Characterization of Complete Normed Quotient Vector Spaces

Theorem
Let $\struct {X, \norm {\, \cdot \,}_X}$ be a normed vector space.

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

Let $\struct {X/N, \norm {\, \cdot \,}_{X/N} }$ be the normed quotient vector space associated with the quotient vector space $X/N$.

Let $\norm {\, \cdot \,}_N$ be the norm on $N$ given by restricting the norm on $X$.

Then $\struct {X, \norm {\, \cdot \,}_X}$ is a Banach space $\struct {N, \norm {\, \cdot \,}_N}$ and $\struct {X/N, \norm {\, \cdot \,}_{X/N} }$ are Banach spaces.

Proof
Let $\pi : X \to X/N$ be the quotient mapping associated with $X/N$.

Necessary Condition
Suppose that $\struct {X, \norm {\, \cdot \,}_X}$ is a Banach space.

From Closed Subspace of Banach Space forms Banach Space, $\struct {N, \norm {\, \cdot \,}_N}$ is a Banach space.

We now show that $\struct {X/N, \norm {\, \cdot \,}_{X/N} }$ is a Banach space.

From Absolutely Convergent Series in Normed Vector Space is Convergent iff Space is Banach, it suffices to show that every absolute convergent series in $X/N$ converges in $X/N$.

That is, we show that if $\sequence {x_n}_{n \mathop \in \N}$ is a sequence in $X$ with:


 * $\ds \sum_{n \mathop = 1}^\infty \norm {\map \pi {x_n} }_{X/N} < \infty$

then:


 * $\ds \sum_{n \mathop = 1}^N \map \pi {x_n} \to \map \pi s$ as $N \to \infty$ for some $s \in X$.

From the definition of the quotient norm, we have:


 * $\ds \norm {\map \pi {x_n} }_{X/N} = \inf_{z \in N} \norm {x_n - z}_X$

Then there exists $z_n \in N$ such that:


 * $\ds \norm {\map \pi {x_n} }_{X/N} \le \norm {x_n - z_n}_X \le \norm {\map \pi {x_n} }_{X/N} + 2^{-n}$

From Sum of Infinite Geometric Sequence, we have:


 * $\ds \sum_{n \mathop = 1}^\infty 2^{-n} = 1$

Hence we have:


 * $\ds \sum_{n \mathop = 1}^\infty \norm {x_n - z_n}_X \le 1 + \sum_{n \mathop = 1}^\infty \norm {\map \pi {x_n} }_{X/N} < \infty$

Since $X$ is a Banach space, there exists $s \in X$ such that:


 * $\ds \lim_{N \mathop \to \infty} \sum_{n \mathop = 1}^N \paren {x_n - z_n} = \sum_{n \mathop = 1}^\infty \paren {x_n - z_n} = s$

by Absolutely Convergent Series in Normed Vector Space is Convergent iff Space is Banach.

Now, for each $N \in \N$:

From Quotient Mapping is Bounded in Normed Quotient Vector Space, $\pi$ is a bounded linear transformation.

So, by Continuity of Linear Transformations, $\pi$ is continuous.

Hence by Sequential Continuity is Equivalent to Continuity in Metric Space:

Hence:


 * $\ds \sum_{n \mathop = 1}^\infty \map \pi {x_n} = \map \pi s$

So every absolute convergent series in $X/N$ converges in $X/N$.

Sufficient Condition
Suppose that $\struct {N, \norm {\, \cdot \,}_N}$ and $\struct {X/N, \norm {\, \cdot \,}_{X/N} }$ are Banach spaces.

We show that every Cauchy sequence converges.

Let $\sequence {x_n}_{n \mathop \in \N}$ be a Cauchy sequence in $X$.

Since $\pi$ is a bounded linear transformation, $\sequence {\map \pi {x_n} }_{n \mathop \in \N}$ is a Cauchy sequence in $X/N$, by Bounded Linear Transformation preserves Cauchy Sequences.

Since $X/N$ is a Banach space, $\sequence {\map \pi {x_n} }_{n \mathop \in \N}$ converges.

That is, there exists $x \in X$ such that $\map \pi {x_n} \to \map \pi x$ in $X/N$ as $n \to \infty$.

We show that there exists $z \in N$ such that:


 * $x_n \to x + z$ in $X$ as $n \to \infty$.

Since $\sequence {\map \pi {x_n} }_{n \mathop \in \N}$ converges, for each $n \in \N$ we can choose $m_n \in \N$ such that:


 * $\norm {\map \pi {x_{m_n} } - \map q x}_{X/N} < 2^{-n}$

so:


 * $\ds \inf_{z \in N} \norm {x_{m_n} - x - z}_X < 2^{-n}$

for each $n \in \N$.

Then for each $n \in \N$, we can pick $z_n \in N$ such that:


 * $\ds \norm {x_{m_n} - x - z_n}_X < 2^{-n}$

We show that $\sequence {z_n}_{n \mathop \in \N}$ is a Cauchy sequence.

It will then converge, since $N$ is a Banach space.

Let $\epsilon > 0$.

We have:

Since $2^{-n} \to 0$ as $n \to \infty$, we can pick $N_1 \in \N$ such that:


 * $\ds 2^{-n} < \frac \epsilon 3$

for $n \ge N_1$.

Since $\sequence {x_n}_{n \mathop \in \N}$ is a Cauchy sequence, we can pick $N_2 \in \N$ such that:


 * $\ds \norm {x_{m_n} - x_{m_k} }_X < \frac \epsilon 3$

for $n, k \ge N_2$.

Now let $N = \max \set {N_1, N_2}$.

Then for $n, k \ge N$ we have:


 * $\norm {z_n - z_k}_X < \epsilon$

So $\sequence {z_n}_{n \in \mathop \in \N}$ is a Cauchy sequence in $N$.

Since $N$ is a Banach space, there exists $z \in N$ such that:


 * $z_n \to z$

We now show that:


 * $x_{m_n} \to x + z$

as $n \to \infty$.

We have:

Taking $n \to \infty$, we have:


 * $2^{-n} + \norm {z_n - z} \to 0$

So, indeed we have:


 * $x_{m_n} \to x + z$ as $n \to \infty$

From Convergent Subsequence of Cauchy Sequence, we can conclude that:


 * $x_n \to x + z$ as $n \to \infty$.

So every Cauchy sequence in $X$ converges.

So $X$ is a Banach space.