Scalar Multiplication on Normed Vector Space is Continuous

Theorem
Let $\struct { K, +, \circ_K }$ be a normed division ring with norm $\norm {\, \cdot \,}_K$.

Let $\struct {X, \norm {\, \cdot \,}_X }$ be a normed vector space over $K$.

Let $\struct {K \times X, \norm {\, \cdot \,}_P }$ be the direct product of $K$ and $X$ with the direct product norm $\norm {\, \cdot \,}_P$.

Let $\circ : K \times X \to X$ be the scalar multiplication defined on $X$.

Then $\circ: K \times X \to X$ is a continuous mapping.

Proof
Let $\tuple { \lambda_0, x_0 } \in K \times X$.

Let $\epsilon' \in \R_{>0}$.

Set $\epsilon = \map \min { \epsilon', 1 }$

To show that $\circ$ is continuous, let $\tuple { \lambda, x } \in K \times X$ such that $\norm { \lambda_0 - \lambda }_K < \dfrac \epsilon { 1 + \norm { \lambda_0 }_K + \norm { x_0 }_X }$, and $\norm { x_0 - x }_X < \dfrac \epsilon { 1 + \norm { \lambda_0 }_K + \norm { x_0 }_X }$.

By definition of direct product norm, it follows that:


 * $\norm { \tuple {\lambda_0, x_0} - \tuple {\lambda, x} }_P = \map \max {\norm {\lambda_0 - \lambda}_K, \norm {x_0 - x}_X } < \dfrac \epsilon { 1 + \norm { \lambda_0 }_K + \norm { x_0 }_X }$.

To show that $\circ$ is continuous at $\tuple {\lambda_0, x_0}$, we calculate:

It follows that $\circ$ is continuous from $K \times X$ to $X$.