Positive Scalar Multiple of Norm on Vector Space is Norm

Theorem

Let $\GF \in \set {\R, \C}$.

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

Let $\alpha > 0$ be a real number.

Define $\norm {\, \cdot \,}' : X \to \R_{\ge 0}$ by:

$\norm x' = \alpha \norm x$

for each $x \in X$.

Then $\norm {\, \cdot \,}'$ is a norm on $X$.

Proof

Norm Axiom $\text N 1$: Positive Definiteness

Suppose that $x \in X$ is such that:

$\norm x' = 0$

Then we have:

$\alpha \norm x = 0$

Since $\alpha > 0$, it follows that:

$\norm x = 0$

From Norm Axiom $\text N 1$: Positive Definiteness, we have $x = {\mathbf 0}_X$.

$\Box$

Norm Axiom $\text N 2$: Positive Homogeneity

Let $\lambda \in \GF$ and $x \in X$.

Then we have:

\(\ds \norm {\lambda x}'\)

\(=\)

\(\ds \alpha \norm {\lambda x}\)

\(\ds \)

\(=\)

\(\ds \alpha \cmod \lambda \norm x\)

Norm Axiom $\text N 2$: Positive Homogeneity for $\norm {\, \cdot \,}'$

\(\ds \)

\(=\)

\(\ds \cmod \lambda \norm x'\)

$\Box$

Norm Axiom $\text N 3$: Triangle Inequality

Let $x, y \in X$.

Then we have:

\(\ds \norm {x + y}'\)

\(=\)

\(\ds \alpha \norm {x + y}\)

\(\ds \)

\(\le\)

\(\ds \alpha \paren {\norm x + \norm y}\)

Norm Axiom $\text N 3$: Triangle Inequality, $\alpha > 0$

\(\ds \)

\(=\)

\(\ds \alpha \norm x + \alpha \norm y\)

\(\ds \)

\(=\)

\(\ds \norm x' + \norm y'\)

$\blacksquare$