Space of Bounded Linear Transformations is Banach Space

== Theorem ==
Let $\GF \in \set {\R, \C}$. 
Let $\struct {X, \norm {\, \cdot \,}_X}$ be a normed vector spaces over $\GF$. 
Let $\struct {Y, \norm {\, \cdot \,}_Y}$ be a Banach space over $\GF$. 
Let $\map B {X, Y}$ be the space of bounded linear transformations from $X$ to $Y$ endowed with pointwise addition and ($\Bbb F$)-scalar multiplication.
Let $\norm {\, \cdot \,}_{\map B {X, Y} }$ be the norm on the space of bounded linear transformations.
Then $\struct {\map B {X, Y}, \norm {\, \cdot \,}_{\map B {X, Y} } }$ is a Banach space.
== Proof ==
Let $\sequence {A_n}_{n \mathop \in \N}$ be a Cauchy sequence in $\map B {X, Y}$. 
Then for each $\epsilon > 0$ there exists $N \in \N$ such that: 
:$\norm {A_n - A_m}_{\map B {X, Y} } < \epsilon$ 
for $n, m \ge N$. 
Then for each $x \in X$ with $\norm x_X \le 1$ we have: 
:$\norm {A_n x - A_m x}_Y < \epsilon$ 
for each $n, m \ge N$. 
So $\sequence {A_n x}_{n \in \N}$ is Cauchy for each $x \in X$ with $\norm x_X \le 1$. 
For $x \in X$ with $\norm x_X > 1$ we can write: 
:$\ds A_n x = \norm x_X \map {A_n} {\frac x {\norm x_X} }$
Then for each $\epsilon > 0$ again pick $N \in \N$ such that: 
:$\ds \norm {\map {A_n} {\frac x {\norm x_X} } - \map {A_m} {\frac x {\norm x_X} } }_Y < \frac \epsilon {\norm x_X}$ 
for $n, m \ge N$. 
So: 
:$\norm {A_n x - A_m x}_Y < \epsilon$ 
So $\sequence {A_n x}_{n \in \N}$ is Cauchy for each $x \in X$. 
Since $\struct {Y, \norm {\, \cdot \,}_Y}$ is a Banach space, for each $x \in X$ there exists $A x \in Y$ such that $A_n x \to A x$ as $n \to \infty$. 
It now needs to be verified that the thus defined $A : X \to Y$ is a bounded linear transformation.
For linearity, let $x, y \in X$ and $\lambda, \mu \in \GF$. 
Then we have: 
Now take $N' \in \N$ such that: 
:$\norm {A_n - A_m}_{\map B {X, Y} } < 1$
for $n, m \ge N'$. 
Then from Fundamental Property of Norm on Bounded Linear Transformation, we have: 
:$\norm {A_n x - A_N x}_Y \le \norm x_X$
for each $x \in X$ and $n \ge N$. 
Taking $n \to \infty$, we obtain: 
:$\norm {A x - A_N x}_Y \le \norm x_X$ 
from Modulus of Limit: Normed Vector Space and Inequality Rule for Real Sequences.
By Reverse Triangle Inequality: Normed Vector Space, we have: 
:$\norm {A x}_Y - \norm {A_N x}_Y \le \norm x_X$
So:
:$\norm {A x}_Y \le \norm {A_N x}_Y + \norm x_X$
From Fundamental Property of Norm on Bounded Linear Transformation, we have: 
:$\norm {A x}_Y \le \paren {\norm {A_N}_{\map B {X, Y} } + 1} \norm x_X$
So $A$ is a bounded linear transformation. 
We conclude by showing $A_n \to A$ in $\struct {\map B {X, Y}, \norm {\, \cdot \,}_{\map B {X, Y} } }$.
Again pick $\epsilon > 0$ and $N \in \N$ such that: 
:$\norm {A_n - A_m}_{\map B {X, Y} } < \epsilon$ 
Then for each $x \in X$ with $\norm x_X \le 1$ we have: 
:$\norm {A_n x - A_m x}_Y < \epsilon$ 
for $n, m \ge N$. 
Taking $m \to \infty$, we have: 
:$\norm {A_n x - A x}_Y < \epsilon$
for $n \ge N$ and $\norm x_X \le 1$. 
That is: 
:$\norm {A_n - A}_{\map B {X, Y} } < \epsilon$
for $n \ge N$. 
So we have $A_n \to A$ in $\struct {\map B {X, Y}, \norm {\, \cdot \,}_{\map B {X, Y} } }$.
So every Cauchy sequence in $\struct {\map B {X, Y}, \norm {\, \cdot \,}_{\map B {X, Y} } }$ converges.
So $\struct {\map B {X, Y}, \norm {\, \cdot \,}_{\map B {X, Y} } }$ is Banach. 
== Sources ==
* : $\text {II}.1.2$, $\text {II}.1$ Exercise $4$
Category:Banach Spaces
Category:Space of Bounded Linear Transformations