# 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 $\GF = \C$ $\ds$ $=$ $\ds \map {p_f} x + \map {p_f} y$

$\blacksquare$