# Quotient Norm is Norm

## Theorem

Let $\Bbb F \in \set {\R, \C}$.

Let $\struct {X, \norm {\, \cdot \,} }$ be a normed vector space over $\Bbb F$.

Let $N$ be a closed linear subspace of $X$.

Let $X/N$ be the quotient vector space of $X$ modulo $N$.

Let $\norm {\, \cdot \,}_{X/N}$ be the quotient norm on $X/N$.

Then $\norm {\, \cdot \,}_{X/N}$ is indeed a norm.

## Proof

### Norm is Well-Defined and Finite

Let $\pi$ be the quotient map associated with $X/N$.

We show that if $x, x' \in X$ have $\map \pi x = \map \pi {x'}$, then:

$\ds \inf_{z \mathop \in N} \norm {x - z} = \inf_{z \mathop \in N} \norm {x' - z}$
$\map \pi {x' - x} = 0$
$x' - x \in N$

Since $N$ is a linear subspace of $X$:

$\paren {x' - x} + N = N$

So, we may manipulate:

 $\ds \inf_{z \mathop \in N} \norm {x - z}$ $=$ $\ds \inf_{z \mathop \in \paren {x' - x} + N} \norm {x - \paren {z - \paren {x' - x} } }$ $\ds$ $=$ $\ds \inf_{z \mathop \in \paren {x' - x} + N} \norm {x' - z}$ $\ds$ $=$ $\ds \inf_{z \mathop \in N} \norm {x' - z}$

and so the quotient norm is well-defined.

Since:

$\norm {x - z} \ge 0$

for all $z \in N$, we also have:

$\ds \inf_{z \mathop \in N} \norm {x - z} \ge 0$

$\Box$

### Proof of Norm Axiom $\text N 1$: Positive Definiteness

First, we can calculate:

 $\ds \norm {0_{X/N} }_{X/N}$ $=$ $\ds \norm {\map \pi 0}_{X/N}$ $\ds$ $=$ $\ds \inf_{z \mathop \in N} \norm z$ $\ds$ $=$ $\ds 0$ since $0 \in N$ and $\norm z \ge 0$ for each $z \in N$

Conversely, suppose that:

$\ds \inf_{z \mathop \in N} \norm {x - z} = 0$

Then by the definition of infimum, for each $n \in \N$ there exists $z_n \in N$ such that:

$\ds \norm {x - z_n} < \frac 1 n$

By the Squeeze Theorem, we then have:

$\ds \lim_{n \mathop \to \infty} \norm {x - z_n} = 0$
$z_n \to x$

Since $N$ is closed:

$x \in N$

So from Kernel of Quotient Mapping:

$\map \pi x = 0_{X/N}$

and hence:

$\norm {\map \pi x}_{X/N} = 0$

So we have:

$\norm {\map \pi x}_{X/N} = 0$ if and only if $x = 0_{X/N}$

$\Box$

### Proof of Norm Axiom $\text N 2$: Positive Homogeneity

Let $t \in \Bbb F$.

Clearly if $t = 0$, we have:

$\norm {t \map \pi x}_{X/N} = 0$

So take $t \ne 0$.

We have:

 $\ds \norm {t \map \pi x}_{X/N}$ $=$ $\ds \norm {\map \pi {t x} }_{X/N}$ Quotient Mapping is Linear Transformation $\ds$ $=$ $\ds \inf_{z \mathop \in N} \norm {t x - z}$

Since $N$ is a linear subspace of $X$:

$\ds \paren {\frac 1 t} N = N$

So:

 $\ds \inf_{z \mathop \in N} \norm {t x - z}$ $=$ $\ds \inf_{z \mathop \in \paren {1/t} N} \norm {t x - t z}$ $\ds$ $=$ $\ds \cmod t \inf_{z \mathop \in \paren {1/t} N} \norm {x - z}$ Norm Axiom $\text N 2$: Positive Homogeneity for $\norm {\, \cdot \,}$ $\ds$ $=$ $\ds \cmod t \inf_{z \mathop \in N} \norm {x - z}$ $\ds$ $=$ $\ds \cmod t \norm {\map \pi x}_{X/N}$

$\Box$

### Proof of Norm Axiom $\text N 3$: Triangle Inequality

We first argue that:

$\ds \inf_{z \mathop \in N} \norm {x - z} = \inf_{z_1, z_2 \mathop \in N} \norm {x - \paren {z_1 + z_2} }$

First, for each $z_1, z_2 \in N$ we have:

$z_1 + z_2 \in N$

so:

$\ds \inf_{z \mathop \in N} \norm {x - z} \le \inf_{z_1, z_2 \mathop \in N} \norm {x - \paren {z_1 + z_2} }$

Conversely, we have $0 \in N$ and so $z, 0 \in N$ have $z + 0 = z$, and so we also get:

$\ds \inf_{z_1, z_2 \mathop \in N} \norm {x - \paren {z_1 + z_2} } \le \inf_{z \mathop \in N} \norm {x - z}$

and hence:

$\ds \inf_{z \mathop \in N} \norm {x - z} = \inf_{z_1, z_2 \mathop \in N} \norm {x - \paren {z_1 + z_2} }$

We now have for $x, y \in X$:

 $\ds \norm {\map \pi x + \map \pi y}_{X/N}$ $=$ $\ds \norm {\map \pi {x + y} }_{X/N}$ Quotient Mapping is Linear Transformation $\ds$ $=$ $\ds \inf_{z \mathop \in N} \norm {\paren {x + y} - z}$ Definition of Quotient Norm $\ds$ $=$ $\ds \inf_{z_1, z_2 \mathop \in N} \norm {\paren {x + y} - \paren {z_1 + z_2} }$ $\ds$ $\le$ $\ds \inf_{z_1, z_2 \mathop \in N} \paren {\norm {x - z_1} + \norm {y - z_2} }$ Norm Axiom $\text N 3$: Triangle Inequality $\ds$ $=$ $\ds \inf_{z_1, z_2 \mathop \in N} \norm {x - z_1} + \inf_{z_1, z_2 \mathop \in N} \norm {y - z_2}$ $\ds$ $=$ $\ds \inf_{z_1 \mathop \in N} \norm {x - z_1} + \inf_{z_2 \mathop \in N} \norm {y - z_2}$ $\ds$ $=$ $\ds \norm {\map \pi x}_{X/N} + \norm {\map \pi y}_{X/N}$

$\blacksquare$