Continuous Linear Transformations form Subspace of Linear Transformations

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Theorem

The space of continuous linear transformations is a subspace of the space of linear transformations.


Proof

Let $\struct {X, \norm \cdot }$ and $\struct {Y, \norm \cdot }$ be normed vector spaces.


Closure under vector addition

Let $S, T \in \map {CL} {X, Y}$.

By Continuity of Linear Transformation between Normed Vector Spaces:

$\exists M_S \in \R : M_S > 0 : \forall x \in X : \norm {S x} \le M_S \norm x$
$\exists M_T \in \R : M_T > 0 : \forall x \in X : \norm {T x} \le M_T \norm x$

Furthremore:

\(\ds \norm {\paren {S + T} x}\) \(=\) \(\ds \norm {S x + T x}\) Linear Transformation Space is Vector Space
\(\ds \) \(\le\) \(\ds \norm {S x} + \norm {T x}\) Definition of Norm on Vector Space
\(\ds \) \(\le\) \(\ds M_S \norm x + M_T \norm x\) Continuity of Linear Transformation between Normed Vector Spaces
\(\ds \) \(=\) \(\ds \paren {M_S + M_T} \norm x\)

By Continuity of Linear Transformation between Normed Vector Spaces:

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


Closure under scalar multiplication

Let $\alpha \in \R$.

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

By Continuity of Linear Transformation between Normed Vector Spaces:

$\exists M \in \R_{> 0} : \forall x \in X : \norm {T x} \le M \norm x$

Hence:

\(\ds \norm {\paren {\alpha T} x}\) \(=\) \(\ds \norm {\alpha \paren {T x} }\) Linear Transformation Space is Vector Space
\(\ds \) \(=\) \(\ds \size \alpha \norm {T x}\) Norm Axiom $\text N 2$: Positive Homogeneity
\(\ds \) \(\le\) \(\ds \size \alpha M \norm x\) Continuity of Linear Transformation between Normed Vector Spaces
\(\ds \) \(=\) \(\ds M' \norm x\) $M' := \size \alpha M$, $M' \in \R_{> 0}$

By Continuity of Linear Transformation between Normed Vector Spaces:

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


Existence of identity element under vector addition

Let $\mathbf 0 : X \to Y$ be the zero mapping.

Then:

\(\ds \forall x \in X : \ \ \) \(\ds \norm {\mathbf 0 x}\) \(=\) \(\ds \norm {\mathbf 0_Y}\)
\(\ds \) \(=\) \(\ds 0\) Norm Axiom $\text N 1$: Positive Definiteness
\(\ds \) \(\le\) \(\ds 1 \cdot \norm x\)

Hence:

$\mathbf 0 \in \map {CL} {X, Y}$

$\blacksquare$


Sources

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