Modulus of Linear Functional on Vector Space is Seminorm

Theorem

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

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

Let $f : X \to \GF$ be a linear functional.

Define $p_f : X \to \R_{\ge 0}$ by:

$\map {p_f} x = \cmod {\map f x}$

for each $x \in X$.

Then $p_f$ is a seminorm.

Proof

Proof of Seminorm Axiom $\text N 2$: Positive Homogeneity

For each $\lambda \in \GF$ and $x \in X$, we have:

\(\ds \map {p_f} {\lambda x}\)

\(=\)

\(\ds \cmod {\map f {\lambda x} }\)

\(\ds \)

\(=\)

\(\ds \cmod {\lambda \map f x}\)

since $f$ is linear

\(\ds \)

\(=\)

\(\ds \cmod \lambda \cmod {\map f x}\)

\(\ds \)

\(=\)

\(\ds \cmod \lambda \map {p_f} x\)

$\Box$

Proof of Seminorm Axiom $\text N 3$: Triangle Inequality

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

\(\ds \map {p_f} {x + y}\)

\(=\)

\(\ds \cmod {\map f {x + y} }\)

\(\ds \)

\(=\)

\(\ds \cmod {\map f x + \map f y}\)

since $f$ is linear

\(\ds \)

\(\le\)

\(\ds \cmod {\map f x} + \cmod {\map f y}\)

Triangle Inequality for Real Numbers if $\GF = \R$ and Triangle Inequality for Complex Numbers if $\GF = \C$

\(\ds \)

\(=\)

\(\ds \map {p_f} x + \map {p_f} y\)

$\blacksquare$