I'm currently studying this theorem:
https://en.wikipedia.org/wiki/Spectral_theorem#Multiplication_operator_version
So the self Adjoint Operator A is unitary equivallent to a multiplication Operator.
My question is, is it possible to determine the function f (see wikipedia article)
So assuming we would know the spectrum of the multiplication Operator and we would know f, we could determine the spectrum of A
We cannot determine $f$ because $f$ is not determined by $A$. For example, the operator of multiplication by $f(t)=t$ on $L^2([0,1])$ is obviously equivalent to multiplication by $f(t)=t/2$ on $L^2([0,2])$.
Given $A$, can we find a $\mu$ and an $f$ that work? Probably at least theoretically, by looking at the proof of the Spectral Theorem.
But in typical applications we don't need to know exactly what $f$ is. For example:
A: By the Spectral Theorem, wlog $A$ is the operator of multiplication by $f$ on $L^2(\mu)$. Now $A$ positive definite implies $f\ge0$, while $||A||\le1$ implies that $||f||_\infty\le1$. So $0\le f\le 1$ almost everywhere. Since $1$ is not an eigenvalue we actually have $0\le f<1$ almost everywhere; now dominated convergence implies that $||f^kg||_2\to0$ for every $g\in L^2$.
Sure enough, the advantage of this version is that multiplication operators are easy to understand, as wikipedia claims. A possible flaw is that it's not canonical, but that doesn't matter in practice; it's clear how various properties of $A$ translate to properties of $f$.