Norm on Bounded Linear Transformation is Submultiplicative
Theorem
Let $\struct {X, \norm \cdot_X}$, $\struct {Y, \norm \cdot_Y}$ and $\struct {Z, \norm \cdot_Z}$ be normed vector spaces.
Let $A : X \to Y$ and $B : Y \to Z$ be bounded linear transformations.
Let $\norm \cdot _{\map B {X,Y} }$ be the norm for bounded linear transformations $X \to Y$.
Let $\norm \cdot _{\map B {Y,Z} }$ be the norm for bounded linear transformations $Y \to Z$.
Let $\norm \cdot _{\map B {X,Z} }$ be the norm for bounded linear transformations $X \to Z$.
Then, we have that:
- $B \circ A$ is a bounded linear transformation
with:
- $\norm {B \circ A} _{\map B {X,Z} } \le \norm B _{\map B {Y,Z} } \norm A _{\map B {X,Y} }$
That is:
- $\norm \cdot$ is submultiplicative.
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Proof
From Composition of Linear Transformations is Linear Transformation, we have:
- $B \circ A$ is a linear transformation
Let $x \in X$.
Then, we have:
\(\ds \norm {\paren {B \circ A} x}_Z\) | \(\le\) | \(\ds \norm B \norm {A x}_Y\) | Fundamental Property of Norm on Bounded Linear Transformation | |||||||||||
\(\ds \) | \(\le\) | \(\ds \norm B \norm A \norm x_X\) | Fundamental Property of Norm on Bounded Linear Transformation |
We can therefore see that:
- $B \circ A$ is a bounded linear transformation.
So, if:
- $\norm x_X = 1$
we have:
- $\norm {\paren {B \circ A} x}_Z \le \norm B \norm A$
By the definition of supremum, we have:
- $\ds \sup_{\norm x_X = 1} \norm {\paren {B \circ A} x}_Z \le \norm B \norm A$
So by the definition of the norm on the space of bounded linear transformations, we have:
- $\norm {B \circ A} \le \norm B \norm A$
$\blacksquare$