Proposition 3

Let A be a unital Banach algebra.

a) If $x\in A$ admits a left inverse (resp. right) $y$, every element $x'\in A$ such that $\|x'-x\|<\|y\|^{-1}$ admits a left inverse (resp. right).

b) The set of elements of A which are invertible (resp. left, resp. right) is open in A.

c) Let $(x_n)$ be a sequence of elements of A admitting left inverses (resp. right)$y_n$, and converging to an element $x\in A$. If the sequence $(y_n)$ is bounded, then $x$ is invertible on the left (resp. right).

It suffices to treat the case of left inverses; that of right inverses follows by considering the opposite algebra.

Let $x, y, x'\in A$ be such that $yx= 1$ and $\|x'-x\|<\|y\|^{-1}$. We have

therefore $yx'$ is invertible: there exists $z\in$ A such that $z(yx') = 1$. Thus the element $x'$ is left invertible with inverse $zy$. This proves assertion a) and assertion b) follows from it immediately.

Let $(x_n)$ be a sequence of elements of A admitting left inverses $y_n$, which converges to an element $x\in$ A, and such that the sequence $(y_n)$ is bounded. If $M\geqslant 1$ is a real number such that $\|y_n\|\leqslant M$ for all $n\geqslant 1$, then we have $\|x_n-x\|<M^{-1}\leqslant \|y_n\|^{-1}$ for $n$ sufficiently large, and therefore $x$ admits a left inverse according to a).