I am looking for a justification for the following claim that was made in this post, regarding a function of an operator $A$.
Considering the following differential equation (Cauchy vector problem) $$ \frac{dx(t)}{dt} = Ax(t),\;\;\; x(0)=x_0. $$ One ends up with a solution operator $x(t)=e^{tA}x_0$. Then one can form functions of $A$ using a calculus related to the Laplace transform: $$ \int_{0}^{\infty}F(t)e^{tA}x_0 dt $$ If the Laplace transform $\mathscr{L}\{F\}$ is $f$, then the above corresponds to $f(A)$.
The closest that I can get is the following:
Using the inverse Laplace transform, we have:
$f(t)=\frac{1}{2\pi i}\int_{\gamma-i\infty}^{\gamma+i\infty}F(s)e^{st}ds$
If I replace $t$ with $A$, I get $f(A)=\frac{1}{2\pi i}\int_{\gamma-i\infty}^{\gamma+i\infty}F(s)e^{sA}ds=\frac{1}{2\pi i}\int_{\gamma-i\infty}^{\gamma+i\infty}F(t)e^{tA}dt$.
For $A$ that is diagonalizable, that is, $A=D^{-1}\Lambda D$, where $\Lambda=\mathrm{diag}(\lambda_1,\dots,\lambda_n)$, one has $$e^{tA}=D^{-1}e^{t\Lambda} D=D^{-1}\mathrm{diag}(e^{t\lambda_1},\dots,e^{t\lambda_n}) D,$$ so plugging into your expression gives $$ D^{-1}\mathrm{diag}(\mathcal{L}\{F\}(\lambda_1),\dots,\mathcal{L}\{F\}(\lambda_n)) D=D^{-1}\mathrm{diag}(f(\lambda_1),\dots,f(\lambda_n))D. $$ which is $f(A),$ by definition. The diagonalizable operators are dense in the space of all operators, thus the general case follows by approximation.