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 . Note that . Hence .
Thus the equality doesn't hold in general.
Anyway, in general we can have
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
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