could anyone provide me a hint on how to prove the following: Let X be a Banach space and $d_H$ be the Hausdorff metric defined on the compact subsets of $\mathbb{C}$, $$ d_H(M,N):=\max ( \sup_{x \in M} \inf_{y \in N} |x-y|, \ \sup_{y \in N} \inf_{x \in M} |x-y| ) $$
Show that for $A,B \in L_b(X)$ (bounded operators on $X$) that commute, i.e., $AB = BA$ the following holds: $d_H ( \sigma(A), \sigma(B) ) \le r(A-B) $, where $r(A-B)$ is the spectral radius of $A-B$
Since you are looking for just a hint:
Hint 1:
Suppose $A,B \in L_b(X)$ with $A$ invertible and such that $A,B$ commute. Then $A + B$ is also invertible, if $r(B) < 1/ r(A^{-1})$.
Hint to Hint 1:
Geometric series! Like here, but using the spectral radius instead of the crude norm estimate for convergence. At this point you may want to invoke the spectral radius formula (depending on the definition of spectral radius you know).
Full Solution for Hint 1:
It is sufficient to show that $A^{-1}(A+B)= I + A^{-1}B$ is invertible. Now $I+ A^{-1}B = I - (-A^{-1}B)$ is invertible if $r(-A^{-1}B)=r(A^{-1}B)<1$, since the radius of convergence of the geometric series is $1$. By definition (or the spectral radius formula depending on your definition of the spectral radius) $$r(A^{-1}B) = \limsup_{n \to \infty} \|(A^{-1}B)^n \|^{1/n}.$$ Now $A^{-1}$ commutes with $B$ since $A$ does. Therefore it is true that $(A^{-1}B)^n = A^{-n} B^n$ and hence $r(A^{-1}B) \leq r(A^{-1}) r(B)$, because the operatornorm is submultiplicative. Now if $r(B) < 1 / r(A^{-1})$, then the above implies $r(A^{-1}B) <1$ and so $A+B$ is invertible.
Hint 2:
Now apply the result to $z I - A$ and $A-B$ and then to $z I -B$ and $B-A$ for $z$ in the respective resolvent set. At this point you may want to invoke the spectral radius formula (depending on the definition of spectral radius you know).
Hint 3: (full solution)
Let $z \in \rho(A)$, where $\rho(A)$ is the resolvent set of $A$. Then $zI - A$ commutes with $B-A$, since $A$ does. By hint 1, $$z I - B= zI - A + (A-B)$$ is invertible if $r(A-B) < 1 / r(R(z))$ where $R(z)$ is the resolvent of $A$ at $z$. Now $$1/r(R(z)) = ( \sup_{\lambda \in \sigma ( R(z)) }| \lambda |)^{-1} =( \sup_{\lambda \in \sigma (A) }| \lambda - z |^{-1})^{-1} = \inf_{\lambda \in \sigma (A) } | \lambda - z|.$$ Therefore $zI-B$ is invertible if $r(A-B) < \inf_{\lambda \in \sigma (A) } | \lambda - z|$. In other words: If $z \in \sigma(B) \cap \rho (A)$, then $$\inf_{\lambda \in \sigma (A) } | \lambda - z| \leq r(A-B).$$ If $z \in \sigma(B) \cap \sigma (A)$ then $\inf_{\lambda \in \sigma (A) } | \lambda - z| =0 \leq r(A-B)$ since the spectral radius is allways $\geq 0$. In total $$ \sup_{z \in \sigma(B) }\inf_{\lambda \in \sigma (A) } | \lambda - z|\leq r(A-B).$$ By swapping $A$ and $B$ in the above we also receive $$ \sup_{z \in \sigma(A) }\inf_{\lambda \in \sigma (B) } | \lambda - z| \leq r(B-A) = r (A-B)$$ and hence $d_H (\sigma(A), \sigma(B) ) \leq r(A-B)$.
