Modulus of Linear Functional on Vector Space is Seminorm

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


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