Set of Points for which Seminorm is Zero is Vector Subspace
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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