= Solution
The constant term in the spin sum cancels. Linearizing the <saddle point> equation gives
$$
M=UM\chi_0+O(M^3),\qquad \chi_0=\int d\epsilon\,\nu(\epsilon)[-\partial_\epsilon n_F(\epsilon)].
$$
The <paramagnetic auxiliary-field curvature> is
$$
\frac{S_{\rm eff}(M)-S_{\rm eff}(0)}{\beta N}=\frac12(U^{-1}-\chi_0)M^2+O(M^4).
$$
Consequently the <Stoner criterion> for the loss of local paramagnetic stability is
$$
\boxed{U>U_c,\qquad U_c\chi_0=1,\qquad U_c\nu(\mu)\simeq1\quad(T\to0).}
$$
The printed $\nu(0)$ uses energy measured from the <Fermi level>. For a smooth <density of states>, $-n_F'$ becomes a narrow unit-area peak at $\mu$; a singular density of states or appreciable temperature requires the integral form. The curvature establishes local instability, not the order or location of every possible thermodynamic transition.
With exact spin-rotation symmetry, <itinerant ferromagnetism> selects a magnetization direction. The two broken transverse generators form \b[one quadratic <type-B Goldstone boson>], a long-wave <ferromagnetic magnon> with $E_q\sim Dq^2$. Longitudinal amplitude motion is not symmetry-protected to be gapless. In a metallic ordered state, particle-hole excitations remain near each surviving exchange-split <Fermi surface>; these are distinct from the collective magnon. The <itinerant spin-flip particle-hole continuum> starts at exchange scale $2|M|$ at $q=0$ in this simple mean-field spectrum, while its finite-$q$ boundaries differ. Spin anisotropy, spin-orbit coupling or an external field can gap the transverse mode; none is present in the stated model.
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