1-) Is the expression
$f(A) = \sum_n \frac{f'(0)}{n!}(A)^n$
always meaningful for any diagonalizable linear operator $A$ and for any analytic function $f$? This seems strange to me because then I could express any non-linear differential operator $f(A)$ in terms of the linear ones $A$ and then solve easily any non-linear differential equation.
2-) Is this problem related to the fact that differential operators are not bounded?
3-) What are the specific conditions that makes the above equation true? I would say that the expansion only makes sense when $f(A)$ is applied to the eigenvectors $v$ of $A$, which in turn would mean that
$f(A)v_n=f(\lambda_n)v_n$.
This seems to make sense since we can define $f(A)$ for diagonalizable $A$ as being the matrix/operator whose diagonal is $f(\lambda_n)$. Is it right?