Norm on Bounded Linear Transformation is Submultiplicative

Theorem
Let $\mathbb F \in \set {\R, \C}$.

Let $\struct {\HH, \innerprod \cdot \cdot_\HH}$, $\struct {\KK, \innerprod \cdot \cdot_\KK}$ and $\struct {\LL, \innerprod \cdot \cdot_\LL}$ be Hilbert spaces over $\mathbb F$.

Let $A : \HH \to \KK$ and $B : \KK \to \LL$ be bounded linear transformations.

Let $\norm \cdot$ be the norm on the space of bounded linear transformations.

Then, we have:


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

That is:


 * $\norm \cdot$ is submultiplicative.

Proof
Let $\norm \cdot_\HH$ be the inner product norm on $\HH$.

Let $\norm \cdot_\KK$ be the inner product norm on $\KK$.

Let $\norm \cdot_\LL$ be the inner product norm on $\LL$.

Let $h \in \HH$.

Then, we have:

So, if:


 * $\norm h_\HH = 1$

we have:


 * $\norm {\paren {B A} h}_\LL \le \norm B \norm A$

By the definition of supremum, we have:


 * $\ds \sup_{\norm h_\HH = 1} \norm {\paren {B A} h}_\LL \le \norm B \norm A$

So by the definition of the norm on the space of bounded linear transformations, we have:


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