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.

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 $\struct {\map {CL} {X, Y}, \norm{\, \cdot \,} }$.