# Metric Defines Norm iff it Preserves Linear Structure

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## Theorem

Let $\struct {k, \norm {\,\cdot\,}_k}$ be a valued field.

Let $V$ be a vector space over the valued field $\struct {k, \norm {\,\cdot\,}_k}$.

Let $d: V \times V \to k$ be a metric on $V$.

Then the function $\norm v := \map d {v, 0}$ is a norm on $V$ if and only if for all $x, y, z \in V$, $\lambda \in k$:

- $(1): \quad \map d {x + z, y + z} = \map d {x, y}$ (homogeneity or translation invariance)

- $(2): \quad \map d {\lambda x, \lambda y} = \norm \lambda_k \map d {x, y}$ (the enlargement property)

## Proof

Suppose first that $d$ satisfies the hypotheses $(1)$ and $(2)$.

From Metric Space Axiom $\text M 4$:

- $\forall u, v \in V: \map d {u, v} \ge 0$

Hence:

- $\forall u \in V: \norm u = \map d {u, 0} \ge 0$

Moreover, from Metric Space Axiom $\text M 1$:

- $\norm u = 0 \implies \map d {u, 0} = 0$

and hence:

- $u = 0$

Now let $\lambda \in K$, $u \in V$.

Then, using the enlargement property of $d$:

\(\ds \norm {\lambda v}\) | \(=\) | \(\ds \map d {\lambda v, 0}\) | Definition of $\norm {\,\cdot\,}$ | |||||||||||

\(\ds \) | \(=\) | \(\ds \map d {\lambda v, \lambda \cdot 0}\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \norm \lambda_k \map d {v, 0}\) | Enlargement Property | |||||||||||

\(\ds \) | \(=\) | \(\ds \norm \lambda_k \norm v\) | Definition of $\norm{\,\cdot\,}$ |

Finally if $u,v \in V$, then we have

\(\ds \norm {u + v}\) | \(=\) | \(\ds \map d {u + v, 0}\) | Definition of $\norm {\,\cdot\,}$ | |||||||||||

\(\ds \) | \(=\) | \(\ds \map d {u, -v}\) | Translation Invariance of $d$ | |||||||||||

\(\ds \) | \(\le\) | \(\ds \map d {u, 0} + d(0,-v)\) | Triangle Inequality | |||||||||||

\(\ds \) | \(=\) | \(\ds \map d {u, 0} + \map d {v, 0}\) | Enlargement Property | |||||||||||

\(\ds \) | \(=\) | \(\ds \norm u + \norm v\) | Definition of $\norm {\,\cdot\,}$ |

$\blacksquare$