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Suppose $b$ were a left inverse of an algebra element $x$, so $bx=1$. Then $b{x}_n\to 1$, and $b{x}_n$ is invertible for all sufficiently large $n$ by the Neumann series. For such $n$, $b=(b{x}_n)x_n^{-1}$ is a product of invertibles, hence invertible. The equation $bx=1$ then gives $x=b^{-1}$, a contradiction.

If instead $xb=1$, then $x_{n}b\to 1$, so $x_{n}b$ is eventually invertible. Now $b=x_n^{-1}(x_{n}b)$ is invertible, and again $x=b^{-1}$, a contradiction. This proves that a noninvertible limit of invertibles has neither type of one-sided inverse. No finite-dimensional assumption is used.

In a unital Banach algebra, the Neumann series gives $(1-u{)}^{-1}={\sum }_{k=0}^{\infty }{u}^k$ whenever $\Vert u\Vert <1$: the series converges in norm by submultiplicativity and completeness, and multiplication of its partial sums on either side gives $1-u^{N+1}\to 1$. Consequently, if $g$ is invertible and $\Vert g^{-1}\Vert \Vert h-g\Vert <1$, writeThe second factor has a Neumann series inverse, so $h$ is invertible. This exhibits a ball around each $g$ in the group of invertible elements of a Banach algebra, proving that group is an open set.

The same series proves continuity of inversion in a Banach algebra, with the useful quantitative estimateThe right side tends to zero as $h\to g$. Since applying inversion twice gives the original element, it is a bijection whose inverse is the same continuous map. Thus inversion is a homeomorphism of the open group of invertibles onto itself.

For the final unheaded request, defineIt is open by the preceding argument and contains $0$. If ${\lambda }_0$ were a boundary point, a sequence ${\lambda }_n\in E$ would converge to ${\lambda }_0$, while openness gives ${\lambda }_0\notin E$. Then $1-{\lambda }_{n}a\to 1-{\lambda }_{0}a$ is a noninvertible limit of invertibles. The one-sided-inverse argument in part (ii) shows that this limit has neither a left inverse of an algebra element nor a right inverse of an algebra element, contradicting the hypothesis. Therefore $E$ has empty boundary; it is both open and closed in the connected complex plane, and $E=\mathbb{C}$.

For every nonzero $\mu$, put $\lambda ={\mu }^{-1}$. The identity $\mu 1-a=\mu (1-\lambda a)$ proves invertibility, so the spectrum of an element $a$ is contained in $\{0\}$. It is nonempty: if the resolvent of an element existed for all $\mu \in \mathbb{C}$, it would be an entire algebra-valued function tending to zero at infinity by the Neumann series. Applying any bounded linear functional and then the Liouville theorem would make it zero; the Hahn-Banach theorem separates points and would force the resolvent itself to be zero, contradicting its inverse equation. Hence

If $x$ is invertible, part (i)'s inverse calculation gives an inverse for $x^{\ast }$, and therefore both $x{x}^{\ast }$ and $x^{\ast }x$ are invertible. Conversely, if both products are invertible, setThen $xr=1$ and $\ell x=1$: $r$ is a right inverse of an algebra element $x$ and $\ell$ is a left inverse of an algebra element $x$. They coincide because $\ell =\ell (xr)=(\ell x)r=r$, providing a two-sided inverse. ThusBoth products are required in a general Banach algebra; a single one-sided inverse need not suffice.

A right inverse of an element $a$ is an element $b$ with $ab=1$. A left inverse of an algebra element and a right inverse of an algebra element, when both exist, coincide. For the unilateral shift operator $S$, $S^{\ast }S=1$ but $S{S}^{\ast }\ne 1$, exhibiting the distinction in the Banach algebra of bounded operators.