Set of Points for which Seminorm is Zero is Vector Subspace

Theorem

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

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

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

Let:

$U = \set {x \in X : \map p x = 0}$

Then $U$ is a vector subspace of $X$.

Proof

From Seminorm Maps Zero Vector to Zero, $\map p {\mathbf 0_X} = 0$.

So $\mathbf 0_X \in U$ and in particular $U \ne \O$.

So we look to apply One-Step Vector Subspace Test.

Let $x, y \in U$ and $\lambda, \mu \in \GF$.

Then we have:

\(\ds \map p {\lambda x + \mu y}\)

\(\le\)

\(\ds \map p {\lambda x} + \map p {\mu y}\)

Seminorm Axiom $\text N 3$: Triangle Inequality

\(\ds \)

\(=\)

\(\ds \cmod \lambda \map p x + \cmod \mu \map p y\)

Seminorm Axiom $\text N 2$: Positive Homogeneity

\(\ds \)

\(=\)

\(\ds 0\)

since $x, y \in U$

Since $\map p {\lambda x + \mu y} \ge 0$, so:

$\map p {\lambda x + \mu y} = 0$

so $\lambda x + \mu y \in U$.

By One-Step Vector Subspace Test, we have that $U$ is a vector subspace of $X$.

$\blacksquare$

Sources

• 1991: Walter Rudin: Functional Analysis (2nd ed.) ... (previous) ... (next): $1.34$: Theorem