Past exam of the analysis course of the University of Oxford / 2011 / 3 / a / iii

If $T$ has a bounded inverse, it is onto and $\Vert x\Vert \le \Vert T^{-1}\Vert \Vert Tx\Vert$, giving the claimed lower bound. Conversely a positive lower bound makes $T$ injective. Together with surjectivity it gives an everywhere-defined linear inverse with norm at most $1/\epsilon$. Hence it is bounded, proving the equivalence.

If $R,S$ are individually invertible, their product has inverse $S^{-1}{R}^{-1}$. Conversely assume $RS=SR$ and put $A=RS$, with $A$ invertible. Both factors commute with $A$ and therefore with $A^{-1}$. Direct multiplication givesThese are bounded two-sided inverses, so an invertible commuting product has both factors invertible. The commutation assumption is used when verifying the second side of each inverse.