Given an element $A$ of a Banach algebra $V$ and $\epsilon > 0$, prove that if $0 \in \textrm{sp}(A)$, then there is $\delta > 0$ such that if $B \in V$ commutes with $A$ and $||A-B|| < \delta$, then there is a $\lambda$ in $\textrm{sp}(B)$ with $|\lambda | < \epsilon$.
My Attempt
My goal was to prove that by making $\delta$ small enough, we can guarantee $r(B)$ (spectral radius of $B$) to be less than $\epsilon$, and then use the fact that the spectrum is non-empty to prove the existence of such a $\lambda$.
Since $A$ and $B$ commutes, then $B-A$ and $A$ commutes, so we have: $$r(B) = r(B-A+A) \leq r(B-A) + r(A) \leq ||B-A|| + r(A).$$ I am not sure how to proceed from here. I do not think this approach will work since I think the RHS of the above expression will not always be less than $\epsilon$ since $r(A)$ can be anything. However, I cannot think of any other approaches to take.
Could someone give a hint on how to approach this question?
Lemma. Given a unital Banach algebra $V$, and a commutative subalgebra $W$, there exists another commutative subalgebra $W_1$, containing $W$, such that for any element $a$ in $W_1$, one has that $a$ is invertible relative to $W_1$ iff $a$ is invertible relative to $V$. In particular, $$ \text{sp}_{V}(a)=\text{sp}_{W_1}(a), $$ for every $a$ in $W_1$.
Proof. For every subset $S\subseteq V$, define the commutant of $S$ by $$ S'=\{a\in V: as=sa: \text{ for all } s\in S\}. $$ It is easy to see that
(1) $S'$ is always a unital subalgebra,
(2) $S$ is commutative iff $S\subseteq S'$,
(3) if $S\subseteq T$ then $S'\supseteq T'$.
(4) if $a\in S'$, and $a$ is invertible, then $a^{-1}\in S'$.
Now, given $W$ as in the statement, we claim that $W_1:= W''$ (that is, the commutant of the commutant of $W$) satisfies all of the required conditions.
First observe that $W\subseteq W''$ by the following very trivial (if clumsy) reason: every element of $W$ commutes with everything that commutes with the elements of $W$.
Since $W$ is commutative, we deduce from (2) that $W\subseteq W'$. Using (3) we get $W'\supseteq W''$, and using (3) again we get $W''\subseteq W'''$. So the converse part of (2) implies that $W''$ is commutative.
Finally the last condition in the statement regarding invertible elements follows immediately from (4). QED
Back to the original question, consider the commutative Banach algebra $W$ generated by $A$ and $B$, and let $W_1$ be as in the Lemma. Then, for every element $a$ in $W_1$, we have that $$ \text{sp}_{V}(a)=\text{sp}_{W_1}(a) = \{\phi(a): \phi\in \text{Hom}(W_1, \mathbb C)\}, $$ so the end result follows easily by the continuity of complex homomorphisms.