Let $A$ be an unbounded operator on $L^2(\mathbb{R}^d)$ unitarily equivalent to a multiplication operator $UAU^*= B $. One thus has $$UAU^*f(x) = B(x)f(x)$$ for a suitable subset of $L^2(\mathbb{R}^d)$. Now, one may define functions of $A$ via the functional calculus. I am interested in continuous $f$. My question is the following: Can one conclude that $Uf(A)U^*$ is again a multiplication operator acting like
$$Uf(A)U^* g(x) = f(B(x))g(x) \text{?}$$ I think it's easy to show for polynomials. Unfortunately, I don't see how I can extend this to general (possibly unbounded over $\mathbb{R}^d$) continuous $f$, since the $L^2$ space is over $\mathbb{R}^d$ and I dont know how to do an approximation result here. Is approximation the wrong approach? I saw something about uniqueness of spectral measures on polynomials (Absolutely continuous spectrum invariant under unitary equivalence), but I don't understand the argument. Other questions I found were only dealing with bounded $f$.
Edit: I think one possibility is the following: One can restrict $g \in L^2$ with compact support. For such $g$ the equality also holds, especially for an arbitrary continuous $f$. Then, use the denseness of such functions to show equality in the Operatornorm. That's it. I will post a full argument tomorrow.