# Scalar Multiplication 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 $\struct {\GF \times X, \tau}$ be the Cartesian product $\GF \times X$ equipped with the product topology.

Let $\circ : \struct {\GF \times X, \tau} \to \struct {X, \PP}$ be the scalar multiplication defined on $X$.

Then $\circ : \struct {\GF \times X, \tau} \to \struct {X, \PP}$ is a continuous mapping.

## Proof

From the definition of the standard topology, 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:

$\circ^{-1} \sqbrk {\map {B_p} {x_0, \epsilon} } \in \tau$ for each $p \in \PP$, $\epsilon > 0$ and $x_0 \in X$.

That is:

$\set {\tuple {\lambda, x} \in \GF \times X : \map p {\lambda x - x_0} < \epsilon}$

For brevity, let:

$U_{x_0} = \set {\tuple {\lambda, x} \in \GF \times X : \map p {\lambda x - x_0} < \epsilon}$

Let $\tuple {\lambda, x} \in U_{x_0}$ and $\tuple {\lambda', x'} \in \GF \times X$.

We then have:

 $\ds \map p {\lambda' x' - x_0}$ $=$ $\ds \map p {\lambda' x' - x_0 + \lambda x - \lambda x + \lambda x' - \lambda x'}$ $\ds$ $=$ $\ds \map p {\lambda x - x_0 - \lambda \paren {x - x'} + \paren {\lambda' - \lambda} x'}$ $\ds$ $\le$ $\ds \map p {\lambda x - x_0} + \cmod \lambda \map p {x - x'} + \cmod {\lambda' - \lambda} \map p {x'}$ Seminorm Axiom $\text N 2$: Positive Homogeneity, Seminorm Axiom $\text N 3$: Triangle Inequality

We want to make:

$\ds \cmod \lambda \map p {x - x'} < \frac {\epsilon - \map p {\lambda x - x_0} } 2$

and:

$\ds \cmod {\lambda' - \lambda} \map p {x'} < \frac {\epsilon - \map p {\lambda x - x_0} } 2$

First suppose that $\lambda = 0$, then the first requirement is satisfied for all $x' \in X$.

Then, for $\lambda' \in \GF$ with:

$\ds \cmod {\lambda'} < \sqrt {\frac {\epsilon - \map p {x_0} } 2}$

and $x' \in X$ with:

$\ds \map p {x'} < \sqrt {\frac {\epsilon - \map p {x_0} } 2}$

we have:

$\ds \cmod {\lambda' - \lambda} \map p {x'} < \frac {\epsilon - \map p {x_0} } 2 = \frac {\epsilon - \map p {\lambda x - x_0} } 2$

So for each $x \in X$, set:

$\ds \map {\epsilon_1} {0, x} = \map {\epsilon_2} {0, x} = \sqrt {\frac {\epsilon - \map p {x_0} } 2}$

Then, for $\tuple {\lambda', x'} \in \map {B_{\map {\epsilon_1} {0, x} } } {0, \GF} \times \map {B_p} {\map {\epsilon_2} {0, x}, x}$, we have:

$\map p {\lambda' x' - x_0} < \epsilon$

where $\map {B_{\map {\epsilon_1} {0, x} } } {0, \GF}$ denotes the open ball of radius $\map {\epsilon_1} {0, x}$ and center $0$.

Now take $\lambda \ne 0$.

Using Seminorm Axiom $\text N 3$: Triangle Inequality, we have:

$\map p {x'} \le \map p {x' - x} + \map p x$

for each $x, x' \in X$.

So it suffices to make:

$\ds \cmod {\lambda - \lambda'} \map p {x' - x} + \cmod {\lambda - \lambda'} \map p x < \frac {\epsilon - \map p {\lambda x - x_0} } 2$

We split further on cases.

Suppose that $\map p x = 0$, so we are aiming to make:

$\ds \cmod {\lambda - \lambda'} \map p {x' - x} < \frac {\epsilon - \map p {\lambda x - x_0} } 2$

and:

$\ds \cmod \lambda \map p {x - x'} < \frac {\epsilon - \map p {\lambda x - x_0} } 2$

Then, if $x' \in X$ has:

$\ds \map p {x - x'} < \frac {\epsilon - \map p {\lambda x - x_0} } {2 \cmod \lambda}$

and $\lambda' \in \GF$ has:

$\ds \cmod {\lambda - \lambda'} < \cmod \lambda$

we have:

$\ds \cmod {\lambda - \lambda'} \map p {x - x'} < \cmod \lambda \times \frac {\epsilon - \map p {\lambda x - x_0} } {2 \cmod \lambda} = \frac {\epsilon - \map p {\lambda x - x_0} } 2$

So define:

$\ds \map {\epsilon_1} {\lambda, x} = \cmod \lambda$

and:

$\ds \map {\epsilon_2} {\lambda, x} = \frac {\epsilon - \map p {\lambda x - x_0} } {2 \cmod \lambda}$

if $\map p x = 0$.

Then for $\tuple {\lambda', x'} \in \map {B_{\map {\epsilon_1} {\lambda, x} } } {\lambda, \GF} \times \map {B_p} {\map {\epsilon_2} {\lambda, x}, x}$, we have:

$\map p {\lambda' x' - x_0} < \epsilon$

Now suppose that $\map p x \ne 0$.

To make:

$\ds \cmod \lambda \map p {x - x'} < \frac {\epsilon - \map p {\lambda x - x_0} } 2$

we can take $x' \in X$ to have:

$\ds \map p {x - x'} < \frac {\epsilon - \map p {\lambda x - x_0} } {2 \cmod \lambda}$

Then it suffices to pick $\lambda'$ such that:

$\ds \cmod {\lambda - \lambda'} \frac {\epsilon - \map p {\lambda x - x_0} } {2 \cmod \lambda} < \frac {\epsilon - \map p {\lambda x - x_0} } 4$

and:

$\ds \cmod {\lambda - \lambda'} \map p x < \frac {\epsilon - \map p {\lambda x - x_0} } 4$

For this, it suffices that:

$\ds \cmod {\lambda - \lambda'} < \frac {\epsilon - \map p {\lambda x - x_0} } {4 \map p x}$

and:

$\ds \cmod {\lambda - \lambda'} < \frac {\cmod \lambda} 2$

So, if:

$\ds \cmod {\lambda - \lambda'} < \min \set {\frac {\epsilon - \map p {\lambda x - x_0} } {4 \map p x}, \, \frac {\cmod \lambda} 2}$

we have:

$\ds \cmod {\lambda - \lambda'} \map p {x' - x} + \cmod {\lambda - \lambda'} \map p x < \frac {\epsilon - \map p {\lambda x - x_0} } 2$

So, for $\map p x \ne 0$ and $\lambda \ne 0$, take:

$\ds \map {\epsilon_1} {\lambda, x} = \min \set {\frac {\epsilon - \map p {\lambda x - x_0} } {4 \map p x}, \, \frac {\cmod \lambda} 2}$

and:

$\ds \map {\epsilon_2} {\lambda, x} = \frac {\epsilon - \map p {\lambda x - x_0} } {2 \cmod \lambda}$

Then for $\tuple {\lambda', x'} \in \map {B_{\map {\epsilon_1} {\lambda, x} } } {\lambda, \GF} \times \map {B_p} {\map {\epsilon_2} {\lambda, x}, x}$, we have:

$\map p {\lambda' x' - x_0} < \epsilon$

To recap, for each $\tuple {\lambda, x} \in U_{x_0}$, we have defined:

$\ds \map {\epsilon_1} {\lambda, x} = \begin{cases}\sqrt {\frac {\epsilon - \map p {x_0} } 2} & \lambda = 0, \\ \cmod \lambda & \lambda \ne 0, \, \map p x = 0 \\ \min \set {\frac {\epsilon - \map p {\lambda x - x_0} } {4 \map p x}, \, \frac {\cmod \lambda} 2} & \lambda \ne 0, \, \map p x \ne 0\end{cases}$

and:

$\ds \map {\epsilon_2} {\lambda, x} = \begin{cases}\sqrt {\frac {\epsilon - \map p {x_0} } 2} & \lambda = 0, \\ \frac {\epsilon - \map p {\lambda x - x_0} } {2 \cmod \lambda} & \lambda \ne 0\end{cases}$

and these are such that if $\tuple {\lambda', x'} \in \map {B_{\map {\epsilon_1} {\lambda, x} } } {\lambda, \GF} \times \map {B_p} {\map {\epsilon_2} {\lambda, x}, x}$, we have:

$\map p {\lambda' x' - x_0} < \epsilon$

so that $\tuple {\lambda', x'} \in U_{x_0}$.

So, we have:

$\map {B_{\map {\epsilon_1} {\lambda, x} } } {\lambda, \GF} \times \map {B_p} {\map {\epsilon_2} {\lambda, x}, x} \subseteq U_{x_0}$

for each $\tuple {\lambda, x} \in U_{x_0}$.

From Union of Subsets is Subset, we have:

$\ds \bigcup_{\tuple {\lambda, x} \in U_{x_0} } \map {B_{\map {\epsilon_1} {\lambda, x} } } {\lambda, \GF} \times \map {B_p} {\map {\epsilon_2} {\lambda, x}, x} \subseteq U_{x_0}$

On the other hand, for each $\tuple {\lambda, x} \in U_{x_0}$, we have:

$\tuple {\lambda, x} \in \map {B_{\map {\epsilon_1} {\lambda, x} } } {\lambda, \GF} \times \map {B_p} {\map {\epsilon_2} {\lambda, x}, x}$

So we have:

$\ds U_{x_0} = \bigcup_{\tuple {\lambda, x} \in U_{x_0} } \map {B_{\map {\epsilon_1} {\lambda, x} } } {\lambda, \GF} \times \map {B_p} {\map {\epsilon_2} {\lambda, x}, x}$

From Natural Basis of Product Topology of Finite Product, we have that:

$\set {\map {B_{\epsilon_1} } {\lambda, x_1} \times \map {B_p} {\epsilon_2, x_2} : p \in \PP, \, \lambda \in \GF, \, x_1, x_2 \in X, \, \epsilon_1, \epsilon_2 > 0}$

forms a basis for $\struct {\GF \times X, \tau}$.

So:

$\map {B_{\map {\epsilon_1} {\lambda, x} } } {\lambda, \GF} \times \map {B_p} {\map {\epsilon_2} {\lambda, x}, x} \in \tau$ for each $\tuple {\lambda, x} \in U_{x_0}$

Since topologies are closed under set union, we have $\circ^{-1} \sqbrk {\map {B_p} {x_0, \epsilon} } = U_{x_0} \in \tau$, which is what we wanted.

$\blacksquare$