Scalar Multiplication on Normed Vector Space is Continuous

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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:

\(\ds \norm { \lambda_0 \circ x_0 - \lambda \circ x }_X\) \(=\) \(\ds \norm { \lambda_0 \circ x - \lambda_0 \circ x_0 - \lambda \circ x + \lambda \circ x_0 + \lambda_0 \circ x_0 - \lambda \circ x_0 + \lambda_0 \circ x_0 - \lambda_0 \circ x}_X\) adding and subtracting the same quantities
\(\ds \) \(=\) \(\ds \norm { \paren{ \lambda_0 - \lambda } \circ \paren { x - x_0 } + \paren{ \lambda_0 - \lambda } \circ x_0 + \lambda_0 \circ \paren { x_0 - x } }_X\) Vector Space Axiom $\text V 5$: Distributivity over Scalar Addition and Vector Space Axiom $\text V 6$: Distributivity over Vector Addition
\(\ds \) \(\le\) \(\ds \norm { \paren{ \lambda_0 - \lambda } \circ \paren { x - x_0 } }_X + \norm { \paren{ \lambda_0 - \lambda } \circ x_0 }_X + \norm { \lambda_0 \circ \paren { x_0 - x } }_X\) Norm Axiom $\text N 3$: Triangle Inequality
\(\ds \) \(=\) \(\ds \norm { \lambda_0 - \lambda }_K \norm { x - x_0 }_X + \norm { \lambda_0 - \lambda }_K \norm { x_0 }_X + \norm { \lambda_0 }_K \norm { x_0 - x }_X\) Norm Axiom $\text N 2$: Positive Homogeneity
\(\ds \) \(<\) \(\ds \paren { \dfrac { \epsilon }{ 1 + \norm { \lambda_0 }_K + \norm { x_0 }_X } }^2+ \dfrac{ \epsilon \norm { x_0 }_X }{ 1 + \norm { \lambda_0 }_K + \norm { x_0 }_X } + \dfrac { \epsilon \norm { \lambda_0 }_K }{ 1 + \norm { \lambda_0 }_K + \norm { x_0 }_X }\) by choice of $\lambda$ and $x$
\(\ds \) \(\leq\) \(\ds \dfrac { \epsilon }{ 1 + \norm { \lambda_0 }_K + \norm { x_0 }_X } + \dfrac{ \epsilon \norm { x_0 }_X }{ 1 + \norm { \lambda_0 }_K + \norm { x_0 }_X } + \dfrac { \epsilon \norm { \lambda_0 }_K }{ 1 + \norm { \lambda_0 }_K + \norm { x_0 }_X }\) as $\epsilon \le 1$. and $1 + \norm { \lambda_0 }_K + \norm { x_0 }_X \ge 1$
\(\ds \) \(=\) \(\ds \dfrac { \epsilon \paren{1 + \norm { \lambda_0 }_K + \norm { x_0 }_X} }{ 1 + \norm { \lambda_0 }_K + \norm { x_0 }_X }\)
\(\ds \) \(=\) \(\ds \epsilon\)
\(\ds \) \(\le\) \(\ds \epsilon'\)


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

$\blacksquare$


Sources

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