= Solution
If $T$ has a bounded inverse, it is onto and $\|x\|\leq\|T^{-1}\|\,\|Tx\|$, 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 $1/\varepsilon$. 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 gives
$$
R(SA^{-1})=I=(SA^{-1})R,qquad
S(RA^{-1})=I=(RA^{-1})S.
$$
These are bounded two-sided inverses, so \b[an invertible commuting product has both factors invertible]. The commutation assumption is used when verifying the second side of each inverse.
Back to article page
