Let $X$ be a Hilbert space having a countable orthonormal base $[e_1, e_2, \cdots]$. Also, suppose $Ax = \sum _{n=1} ^{\infty} \alpha_n \langle x,e_n \rangle e_n $, where $[\alpha_n]$ is a real bounded sequence. Find the conditions under which $A^{-1}$ exists as a bounded linear operator
Now, first, obviously, $\alpha_n$s are the eigenvalues of $A$, and $\frac{1}{\alpha_n}$ would be the corresponding eigenvalue for $A^{-1}$. Therefore, one condition for the existence of $A^{-1}$ is that $\alpha_n \neq 0, \forall n.$ But what is the general way to approach this problem? And what about the boundedness of $A^{-1}$. I'd appreciate any help/hints. Thanks.
Assuming that $u_n=e_n$ (otherwise the $\alpha_n$ are not eigenvalues), you have $$ \|A\|=\sup\{|\alpha_n|:\ n\}. $$ So $A^{-1}$ will be bounded if $$ \sup\{|\alpha_n|^{-1}:\ n\}<\infty. $$