Necessary and Sufficient Conditions for Continuous Linear Transformation Space to be Banach Space

== Theorem ==
Let $\struct {X, \norm {\, \cdot \,}_X}$ and $\struct {Y, \norm {\, \cdot \,}_Y}$ be normed vector spaces.
Let $\struct{\map {CL} {X, Y}, \norm{\, \cdot \,}}$ be the continuous linear transformation space equipped with the supremum operator norm.
Then $\struct {\map {CL} {X, Y}, \norm{\, \cdot \,} }$ is a Banach Space $\struct {Y, \norm{\, \cdot \,}_Y}$ is a Banach Space.
== Proof ==
=== Necessary Condition ===
Let $Y$ be a Banach space.
Let $\sequence {T_n}_{n \mathop \in \N} \in \map {CL} {X, Y}$ be a Cauchy sequence.
Let $x \in X$.
Let $\norm {\, \cdot \,}$ be the supremum operator norm.
==== $\sequence {T_n x}_{n \mathop \in \N}$ is a Cauchy sequence in $\struct {Y, \norm {\, \cdot \,}_Y}$ ====
We have that:
By definition of Cauchy sequence:
:$\forall \epsilon \in \R_{>0}: \exists N \in \N: \forall m, n \in \N: m, n \ge N: \norm {T_n - T_m} < \epsilon$
Suppose $m, n \ge N$.
Then:
:$\forall \epsilon \in \R_{>0} : \forall x \in X : \norm {T_n x - T_m x}_Y < \epsilon \norm x_X$
Let $\epsilon' = \epsilon \norm x_X$
$\epsilon \in \R_{>0}$ and $x \in X$ were arbitrary.
Hence $\epsilon' \in \R_{> 0}$ is also arbitrary.
Therefore:
:$\forall \epsilon' \in \R_{> 0} : \exists N \in \N: \forall m, n \in \N: m, n \ge N: \norm {T_n x - T_m x}_Y < \epsilon'$
By definition, $\sequence {T_n x}_{n \mathop \in \N}$ is a Cauchy sequence in $\struct {Y, \norm {\, \cdot \,}_Y}$.
==== $\sequence {T_n x}_{n \mathop \in \N}$ converges in $\struct {Y, \norm {\, \cdot \,}_Y}$ ====
$Y$ is Banach.
$\sequence {T_n x}_{n \mathop \in \N}$ is a Cauchy sequence in $\struct {Y, \norm {\, \cdot \,}_Y}$.
Hence, $\sequence {T_n x}_{n \mathop \in \N}$ converges in $Y$ with limit, say, $Tx \in Y$.
==== $T$ is a linear transformation ====
Let $x_1, x_2 \in X$.
Then:
:$\ds \lim_{n \mathop \to \infty} \paren{T_n x_1} = T x_1$
:$\ds \lim_{n \mathop \to \infty} \paren{T_n x_2} = T x_2$
$Y$ is a vector space.
Thus, $x_1 + x_2 \in Y$.
Hence:
:$\ds \lim_{n \mathop \to \infty} \paren {T_n \paren {x_1 + x_2}} = T \paren {x_1 + x_2}$
By combination of limits:
:$\ds \lim_{n \mathop \to \infty} \paren{T_n x_1 + T_n x_2} = T x_1 + T x_2$
By linearity of $T_n$:
:$\sequence {T_n x_1 + T_n x_2}_{n \mathop \in \N} = \sequence {T\paren {x_1 + x_2}}_{n \mathop \in \N}$
By uniqueness of limits:
:$T \paren {x_1 + x_2} = T x_1 + T x_2$.
Let $\alpha \in \set {\R, \C}$.
Let $x \in X$.
Then:
:$\ds \lim_{n \mathop \to \infty} \paren {T_n x} = T x$
By Multiple Rule for Sequences:
:$\ds \lim_{n \mathop \to \infty} \paren {\alpha \cdot T_n x} = \alpha \cdot T x$
By linearity of $T_n$:
:$\sequence{\alpha \cdot \paren{T_n x}}_{n \mathop \in \N} = \sequence{T_n\paren{\alpha \cdot x}}_{n \mathop \in \N}$
Since $X$ is a vector space:
:$\alpha \cdot x \in X$
Then:
:$\ds \lim_{n \mathop \to \infty} \paren {T_n\paren{\alpha \cdot x}}_{n \mathop \in \N} = T \paren {\alpha \cdot x}$.
Altogether:
:$\alpha \cdot \map T x = \map T {\alpha \cdot x}$.
Altogether, by definition of linear transformation:
:$T \in \map \LL {X, Y}$
==== $T$ is a continuous transformation ====
Let $\sequence {T_n}_{n \mathop \in \N} \in \map {CL} {X, Y}$ be a Cauchy sequence.
Then:
:$\exists N \in \N : \forall m, n \in \N : m, n > N : \norm {T_n - T_m} < \epsilon$
Hence:
:$\exists N \in \N : \forall n > N : \norm {T_n - T_{N + 1} } < \epsilon$
Therefore:
Take the limit $n \to \infty$.
Then: 
:$\forall x \in X : \norm {T x - T_{N + 1} x} < \epsilon \norm x_X$
Thus:
By continuity of linear transformations:
:$T \in \map {CL} {X, Y}$
==== $T_n$ converges to $T$ in $\struct {\map {CL} {X, Y}, \norm {\, \cdot \,}}$ ====
By definition of Cauchy sequence.
:$\forall \epsilon \in \R_{>0} : \exists N \in \N : \forall n, m \in \N : n, m > N \implies \norm {T_n - T_m} < \epsilon$
Hence:
We have that Norm on Vector Space is Continuous Function.
Take the limit $m \to \infty$.
By Limit of Composite Function:
Hence:
:$\forall \epsilon \in \R_{>0} : \exists N \in \N : \forall n > N : \norm {T_n - T} < \epsilon$
By definition, $T$ is continuous.
Altogether, a Cauchy sequence $\sequence {T_n}_{n \mathop \in \N}$ converges to a linear and continuous mapping $T$ in $\struct {\map {CL} {X, Y}, \norm{\, \cdot \,} }$.
=== Sufficient Condition ===
Let $\sequence {y_n}$ be a Cauchy sequence in $Y$.
Given $y\in Y$, define the mapping $f_y: X \to Y$ that
:$\map {f_y} x = \map \varphi x y$
where $\varphi \in X'$ is a fixed nonzero element of the dual space $X'$.
Note that
:$\norm {\map {f_y} x} = \size {\map \varphi x} \norm {y}$
so
:$\norm {f_y} = \norm {\varphi} \norm {y}$
so $f_y \in \map {CL} {X,Y}$.
Since
:$\sequence {y_n}$ is a Cauchy sequence in $Y$,
and
:$\norm {f_y} = \norm {\varphi} \norm {y}$,
the sequence $\sequence {f_{y_n}}$ is a Cauchy sequence in $\map {CL} {X,Y}$.
By assumption, $\sequence {f_{y_n}}$ converges. Let $f \in \map {CL} {X,Y}$ be such that $f_{y_n} \to f$.
Let $x_0 \in X$ be such that $\map \varphi {x_0} = 1$.
Then:
So $\sequence {y_n}$ converges to $\map f {x_0} \in Y$.
So $\struct {Y, \norm{\, \cdot \,}_Y}$ is a Banach Space.
== Sources ==
* : Chapter $\S 2.3$: The normed space $\map {CL} {X, Y}$. When is $\map {CL} {X, Y}$ complete?
Category:Necessary and Sufficient Conditions for Continuous Linear Transformation Space to be Banach Space
Category:Normed Vector Spaces
Category:Banach Spaces
Category:Continuous Linear Transformation Spaces