Pseudometric Induced by Seminorm is Pseudometric

Definition

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

Let $X$ be a vector space over $\GF$.

Let $p$ be a seminorm on $X$.

Let $d_p$ be the pseudometric induced by $p$.

Then $d_p$ is a pseudometric.

Proof

Proof of Metric Space Axiom $(\text M 1)$

For each $x \in X$ we have:

$\map {d_p} {x, x} = \map p {x - x} = \map p { {\mathbf 0}_X}$

From Seminorm Maps Zero Vector to Zero, we therefore have:

$\map {d_p} {x, x} = 0$

$\Box$

Proof of Metric Space Axiom $(\text M 2)$: Triangle Inequality

For $x, y, z \in X$ we have:

\(\ds \map {d_p} {x, z}\)

\(=\)

\(\ds \map p {x - z}\)

Definition of Pseudometric Induced by Seminorm

\(\ds \)

\(=\)

\(\ds \map p {\paren {x - y} + \paren {y - z} }\)

\(\ds \)

\(\le\)

\(\ds \map p {x - y} + \map p {y - z}\)

Seminorm Axiom $\text N 3$: Triangle Inequality

\(\ds \)

\(=\)

\(\ds \map {d_p} {x, y} + \map {d_p} {y, z}\)

Definition of Pseudometric Induced by Seminorm

$\Box$

Proof of Metric Space Axiom $(\text M 3)$

For $x, y \in X$ we have:

\(\ds \map {d_p} {y, x}\)

\(=\)

\(\ds \map p {y - x}\)

Definition of Pseudometric Induced by Seminorm

\(\ds \)

\(=\)

\(\ds \cmod {-1} \map p {x - y}\)

Seminorm Axiom $\text N 2$: Positive Homogeneity

\(\ds \)

\(=\)

\(\ds \map p {x - y}\)

\(\ds \)

\(=\)

\(\ds \map {d_p} {x, y}\)

Definition of Pseudometric Induced by Seminorm

$\blacksquare$