$||AB|| \leq ||A||||B||$

Consider the following theorem in functional analysis.

Let $X$ be a Banach space. Suppose that $S,T \in \mathscr{B}(X)$, $T$ is invertible, and  $||T-S||<||T^{-1}||^{-1}$. Then $S$ is invertible in $\mathscr{B}(X)$.

We see two notes about this theorem.

 

NOTE 1:

At first glance, one may ask that   "Can we write$||T^{-1}||^{-1}$ as ||T||?". Because $TT^{-1}=I$, $||I||=1$ and "norm is generalization of modulus function |.| for which we have $|AB|=|A||B|$." So it is natural to have such a thought.

But it is interesting to note that  $||AB|| \neq ||A||||B||$, but $||AB|| \leq ||A||||B||$.

For example consider the matrix  $\begin{bmatrix}0&\frac{1}{2}\\2&0 \end{bmatrix}$. then 𝑀−1=𝑀. Note that  ‖𝑀‖≥‖𝑀𝑒1‖/‖𝑒1‖=2. Hence ‖𝑀−1‖=‖𝑀‖≥2 . 

Thus the equality ‖𝑀−1‖=‖𝑀‖−1 doesn't hold in general.

Anyway, in general we can have ‖𝑇−1‖≥1‖𝑇‖

NOTE 2:

The theorem helps us to prove set of all invertible operators form an open set. 

 

 

References:

[1] https://math.stackexchange.com/questions/525970/norm-of-an-inverse-operator-t-1-t-1