Seminorm on Vector Space induces Norm on Quotient

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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:

$N = \set {x \in X : \map p x = 0}$

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.

$\blacksquare$

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