is:

$f \circ s = f \circ \pr_1 + f \circ \pr_2$

From the definition of the product topology:

$\pr_1 : \struct {X, w} \times \struct {X, w} \to \struct {X, w}$

and:

$\pr_2 : \struct {X, w} \times \struct {X, w} \to \struct {X, w}$

are continuous.

From the definition of the weak topology, $f : \struct {X, w} \to K$ is continuous.

From Composite of Continuous Mappings is Continuous, $f \circ \pr_1 : \struct {X, w} \times \struct {X, w} \to K$ and $f \circ \pr_2 : \struct {X, w} \times \struct {X, w} \to K$ are continuous.

From Sum of Continuous Functions on Topological Ring is Continuous, $f \circ \pr_1 + f \circ \pr_2 : \struct {X, w} \times \struct {X, w} \to K$ is continuous.

So we have that $f \circ s : \struct {X, w} \to K$ is continuous for each $f \in X^\ast$.

So from Continuity in Initial Topology, $s : \struct {X, w} \times \struct {X, w} \to \struct {X, w}$ is continuous in the weak topology.

$\blacksquare$