Seminorm on Vector Space induces Norm on Quotient
Theorem
Let $\GF \in \set {\R, \C}$.
Let $X$ be a vector space over $\GF$.
Let $p$ be a seminorm on $X$.
Let:
- $N = \set {x \in X : \map p x = 0}$
From Set of Points for which Seminorm is Zero is Vector Subspace, $N$ is a vector subspace.
Let $X/N$ be the quotient vector space of $X$ modulo $N$.
Let $\pi : X \to X/N$ be the quotient mapping.
Then the mapping $\norm {\, \cdot \,} : X/N \to \hointr 0 \infty$ defined by:
- $\norm {\map \pi x} = \map p x$
is well-defined and a norm on $X/N$.
Proof
We first want to show that if $\map \pi x = \map \pi y$ for $x, y \in X$, then $\map p x = \map p y$.
From Quotient Mapping is Linear Transformation and Kernel of Quotient Mapping, it is enough to show that:
- if $y - x \in N$ then $\map p x = \map p y$.
It is therefore enough to show that if $x \in X$ and $z \in N$, then:
- $\map p {x + z} = \map p x$
Let $x \in X$ and $z \in N$.
Then $\map p z = 0$.
From Reverse Triangle Inequality: Seminormed Vector Space, we have:
- $\map p x = \size {\map p x - \map p z} \le \map p {x + z}$
From Norm Axiom $\text N 3$: Triangle Inequality, we also have:
- $\map p {x + z} \le \map p x + \map p z = \map p x$
Hence we have $\map p {x + z} = \map p x$.
Hence $\norm {\, \cdot \,}$ is well-defined.
We now prove the norm axioms.
Proof of Norm Axiom $\text N 1$: Positive Definiteness
Let $x \in X$.
We have $\norm {\map \pi x} = 0$ if and only if $\map p x = 0$.
Hence by the definition of $N$, we have $\norm {\map \pi x} = 0$ if and only if $x \in N$.
From Kernel of Quotient Mapping, we have $\map \pi x = {\mathbf 0}_{X/N}$.
So for $x \in X$ we have $\norm {\map \pi x} = 0$ if and only if $\map \pi x = {\mathbf 0}_{X/N}$.
Hence we have proven Norm Axiom $\text N 1$: Positive Definiteness.
$\Box$
Proof of Norm Axiom $\text N 2$: Positive Homogeneity
Let $x \in X$ and $\lambda \in \GF$.
Then, we have:
\(\ds \norm {\lambda \map \pi x}\) | \(=\) | \(\ds \norm {\map \pi {\lambda x} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \map p {\lambda x}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \cmod \lambda \map p x\) | Norm Axiom $\text N 2$: Positive Homogeneity | |||||||||||
\(\ds \) | \(=\) | \(\ds \cmod \lambda \norm {\map \pi x}\) |
So we have proven Norm Axiom $\text N 2$: Positive Homogeneity.
$\Box$
Proof of Norm Axiom $\text N 3$: Triangle Inequality
Let $x, y \in X$.
Then, we have:
\(\ds \norm {\map \pi x + \map \pi y}\) | \(=\) | \(\ds \norm {\map \pi {x + y} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \map p {x + y}\) | ||||||||||||
\(\ds \) | \(\le\) | \(\ds \map p x + \map p y\) | Norm Axiom $\text N 3$: Triangle Inequality | |||||||||||
\(\ds \) | \(=\) | \(\ds \norm {\map \pi x} + \norm {\map \pi y}\) |
Hence we have proven Norm Axiom $\text N 3$: Triangle Inequality.
$\blacksquare$