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

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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 if and only if $\struct {Y, \norm{\, \cdot \,}_Y}$ 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:

\(\ds \forall x \in X : \forall n, m \in \N: \, \) \(\ds \norm {T_n x - T_m x}_Y\) \(=\) \(\ds \norm {\paren {T_n - T_m} x}_Y\) Linear Mappings between Vector Spaces form Vector Space
\(\ds \) \(\le\) \(\ds \norm {T_n - T_m} \norm x_X\) Supremum Operator Norm as Universal Upper Bound

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 $\struct {Y, \norm {\, \cdot \,}_Y}$.

$\Box$


$\sequence {T_n x}_{n \mathop \in \N}$ 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, $Tx \in Y$.

$\Box$


$T$ 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:

$T \in \map \LL {X, Y}$

$\Box$


$T$ 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:

\(\ds \forall n > N : \forall x \in X: \, \) \(\ds \norm {T_n x - T_{N + 1} x}_Y\) \(\le\) \(\ds \norm {T_n - T_{N + 1} } \norm x_X\) Supremum Operator Norm as Universal Upper Bound
\(\ds \) \(<\) \(\ds \epsilon \cdot \norm x_X\)

Take the limit $n \to \infty$.

Then:

$\forall x \in X : \norm {T x - T_{N + 1} x} < \epsilon \norm x_X$

Thus:

\(\ds \forall x \in X: \, \) \(\ds \norm {Tx}_Y\) \(=\) \(\ds \norm {Tx - T_{N+1} x + T_{N+1} x }_Y\)
\(\ds \) \(\le\) \(\ds \norm {Tx - T_{N + 1} x}_Y + \norm {T_{N + 1} x}_Y\) Definition of Norm on Vector Space
\(\ds \) \(<\) \(\ds \epsilon \norm x_X + \norm {T_{N + 1} } \norm x_X\) Supremum Operator Norm as Universal Upper Bound
\(\ds \) \(=\) \(\ds \paren {\epsilon + \norm {T_{N + 1} } } \norm x_X\)
\(\ds \) \(=\) \(\ds M \norm x_X\) $\epsilon + \norm {T_{N + 1} } = M \in \R$

By continuity of linear transformations:

$T \in \map {CL} {X, Y}$

$\Box$


$T_n$ 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:

\(\ds \forall \epsilon \in \R_{>0} : \exists N \in \N : \forall n, m > N : \forall x \in X: \, \) \(\ds \norm {T_n x - T_m x}_Y\) \(\le\) \(\ds \norm {T_n - T_m} \norm x_X\) Supremum Operator Norm as Universal Upper Bound
\(\ds \) \(<\) \(\ds \epsilon \norm x_X\)

We have that Norm on Vector Space is Continuous Function.

Take the limit $m \to \infty$.

By Limit of Composite Function:

\(\ds \forall \epsilon \in \R_{>0} : \exists N \in \N : \forall n > N : \forall x \in X: \, \) \(\ds \norm {T_n x - T x}_Y\) \(\le\) \(\ds \norm {T_n - T} \norm x_X\)
\(\ds \) \(<\) \(\ds \epsilon \norm x_X\)

Hence:

$\forall \epsilon \in \R_{>0} : \exists N \in \N : \forall n > N : \norm {T_n - T} < \epsilon$

By definition, $T$ is continuous.

$\Box$


Altogether, a Cauchy sequence $\sequence {T_n}_{n \mathop \in \N}$ converges to a linear and continuous mapping $T$ in $\struct {\map {CL} {X, Y}, \norm{\, \cdot \,} }$.

$\Box$


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:

\(\ds \map f {x_0}\) \(=\) \(\ds \lim_{n \to \infty} \map{f_{y_n} } {x_0}\)
\(\ds \) \(=\) \(\ds \lim_{n \to \infty} \map \varphi {x_0} y_n\)
\(\ds \) \(=\) \(\ds \lim_{n \to \infty} y_n\)

So $\sequence {y_n}$ converges to $\map f {x_0} \in Y$.

So $\struct {Y, \norm{\, \cdot \,}_Y}$ is a Banach Space.


$\blacksquare$


Sources

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