This itself does not produce a solution of (1), but an additional hypothesis such as the Palais-Smale condition does provide such a solution.
We have to show that M itself is an algebra. [Or: M is itself an algebra.]
But H itself can equally well be a member of S.
Now f is independent of the choice of γ (although the integral itself is not).
Thus in G itself there can be no such equivalences with f(r)=2.