 Solution

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.