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||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 βˆ’ 1 = 𝑀 . Note that  β€– 𝑀 β€– β‰₯ β€– 𝑀 𝑒 1 β€– / β€– 𝑒 1 β€– = 2 . Hence β€– 𝑀 βˆ’ 1 β€– = β€– 𝑀 β€– β‰₯ 2 .  Thus the equality β€– 𝑀 βˆ’ 1 β€– = β€– 𝑀 β€– βˆ’ 1 doesn't hold in general. Anyway, in general we can have β€– 𝑇 βˆ’ 1 β€– β‰₯ 1 β€–...