I'm stuck with the following problem:
Suppose $A$ is a self-adjoint operator and $\Vert B-z_0\Vert\leq r$. Show that $\sigma(A+B)\subset \sigma(A)+\overline{B_r(z_0)}$, where $B_r(z_0)$ is the ball of radius $r$ around $z_0$ and $\sigma(A)$ denotes the spectrum of $A$.
And there's a hint that suggests using a previously solved exercise. The exercise goes like this:
Suppose $A$ is closed and has bounded inverse. Let $B$ be bounded. Then $A+B$ has bounded inverse if $\Vert B\Vert <\Vert A^{-1}\Vert^{-1}$.
Letting $C=\sigma(A)+\overline{B_r(z_0)}$, I tried to show that $\mathbb{C}\setminus C\subset \rho(A+B)$, where $\rho(A+B)$ is the resolvent of $A+B$. By definition of $C$, this means that for $w\in \mathbb{C}\setminus C$ and every $z$ in the ball, the point $w-z$ is in the resolvent of $A$. If we want to show that $w\in \rho(A+B)$, then we have to show that $A+B-w$ has a bounded inverse. Writing $A+B-w=(A-(w-z))+(B-z)$ for $z$ in the ball, I would like to show that the condition of the hint is satisfied with $A-(w-z)$ in place of $A$ and $B-z$ in place of $B$.
Since $A$ is self-adjoint, $$ \Vert (A-(w-z))^{-1}\Vert^{-1}=\mathrm{dist}(w-z,\sigma(A)) $$ And I need only to find a $z$ such that $\Vert B-z\Vert < \mathrm{dist}(w-z,\sigma(A))$. But this is where I'm stuck. As the ball is compact, it would suffice to be able to make $\Vert B-z\Vert$ as small as we want, but so far the only conclusion that I can make is that $\Vert B-z\Vert \leq 2r$ (which is pretty obvious).
Any hint or help would be appreciated. Thanks in advance