Let $A$ be a Banach algebra, $G(A)$ denotes the set of all invertible elements in $A$. Let $\mathbf 1$ be the identity element of $A$.
Suppose that $x\in\partial G(A)$. Is it true that for any $r>0$, there exists $\varepsilon\in\Bbb C$ such that $|\varepsilon|<r$ and $ x+\varepsilon\mathbf 1\in G(A) $?
I know that this is true for $A=M_n(\Bbb C)$ and $G(A)=GL(n,\Bbb C)$. The proof follows from the discreteness of the roots of characteristic polynomials. I can't think of a proof or a counter example for infinite dimensional cases.
Let $D\subset\mathbb{C}$ be the unit disk, let $A=L^\infty(D)$, and let $x$ be the identity function $x(z)=z$. Note that $x\in \partial G(A)$, since for any $\epsilon>0$, if you define $x_\epsilon(z)=z$ if $|z|>\epsilon$ and $x_\epsilon(z)=\epsilon$ if $|z|\leq \epsilon$, then $\|x-x_\epsilon\|=2\epsilon$ and $x_\epsilon\in G(A)$.
However, if $|\epsilon|<1$, then $x+\epsilon\mathbf{1}\not\in G(A)$, since an inverse to $x+\epsilon\mathbf{1}$ would have to be unbounded near $z=-\epsilon$.