Composition of Continuous Linear Transformations is Continuous Linear Transformation

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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

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