= Solution
Factor the perturbed operator on $D(A)$ as
$$
A+B=(I+BA^{-1})A.
$$
Since
$$
\|BA^{-1}\|\leq\|B\|\,\|A^{-1}\|<1,
$$
the <Neumann series> makes $I+BA^{-1}$ invertible. Therefore $0\notin\operatorname{Sp}(A+B)$ and
$$
(A+B)^{-1}=A^{-1}(I+BA^{-1})^{-1}.
$$
The geometric-series bound gives
$$
\boxed{
\|(A+B)^{-1}\|
\leq\frac{\|A^{-1}\|}
{1-\|B\|\,\|A^{-1}\|}}.
$$
Now choose a bounded open neighborhood $U$ of the isolated spectral component $X$ such that
$$
X\subset U\subset\{z:\operatorname{dist}(z,X)<\epsilon\},
\qquad
\partial U\subset\rho(A),
$$
and $\overline U$ meets no other component of the spectrum. Compactness of $\partial U$ gives
$$
M=\sup_{z\in\partial U}\|(A-zI)^{-1}\|<\infty.
$$
For all sufficiently large $n$, $\|B_n\|M<1$. Applying the first part to $A-zI$ shows uniformly that $\partial U\subset\rho(A+B_n)$.
The corresponding <Riesz projections> are
$$
P=\frac1{2\pi i}\int_{\partial U}(zI-A)^{-1}\,dz,
\qquad
P_n=\frac1{2\pi i}\int_{\partial U}(zI-A-B_n)^{-1}\,dz.
$$
The resolvent identity and the uniform Neumann bound imply $\|P_n-P\|\to0$. The projection $P$ is nonzero because $U$ contains the nonempty spectral component $X$. Projections at distance less than one have isomorphic ranges, so $P_n\ne0$ for large $n$. Therefore $A+B_n$ has spectrum inside $U$, and any such point $z$ satisfies $\operatorname{dist}(z,X)<\epsilon$. Thus
$$
\boxed{
\inf_{z\in\operatorname{Sp}(A+B_n)}
\operatorname{dist}(z,X)<\epsilon}
$$
for every sufficiently large $n$.
Back to article page
