Resolvent set for $A: l^2(\mathbb{C}) \rightarrow l^2(\mathbb{C})$, $A(x_1,x_2,x_3,...) = (a_1x_1,a_2x_2,a_3x_3,...)$ where $a=(a_k)\in l^\infty$

Question

I wish to show that the resolvent set for the bounded linear operator $A: l^2(\mathbb{C}) \rightarrow l^2(\mathbb{C})$, $A(x_1,x_2,x_3,...) = (a_1x_1,a_2x_2,a_3x_3,...)$ where $a=(a_k)\in l^\infty(\mathbb{C})$ is given by $R(A) = \mathbb{C} \setminus \overline{\sigma_p(A)}$.

I understand that $\sigma_p(A)\subset\sigma(A)$, i.e. the point spectrum is contained in the spectrum of $A$, and that $\sigma(A)$ is closed, meaning that $\overline{\sigma_p(A)}\subset\sigma(A)$, and so $R(A)\subset\mathbb{C} \setminus \overline{\sigma_p(A)}$, but I have no idea how to prove the reverse containment.

I have the definitions of the resolvent set but am unsure as to how to prove that $A-\lambda I$ has a bounded linear inverse for each $\lambda \in \mathbb{C} \setminus \overline{\sigma_p(A)}$.

Answer 1

One approach is to find a formula for the inverse. For $\lambda$ outside the closure of the point-spectrum, define $B: \ell^2(\Bbb C) \to \ell^2(\Bbb C)$ by $$ B(x_1,x_2,x_3,\dots) = \left(\frac{x_{1}}{a_1 - \lambda}, \frac{x_{2}}{a_2 - \lambda}, \frac{x_{3}}{a_3 - \lambda},\dots\right) $$ It now suffices to do the following.

• Verify that the sequence $([a_k - \lambda]^{-1})_{k \in \Bbb N}$ is in $\ell^\infty$ as long as $\lambda \notin \overline{\sigma_p(A)}$ (which means that $B$ is a bounded linear operator just like $A$),
• Verify that $(A - \lambda I)B = B(A - \lambda I) = I$.