Metric Defines Norm iff it Preserves Linear Structure

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Let $V$ be a vector space over a valued field $k$.

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

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

$(1): \quad d(x+z,y+z) = d(x,y)$ (homogeneity or translation invariance)
$(2): \quad d(\lambda x, \lambda y) = | \lambda | d(x,y)$ (the enlargement property)


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

Since $d(u,v) \geq 0$ for all $u,v \in V$, $\| u \| = d(u,0) \geq 0$ for all $u \in V$.

Moreover if $\| u \| = 0$ then $d(u,0) = 0$, so $u = 0$.

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

Then, using the enlargement property of $d$:

\(\displaystyle \Vert \lambda v \Vert\) \(=\) \(\displaystyle d(\lambda v, 0 )\) $\quad$ By the definition of $\vert \cdot \vert$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle d(\lambda v, \lambda \cdot 0)\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \vert\lambda\vert d(v,0)\) $\quad$ By the enlargement property $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \vert\lambda\vert \Vert v \Vert\) $\quad$ By the definition of $\Vert \cdot \Vert$ $\quad$

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

\(\displaystyle \Vert u + v \Vert\) \(=\) \(\displaystyle d(u + v, 0)\) $\quad$ By the definition of $\Vert \cdot \Vert$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle d(u, -v)\) $\quad$ By translation invariance of $d$ $\quad$
\(\displaystyle \) \(\leq\) \(\displaystyle d(u, 0) + d(0,-v)\) $\quad$ By the triangle inequality for metrics. $\quad$
\(\displaystyle \) \(=\) \(\displaystyle d(u, 0) + d(v,0)\) $\quad$ By the enlargement property $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \Vert u \Vert + \Vert v \Vert\) $\quad$ By the definition of $\Vert \cdot \Vert$ $\quad$