Derivative of Product of Operator-Valued Functions

Theorem
Let $\struct {X, \norm \cdot_X}$, $\struct {Y, \norm \cdot_Y}$, and $\struct {Z, \norm \cdot_Z}$ normed vector spaces.

Let $\map B {X, Y}$, $\map B {Y, Z}$ and $\map B {X, Z} $ denote the space of bounded linear transformations between $X$ and $Y$, between $Y$ and $Z$, and between $X$ and $Z$, respectively.

Let $A : \R \to \map B {X, Y}$ and $B : \R \to \map B {Y, Z}$ be differentiable mappings whose images are bounded linear transformations.

The product $A B: \R \to \map B {X, Z}: x \mapsto \map A x \map B x$ is differentiable and its derivative at any $x \in \R$ is given by $\map {A'} x \map B x + \map A x \map {B'} x$.

Proof
Given arbitrary $x, h \in X$ such that $h \ne 0$ we compute:
 * $\norm {\dfrac {\map {\paren {A B} } {x + h} - \map {\paren {A B} } x} h - \map {A'} x \map B x - \map A x \map {B'} x}_{\map B {X, Z} }$

to be:

$\norm {\dfrac {\map A {x + h} - \map A x} h - \map {A'} x}_{\map B {X, Y} }$ vanishes as $h \to 0$ because $\map {A'} x$ is the derivative of $A$ at $x$ which exists by assumption.

$\norm {\dfrac {\map B {x + h} - \map B x} h - \map {B'} x}_{\map B {Y, Z} }$ vanishes as $h \to 0$ because $\map {B'} x$ is the derivative of $B$ at $x$ which exists by assumption.

$\norm {\map B {x + h} - \map B x }_{\map B {Y, Z} }$ vanishes as $h \to 0$ because:

By the Reverse Triangle Inequality:
 * $\size {\norm {\map B {x + h} }_{\map B {Y, Z} } - \norm {\map B x}_{\map B {Y, Z} } } \le \norm {\map B {x + h} - \map B x}_{\map B {Y, Z} } \to 0$

as $h \to 0$.

Together this shows:
 * $\ds \lim_{h \mathop \to 0} \norm {\frac {\map {\paren {A B} } {x + h} - \map {\paren {A B} } x} h - \map {A'} x \map B x - \map A x \map {B'} x}_{\map B {X, Z} } = 0$

Therefore the derivative of $A B$ at $x$ is:
 * $\map {A'} x \map B x + \map A x \map {B'} x$