Scalar Multiplication is Continuous in Weak Topology

Theorem

Let $K$ be a topological field.

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

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

$\map m {\lambda, x} = \lambda x$

for each $\lambda \in K$, $x \in X$.

Then $m$ is continuous.

That is, scalar multiplication 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:

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

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

Then for each $\lambda \in K$ and $x \in X$ we have:

\(\ds \map {\paren {f \circ m} } {\lambda, x}\)

\(=\)

\(\ds \map f {\lambda x}\)

\(\ds \)

\(=\)

\(\ds \lambda \map f x\)

Definition of Linear Functional

\(\ds \)

\(=\)

\(\ds \map {\pr_1} {\tuple {\lambda, x} } \map f {\map {\pr_2} {\tuple {\lambda, x} } }\)

That is:

$f \circ m = \pr_1 \cdot \paren {f \circ \pr_2}$

From the definition of the product topology:

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

and:

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

are continuous.

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

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

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

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

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

$\blacksquare$