= Solution
The spin-independent term cancels in the signed sum. The linearized stationary equation becomes
$$
M=U\chi_0(T)M+O(M^3),\qquad
\chi_0(T)=\int d\epsilon\,\nu(\epsilon)[-\partial_\epsilon n_F(\epsilon)]>0.
$$
The even powers vanish because the equation is odd under $M\mapsto-M$. Equivalently, the mean-field action has curvature
$$
\mathcal S_{\rm eff}(M)=\mathcal S_{\rm eff}(0)+\frac{\beta N}{2}[U^{-1}-\chi_0(T)]M^2+O(M^4).
$$
Thus the unpolarized stationary point becomes unstable when
$$
\boxed{U\chi_0(T)>1,\qquad U_c\chi_0(T)=1.}
$$
At low temperature, the derivative of the <Fermi distribution> samples a narrow window around $\mu$, so $\chi_0\simeq\nu(\mu)$ for a smooth <density of states>. Measuring energy from the Fermi level gives \b[$U_c\nu(0)\simeq1$], the <Stoner criterion> in the spin-summed convention fixed above. This identifies local instability of the paramagnetic state; the detailed transition order also depends on higher terms in the action.
In the isotropic ordered phase of <itinerant ferromagnetism>, continuous spin-rotation symmetry leaves a degenerate direction of magnetization. Its transverse fluctuations produce one gapless, quadratically dispersing <mean-field itinerant ferromagnetic Goldstone mode>, $E(\mathbf q)\propto q^2$. The two broken spin generators form a canonical pair, as for the <type-B Goldstone boson> of the localized ferromagnet. The system also remains metallic, with exchange-split <Fermi surfaces> and low-energy spin-conserving particle-hole excitations. Longitudinal-amplitude fluctuations and the spin-flip continuum are distinct from the long-wavelength magnon; their energy and damping scales depend on the band structure. Spin anisotropy, if added, could gap the transverse mode, but it is absent from the stated Hamiltonian.
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