Quotient Norm is Norm

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

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

Then $\norm {\, \cdot \,}$ is indeed a norm.

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:


 * $\ds \inf_{z \in N} \norm {x - z} = \inf_{z \in N} \norm {x' - z}$

From Quotient Mapping is Linear Transformation, we have $\map \pi {x' - x} = 0$.

From Kernel of Quotient Mapping, we have $x' - x \in N$.

Since $N$ is a linear subspace of $X$, we have:


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

So, we may manipulate:

and so the quotient norm is well-defined.

Since:


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

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


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

Proof of
First, we can calculate:

Conversely, suppose that:


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

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:


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

From Sequence in Normed Vector Space Convergent to Limit iff Norm of Sequence minus Limit is Null Sequence, we then have:


 * $z_n \to x$

Since $N$ is closed, we have $x \in N$.

So from Kernel of Quotient Mapping we have $\map \pi x = 0_{X/N}$ and hence $\norm {\map \pi x}_{X/N} = 0$.

So we have:


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

Proof of
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:

Since $N$ is a linear subspace of $X$, we have:


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

So we have:

Proof of
We first argue that:


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

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


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

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


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

and hence:


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

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