In the context of the Hamiltonian mechanics, I am trying to demonstrate the following statement:
For any scalar function $f$, just as the dot product $\boldsymbol{q}·\boldsymbol{p}$, the Poisson Brackets with the components of the angular momentum vanishes: $ \left[L_{i}, f\right] = 0$
My attempt
For the scalar product $\boldsymbol{q}·\boldsymbol{p}$, making use of the expression of the angular momentum in terms of the Levi-Civita symbol, $L_i =\epsilon_{i r s} q_{r} p_{S}$, I can see that the statement is true:
$$ \left[L_{i}, q_{j} p_{j}\right]=\frac{\partial L_{i}}{\partial q_{k}} q_{j} \frac{\partial p_{j}}{\partial p_{k}}-\frac{\partial L_{i}}{\partial p_{k}} p_{j} \frac{\partial q_{j}}{\partial q_{k}}=q_{j} \frac{\partial L_{i}}{\partial q_{j}}-p_{j} \frac{\partial L_{i}}{\partial p_{j}} $$
$$ =q_{j} \frac{\partial \epsilon_{i r s} q_{r} p_{s}}{\partial q_{j}}-p_{j} \frac{\partial \epsilon_{i r s} q_{r} p_{s}}{\partial p_{j}}=\epsilon_{i j s} q_{j} p_{s}-\epsilon_{i r s} q_{r} p_{s} = 0 $$
However, I cannot find a way to prove this for the case of a general scalar function $f$. How could the statement be reasoned?