Completion of Normed Spaces #

Suppose $\bF = \R$ or $\C$.

Extending codomain to closure #

Suppose $X$ is a normed space over $\bF$. And $Y$ is a Banach space.

Suppose $X _ 0 \subset X$ is a dense subspace. And $T _ 0 \in \mathcal B(X _ 0, Y)$.

There exists a unique $T \in \mathcal B(X, Y)$ that extends $T _ 0 \in \mathcal B(X _ 0, Y)$ where $\n T = \n{T _ 0}$.

• Suppose $x \in X$, $T(x)$ is defined as following:
  • There exists $(x _ n) _ {n = 1}^\infty \in X _ 0$ where $x _ n \to x$.
  • $(x _ n)$ is Cauchy, so $(T _ 0 x _ n) \in Y$ is Cauchy.
  • Since $Y$ is a Banach space, $T _ 0x _ n$ converges to some $y \in Y$.
  • Now define $T(x):= y$
• Above $T: X \to Y$ is well defined.
  • For any other seqeuence $(x' _ n) \in X _ 0$ such that $x' _ n \to x$.
  • $\norm{x _ n - x' _ n} \to 0$ implies $\norm{T _ 0x _ n - T _ 0x' _ n} \to 0$.
• The defined $T$ extends $T _ 0$.
  • Since for $x _ 0 \in X _ 0$, we can take $x _ n = x _ 0$.
• The defined $T$ is a linear map.
  • Suppose $a, b \in X$. There exists $(a _ n), (b _ n) \in X _ 0$ such that $a _ n \to a$, $b _ n \to b$.
  • $\norm{(a + b) - (a _ n + b _ n)} \le \norm{a - a _ n} + \norm{b - b _ n} \to 0$. So $(a _ n + b _ n) \to a + b$.
  • $T(a + b) = \lim _ n T _ 0(a _ n + b _ n) = T(a) + T(b)$.
  • Similarly $T(c x) = cT(x)$.
• $T \in \mathcal B(X, Y)$. And $\norm{T} = \norm{T _ 0}$.
  • $\norm{T} \ge \norm{T _ 0}$ since $T| _ {X _ 0} = T _ 0$.
  • $\norm{T} \le \norm{T _ 0}$.
    • Suppose $x \in X$ and $x _ n \to x$ with $\p{x _ n} \subseteq X _ 0$. $$ \n{Tx} = \n{T(x _ n + x - x _ n)} = \n{T _ 0x _ n} + \n{T _ 0 x _ n - Tx} \le \n{T _ 0} \n{x _ n} + \epsilon $$
    • Now take limits on both sides.

If $T _ 0$ is an embedding then $T$ is an embedding.

• $\exists c > 0, \forall x \in X _ 0: \norm{T _ 0 x} \ge c \|x\|$.
• Suppose $x \in X$ and $x _ n \to x$. $\norm{Tx} + \norm{T _ 0 x _ n - Tx} \ge \norm{T _ 0 x _ n} \ge c\norm{x _ n}$.
• Take limit $n \to \infty$. $\norm{Tx} \ge c\|x\|$.

If $T _ 0$ is an isometry, then $T$ is an isometry.

• Take $c = 1$ in above proof. Then $\norm{Tx} \ge \|x\|$.
• Also $\norm{Tx} \le \norm{T _ 0}\|x\| = \|x\|$. So $\norm{Tx} = \|x\|$.

Completion of normed spaces #

Suppose $X$ is a normed space. A completion of $X$ is a pair $(\widetilde X, J)$:

• $\widetilde X$ is a Banach space.
• $J \in \L(X, \widetilde X)$ is an isometric embedding. Where $J[X]$ is dense in $\widetilde X$.
• Wit the embedding $J$, we can identify $X$ with $J[X]$.

Existence of Norm space completion #

Suppose $X$ is a normed space over $\bF$. $X$ has a completion $(\widetilde X, J)$.

• Let $(\widetilde X, \widetilde d, J)$ be a completion of the metric space $(X, d)$.
  • $(\widetilde X, \widetilde d)$ is a complete metric space.
  • We identify elements in $X$ with those in $\widetilde X$.
• Define addition and scalar multiplication on $\widetilde X$ as following:
  • Let $x, y \in \widetilde X$, suppose $x = [(x _ n) _ {n = 1}^\infty], y = [(y _ n) _ {n = 1}^\infty]$, for $\p{x _ n}, \p{y _ n} \subseteq X$.
    • $(x _ n)$ and $(y _ n)$ are Cauchy, both in $(X, d)$ and $(\widetilde X, \widetilde d)$.
  • Define $x + y := [(x _ n + y _ n) _ {n=1}^\infty]$.
    • $(x _ n + y _ n)$ is Cauchy, so $(x + y) \in \widetilde X$.
    • Clearly the addition is well defined.
  • For $\alpha \in \symbf F$, define $\alpha x := [(\alpha x _ n)]$.
    • Clearly the scalar-multiplication is well defined.
• Define the norm on $\widetilde X$ as $x \mapsto \widetilde d(x, 0)$.
• The operations and norm extends original operations on $X$.
• The defined $(\widetilde X, + ,\cdot)$ is a normed vector space over $\bF$.
  • Just verify the axioms of vector spaces.
• $J$ is an isometric homeomorphism.

Completion of bounded linear maps #

Suppose $X$ is a normed space, and $Y$ is a Banach space over $\bF$.

• There exists a completion $(\widetilde X, J)$ of the normed space $X$.

Suppose $T \in \mathcal B(X, Y)$.

• There exists a unique $\widetilde T \in \mathcal B(\widetilde X, Y)$ extending $T$, and further $\|{\widetilde T}\| = \norm{T}$.
• $T$ is an embedding / isometry iff $\widetilde T$ is an embedding / isometry.

The map $\varphi: \mathcal B(\widetilde X, Y) \to \mathcal B(X, Y)$ defined by $\widetilde T \mapsto \widetilde T \circ J = T$.

• $\varphi$ is bijective. Since completion of $T$ is unique.
• $\varphi$ is an isometric homeomorphism.

Uniqueness of norm space completion #

All completions of a norm space are the same up to isometric homeomorphisms.

Suppose $X$ is a normed space with completion $(X _ 1, J _ 1)$ and $(X _ 2, J _ 2)$.

• There is a unique isometry $I _ {12} \in \mathcal B(X _ 1, X _ 2)$ extending $J _ 2 \in \mathcal B(X, X _ 2)$.
• There is a unique isometry $I _ {21} \in \mathcal B(X _ 2, X _ 1)$ extending $J _ 1 \in \mathcal B(X, X _ 1)$.
• There is a unique isometry $I _ {11} \in \mathcal B(X _ 1, X _ 1)$ extending $J _ 1 \in \mathcal B(X, X _ 1)$.
• There is a unique isometry $I _ {22} \in \mathcal B(X _ 2, X _ 2)$ extending $J _ 2 \in \mathcal B(X, X _ 2)$.
• $I _ {11}$ and $I _ {22}$ are identities on $X _ 1$ and $X _ 2$.

Consider the map $I _ {21}\circ I _ {12} \in \mathcal B(X _ 1, X _ 1)$

• It extends $J _ 1 \in \mathcal B(X, X _ 1)$.
  • For any $x \in X$, $I _ {21}\circ I _ {12}(x) = I _ {21}(J _ 2(x)) = J _ 1(x)$.
• By uniqueness of extension $I _ {21}\circ I _ {12} = I _ {11}$. Similarly $I _ {12}\circ I _ {21} = I _ {22}$.

$I _ {12}$ and $I _ {21}$ are isometric homeomorphisms!

Quotient Normed Spaces #

We have already seen quotient vector spaces, these are the new results for normed spaces:

• Consider $X / X _ 0$. The quotient norm is generally a semi-norm. And only when $X _ 0$ is closed, it is a true norm.
• The map $Q: x \mapsto x + X _ 0$ is a coisometry when $X _ 0$ is closed.

We have already seen quotient linear maps, these are the new results:

• Suppose $T \in \mathcal B(X, Y)$ and $X _ 0 \subseteq \null T$ is closed.
• $T _ {X _ 0} \in \mathcal B(X/X _ 0, Y)$ and $\norm{T _ {X _ 0}} = \norm{T}$.

Quotient normed space #

Suppose $X$ is a normed space and $X _ 0$ is a subspace.

Then $X / X _ 0$ is a vector space. Consider $Q \in \L(X, X / X _ 0)$ where $Q: x \mapsto x + X _ 0$.

Define the quotient norm as $$ \norm{x + X _ 0} := \inf\{\norm{x - y}: y \in X _ 0\} = \operatorname{dist}(x, X _ 0) $$

• $\n{\cdot}$ is a seminorm on $X / X _ 0$.
  • Clearly $\norm{x + X _ 0} \ge 0$ and $\norm{0 + X _ 0} = 0$.
  • $\norm{\alpha x + X _ 0} = \alpha\norm{x + X _ 0}$.
  • Triangle inequality is true.
    • Hint: let $(x _ n) _ {n = 1}^\infty \in X _ 0$ such that $\norm{x - x _ n} \to \norm{x + X _ 0}$.
• The following are equivalent:
  1. $X _ 0$ is a closed subspace of $X$.
  2. $\forall x\notin X _ 0: \operatorname{dist}(x, X _ 0) > 0$.
  3. $\forall x \notin X _ 0: \norm{x + X _ 0} > 0$.
  4. $\n{\cdot}$ is a norm on $X / X _ 0$.

Suppose $X _ 0$ is a closed subspace. Then $Q$ is a coisometry.

• $\norm{Q} \le 1$ since $\|x\| = \inf\{\norm{x + y}: y \in \{0\}\} \ge \norm{x + X _ 0}$.
• So $QB _ X(0; 1) \subseteq B _ {X/X _ 0}(0; 1)$.
• Suppose $x + X _ 0 \in B _ {X / X _ 0}(0; 1)$, for some $y \in X _ 0$, $\norm{x + y} < 1$.
• So $QB _ X(0; 1) \supseteq B _ {X/X _ 0}(0; 1)$.

Quotient bounded linear maps #

Suppose $X, Y$ are nonzero normed space over $\bF$. And $T \in \mathcal B(X, Y)$.

• Suppose $X _ 0 \subseteq \null T$ is a closed subspace of $X$. ($\null T$ is also closed.)
• Define $Q _ {X _ 0}(x) = x + X _ 0$. $Q _ {X _ 0}$ is a coisometry by (Quotient normed space).
• By (Quotient linear maps), $T _ {X _ 0} \in \L(X/X _ 0, Y)$ uniquely exists such that $T _ {X _ 0} \circ Q _ {X _ 0} = T$.

In this case, we have the following properties:

• Since $T _ {X _ 0}[B _ {X/X _ 0}(0; 1)] = T[B _ X(0; 1)]$, $( * )$.
  • $T _ {X _ 0} \in \mathcal B(X/X _ 0, Y)$ and $\norm{T _ {X _ 0}} = \norm{T}$.
    • Suppose $Y \neq \{0\}$, then $\norm{T / X _ 0} = \norm{T}$.
  • $T _ {X _ 0}$ is open iff $T$ is open.
  • $T _ {X _ 0}$ is coisometry iff $T$ is coisometry.

Consider space $V = \c{T \in \mathcal B(X, Y): X _ 0 \subseteq \null T}$. $V$ is a vector space over $\bF$.

• Consider the map $T \mapsto T _ {X _ 0}$, this is an isometric homeomorphism from $V$ to $\mathcal B(X / X _ 0, Y)$.

Corollary: double quotient of bounded linear maps #

Suppose $X, Y$ are normed spaces with $X _ 0, Y _ 0$ being subspaces.

Suppose $T \in \mathcal B(X, Y)$ where $T[X _ 0] \subseteq Y _ 0$. Then there exists a unique $\widehat T\in \mathcal B(X / X _ 0, Y / Y _ 0)$ such that $\widehat T \circ Q _ {X _ 0} = Q _ {Y _ 0} \circ T$. Further $\n{\widehat T} = \n{Q _ {Y _ 0}\circ T} \le \n{T}$.

• Apply (Quotient bounded linear maps) to $Q _ {Y _ 0}\circ T$, demonstrates the existence and uniqueness.
• $\n{Q _ {Y _ 0}\circ T} \le \n{T}$ by definition.

Corollary: fundamental theorem of bounded linear homomorphisms #

Suppose $X, Y$ are normed spaces over $\bF$. And $T \in \mathcal B(X, Y)$.

• By (Fundamental theorem of linear homomorphisms) $\widetilde T = T _ {\null T}$ is the unique isomorphism such that $\widetilde T \circ Q _ {\null T} = T$.

We extend above theorem in following ways:

• $\widetilde T$ is a topological isomorphism if and only if $T$ is open.
  • Since $\widetilde T$ is open iff $T$ is open. So $\widetilde T^{-1}$ is continuous iff $\widetilde T$ is top. iso. iff $T$ is open.
• $\widetilde T$ is an isometric isomorphism if and only if $T$ is an coisometry.
  • $T$ is a coisometry iff $\widetilde T$ is a coisometry iff $\widetilde T$ is an isometry by (Coisometry).

Completeness of quotient normed spaces #

Suppose $X$ is a Banach space. $X _ 0 \subseteq X$ is a closed subspace. Then $X / X _ 0$ is a Banach space.

• Suppose $(u _ n) _ {n = 1}^\infty \in X / X _ 0$ and $\sum _ n \norm{u _ n} < \infty$.
• Pick $(x _ n) _ {n = 1}^\infty \in X$ such that $\norm{x _ n} \le \norm{u _ n} + 1/2^n$. So $(x _ n)$ is absolutely convergent.
• Suppose $\sum _ n x _ n = x \in X$. Then $Q(x) = Q(\sum _ n x _ n) = \sum _ n u _ n$ as $Q$ is continuous.
• So absolutely convergent series in $X / X _ 0$ are convergent.