Composition of Continuous Linear Transformations is Continuous Linear Transformation
Jump to navigation
Jump to search
Theorem
Let $K$ be a field.
Let $\struct {X, \norm {\, \cdot \,}_X}$, $\struct {Y, \norm {\, \cdot \,}_Y}$, $\struct{Z, \norm {\, \cdot \,}_Z}$ be normed vector spaces over $K$.
$\map {CL} {X, Y}$ be the continuous linear transformation space.
Let $\norm {\, \cdot \,}$ be the supremum operator norm.
Let $S \circ T : X \to Z$ be the composition of mappings such that:
- $\forall T \in \map {CL} {X, Y} : \forall S \in \map {CL} {Y, Z} : \forall x \in X : \map {S \circ T} x := \map S {\map T x}$
Then $S \circ T \in \map {CL} {X, Z}$.
Proof
Linearity
Follows from Composition of Linear Transformations is Linear Transformation.
Continuity
We have that:
\(\ds \forall x \in X: \, \) | \(\ds \norm{\map {S \circ T} x}_Z\) | \(=\) | \(\ds \norm {\map S {\map T x} }_Z\) | |||||||||||
\(\ds \) | \(\le\) | \(\ds \norm S \norm {\map T x}_Y\) | Supremum Operator Norm as Universal Upper Bound | |||||||||||
\(\ds \) | \(\le\) | \(\ds \norm S \norm T \norm x_X\) | Supremum Operator Norm as Universal Upper Bound | |||||||||||
\(\ds \) | \(=\) | \(\ds M \norm x_X\) | $\R \ni M := \norm S \norm T$ |
By Continuity of Linear Transformation between Normed Vector Spaces, $S \circ T$ is continuous.
$\blacksquare$
Sources
- 2017: Amol Sasane: A Friendly Approach to Functional Analysis ... (previous) ... (next): Chapter $\S 2.4$: Composition of continuous linear transformations