quantum mechanics - Prove: $A$ and $B$ commute, therefore functions $f(A)$ and $g(B)$ will always commute with one another

How do I / can I actually prove the relationship

$[a,b]=0 \Rightarrow [f(a),g(b)]=0$ for all functions $f,g$.

I'm asking because the following sentence in the solution to my quantum mechanics homework irritates me:

For $i \neq j$ , the $\hat{n}_i$ commute with one another, and therefore functions of the $\hat{n}_i$ always commute with one another.

Where $\hat{n}_i = \hat{a}_i^\dagger \hat{a}_i $ with the Bose-Operators $\hat{a}_i^\dagger ,\hat{a}_i $. It is not my task to prove that relation, but the relation itself was required for being able to solve the exercise.

Answer

For normal elements in a C*-algebra you can do continuous functional calculus, that is, if $a$ is a normal operator, then $f(a)$ is well-defined for any $f\in C(\sigma(a))$. Since $\sigma(a)$ is always compact you can use Stone-Weierstrass to write $f$ as a uniform limit of polynomials in one complex variable and its complex conjugate. Hence you can verify what you need on polynomials. If $a$ and $b$ commute, then $a^2$ and $b^2$ commute and so on. Hence $f(a)$ and $g(b)$ commute for any $f\in C(\sigma(a))$ and $g\in C(\sigma(b))$. For von Neumann algebras one can push this argument to Borel functions.