So, here is my next edited attempt to solve this problem. conditions for the existence of the inverse of spectral operator as a bounded linear operator
Let $X$ be a Hilbert space with a countable orthonormal base $[e_1, e_2, \cdots]. $ Suppose $Ax = \sum_{n=1}^{\infty} \alpha_n \langle x,e_n \rangle e_n$. Find the conditions under which $A^{-1}$ exists as a bounded linear operation.
First, $e_n$ is an eigenvector of $A$ correspoding the eigenvalue $\alpha_n$. Now we must have $A^{-1}(A(x)) = x, \forall x \in X.$ Picking $x= e_n$ leads to $A^{-1}(\alpha_n e_n)=e_n$. Since $A^{-1}$ is supposed to be linear, we get $A^{-1}(e_n) = \alpha_n^{-1}e_n$, which makes $e_n$ an eigenvector of $A^{-1}$ as well, correspoding to $\alpha_n^{-1}$.
Now, to bound the norm of $A^{-1}$, a user suggested the following in the original post:
$$ \|A\|=\sup\{|\alpha_n|:\ n\}. \quad (*) $$ So $A^{-1}$ will be bounded if $$ \sup\{|\alpha_n|^{-1}:\ n\}<\infty. \quad (**) $$
Now, I understand the procedure leading to (*) as we have a spectral form for $A$. But where does (**) come from?
So far, we have found the eigenvalues of $A^{-1}$. How does it bound the norm of $A^{-1}$ from above? I mean, as far as I know, in general, the norm of an operator is greater than or equal to the largest absolute value of its eigenvalues. So, at the end of the day, I guess, I must prove $||A^{-1}|| \leq \sup\{|\alpha_n|^{-1}:\ n\}$. How is it done? I'd appreciate some elaboration on this.
Edit: From the very definition of the norm operator, I can see that $$||A^{-1} || \geq |\alpha_n^{-1}| \Rightarrow ||A^{-1} || \geq \sup \{ |\alpha_n^{-1}| \} $$ But then bounding $|\alpha_n^{-1}| $ from above does not necessarily bound $||A^{-1} ||.$ So, I guess I need to show the reverse inequality. But how?
As a consequence of the "Bounded inverse theorem" it suffices to prove that $A$ is bounded and bijective.
Boundedness: $$||Ax||^2 = \sum_{n = 1}^\infty |\alpha_n|^2|\langle x,e_n \rangle|^2 \leq (\sup_n|\alpha_n|)^2 ||x||^2$$ So $A$ is bounded if $\{\alpha_n\}_n$ is bounded.
Injectivity: $$||Ax||^2 = 0 = \sum_{n = 1}^\infty |\alpha_n|^2|\langle x,e_n \rangle|^2$$ gives $|\alpha_n||\langle x,e_n \rangle| = 0$ for every $n$ and $x$ = 0 if $\alpha_n \neq 0$ for every $n$.
Surjectivity: let $y \in X$ then we want to define $x \in X$ such that $Ax = y$ as follow $$x := \sum_{i = 1}^\infty \frac{1}{\alpha_n} \langle y,e_n \rangle e_n$$ which make sense if $\{\frac{1}{\alpha_n}\}_n$ is bounded.
Summarizing you need $\{\alpha_n\}_n$ to be a bounded sequence that never vanish for which $0$ is not an accumulation point.