Let $f,g\in L(X,Y)$ where $X,Y$ are Banach spaces and let $f^{-1},g^{-1}\in L(Y,X)$.
How to prove that $$\| g^{-1}-f^{-1}\|\leq 2\|{f^{-1}}\|^2\|g-f\|$$ if $\|g-f\|\leq\frac{1}{2\| f^{-1}\| }$?
2026-04-03 15:12:13.1775229133
Proving $\| g^{-1}-f^{-1}\|\leq 2\|{f^{-1}}\|^2\|g-f\|$
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$\|gf^{-1}-I\| =\|(g-f)f^{-1}\|\leq \frac 1 2$. The series $ \sum\limits_{k=0}^{\infty} (I-gf^{-1})^{k}$ converges to $(gf^{-1})^{-1}=fg^{-1}$. Hence $\|fg^{-1}\| \leq \sum\limits_{k=0}^{\infty}\frac 1 {2^{k}}=2$. It follows that $\|g^{-1}\| \leq 2\|f^{-1}\|$. Now by the identity in the comment above by Alex we get $\|g^{-1}-f^{-1}\| \leq \|g^{-1}\|\|f^{-1}\||\|g-f\|\leq 2\|f^{-1}\|^{2}|\|g-f\|$.