Space of Bounded Linear Transformations is Banach Space

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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:

\(\ds \map A {\lambda x + \mu y}\) \(=\) \(\ds \lim_{n \mathop \to \infty} \map {A_n} {\lambda x + \mu y}\)
\(\ds \) \(=\) \(\ds \lim_{n \mathop \to \infty} \paren {\lambda A_n x + \mu A_n y}\)
\(\ds \) \(=\) \(\ds \lim_{n \mathop \to \infty} \lambda A_n x + \lim_{n \mathop \to \infty} \mu A_n y\) Sum Rule for Sequence in Normed Vector Space
\(\ds \) \(=\) \(\ds \lambda \lim_{n \mathop \to \infty} A_n x + \mu \lim_{n \mathop \to \infty} A_n y\) Multiple Rule for Sequence in Normed Vector Space
\(\ds \) \(=\) \(\ds \lambda A x + \mu A y\)

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.

$\blacksquare$


Sources

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