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The Hamiltonian perturbation preserving one regular level is explicit:Linearity of the Hamiltonian vector field in the differential of a smooth map givesIt vanishes on ${\Sigma }_0$. At a point of ${\Sigma }_c$, its contraction with the symplectic form isSince the differentials are independent there, this vanishes exactly when $c=0$. Thus on every nonempty nonzero regular fibre the two Hamiltonian vector fields differ at every point, not merely as functions somewhere:Mutual Poisson commutation of the $G_j$ is unnecessary; the given first-integral property does not by itself make the perturbed flow preserve every common level, nor is that requested.

There is a genuine missing nonemptiness hypothesis in the literal statement. Independent differentials on an empty set is a vacuous condition, and any two vector-field restrictions to an empty set are equal. For example, on ${\mathbb{R}}^2$ let $r=1$, $G_1=H=e^q$, and $\rho =1$. Every nonempty ${\Sigma }_c$, $0<c<1$, is regular, while ${\Sigma }_c=\varnothing$ for $-1<c\le 0$. In particular, the restrictions at $c=-1/2$ can never be unequal for any $K$. Therefore the printed universal inequality requires either all the compared fibres to be nonempty, or the pointwise formulation displayed above. The explicit $K$ fully proves either intended qualification.