Vector Addition on Normed Vector Space is Continuous

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Theorem

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

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

Let $+_{\scriptscriptstyle X} : X \times X \to X$ be the vector addition defined on $X$.


Then $+_{\scriptscriptstyle X} : X \times X \to X$ is a continuous mapping.


Proof

Let $x_0, y_0 \in X$.

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

For $a, b \in X$, let $a-_{\scriptscriptstyle X} b$ denote the sum $a +_{\scriptscriptstyle X} \paren { -b }$, where $-b$ is the inverse vector of $b$ in $X$.

To show that $+_{\scriptscriptstyle X}$ is continuous, let $x, y \in X$ such that $\norm { x_0 -_{\scriptscriptstyle X} x }_X < \dfrac \epsilon 2$, and $\norm { y_0 -_{\scriptscriptstyle X} y }_X < \dfrac \epsilon 2$.

By definition of direct product norm, it follows that:

$\norm { \tuple {x_0,y_0} -_{\scriptscriptstyle {X \times X} } \tuple {x,y} }_P = \map \max {\norm {x_0 -_{\scriptscriptstyle X} x}_X, \norm {y_0 -_{\scriptscriptstyle X} y}_X } < \dfrac \epsilon 2$


To show that $+_{\scriptscriptstyle X}$ is continuous at $\tuple {x_0, y_0}$, we calculate:

\(\ds \norm { \paren {x_0 +_{\scriptscriptstyle X} y_0} -_{\scriptscriptstyle X} \paren { x +_{\scriptscriptstyle X} y } }_X\) \(=\) \(\ds \norm { \paren {x_0 +_{\scriptscriptstyle X} y_0} -_{\scriptscriptstyle X} x -_{\scriptscriptstyle X} y }_X\)
\(\ds \) \(=\) \(\ds \norm { x_0 -_{\scriptscriptstyle X} x +_{\scriptscriptstyle X} y_0 -_{\scriptscriptstyle X} y }_X\) Vector Space Axiom $(\text V 1)$: Commutativity and Vector Space Axiom $(\text V 2)$: Associativity
\(\ds \) \(\le\) \(\ds \norm { x_0 -_{\scriptscriptstyle X} x }_X + \norm { y_0 -_{\scriptscriptstyle X} y }_X\) Norm Axiom $\text N 3$: Triangle Inequality
\(\ds \) \(<\) \(\ds \dfrac \epsilon 2 + \dfrac \epsilon 2\) by assumption
\(\ds \) \(=\) \(\ds \epsilon\)


It follows that $+_{\scriptscriptstyle X}$ is continuous from $X \times X$ to $X$.

$\blacksquare$


Sources

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