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
- 2017: Amol Sasane: A Friendly Approach to Functional Analysis ... (previous) ... (next): Chapter $\S 2.3$: The normed space $\map {CL} {X, Y}$. Operator norm and the normed space $\map {CL} {X, Y}$