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.

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:

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:


 * $\norm {B \circ A} \le \norm B \norm A$