= Solution
Use the <connected correlation function>
$$
G(r)=\langle\sigma_0\sigma_r\rangle-\langle\sigma_0\rangle\langle\sigma_r\rangle.
$$
Away from criticality its large-distance decay is exponential, $G(r)\sim r^{-p}e^{-r/\xi}$ with a model-dependent algebraic prefactor; the exponential rate defines the <correlation length> $\xi$. At criticality $\xi$ diverges and the decay becomes a power law. Subtracting the disconnected product is essential in the ordered phase: the unconnected <correlation function> tends to the square of the spontaneous <magnetization> rather than to zero.
For total <free energy> $F=-\beta^{-1}\ln Z$, at fixed temperature and a field-independent additive constant,
$$
M=-\frac1N\frac{\partial F}{\partial h},\qquad\chi=-\frac1N\frac{\partial^2F}{\partial h^2}.
$$
Writing $S=\sum_r\sigma_r$, differentiation of the <Boltzmann weight> gives $\partial_h\langle S\rangle=\beta(\langle S^2\rangle-\langle S\rangle^2)$. Consequently
$$
\chi=\frac\beta N\sum_{r,s}\left(\langle\sigma_r\sigma_s\rangle-\langle\sigma_r\rangle\langle\sigma_s\rangle\right).
$$
For a translation-invariant state, each separation occurs $N$ times, so \b[$\chi=\beta\sum_rG(r)$]. Without translation invariance, the double-sum formula remains the correct relation.
\b[Gaussian calculation and thinning.] Interpret the displayed local expression as a <Hamiltonian> density, to be integrated over space. For $\kappa>0$ and $m^2>0$, completing the square in Fourier modes gives the <connected correlation function>
$$
\widetilde G(k)=\frac1{\beta(\kappa^{-1}k^2+m^2)}=\frac\kappa{\beta(k^2+\kappa m^2)}.
$$
The uniform field changes the mean to $-h/m^2$ because of the source's $+h\phi$ convention, but does not change this covariance. The nearest pole has inverse length $\sqrt{\kappa m^2}$, equivalently the Green function solves $(-\nabla^2+\kappa m^2)G=\kappa\delta/\beta$. Thus
$$
\boxed{\xi=\frac1{\sqrt{\kappa m^2}}.}
$$
Here $m^2$ is the positive mass coefficient; if $m$ denotes its positive square root, $\xi=1/(\sqrt\kappa\,m)$.
For a <Gaussian thinning transformation>, split modes into $|k|<\Lambda/b$ and $\Lambda/b<|k|<\Lambda$. The modes are independent, so integrating out the shell produces only an additive <free-energy density> and leaves the low-mode quadratic action unchanged. Restore the cutoff by $x=b x'$ and $\phi(x)=b^{-(D-2)/2}\phi'(x')$. Substituting into the kinetic, mass and source terms gives
$$
\kappa'=\kappa,\qquad (m^2)'=b^2m^2,\qquad h'=b^{(D+2)/2}h.
$$
Hence the thermal and field scaling eigenvalues are $y_t=2$ and $y_h=(D+2)/2$. Applying the exponent relations derived in part (f) gives the requested formal <Gaussian critical exponents>
$$
\boxed{\alpha=2-\frac D2=\frac{4-D}{2},\qquad\beta_m=\frac{D-y_h}{y_t}=\frac{D-2}{4}.}
$$
The heat-capacity result can also be checked directly: the quadratic determinant contributes $f_s\sim(m^2)^{D/2}$, after removing analytic ultraviolet terms, and two derivatives with respect to $m^2\propto t$ give $|t|^{D/2-2}$. At $D=2$, the free-energy power is accompanied by a logarithm, $f_s\sim m^2\ln m^2$, whose second derivative still has index $\alpha=1$. At $D=4$ this becomes a logarithmic <heat capacity> singularity, with index $\alpha=0$, rather than a strictly constant singular term.
There is an important <ordered-phase obstruction in a Gaussian scalar model>. A purely quadratic potential is unbounded below for $m^2<0$, so the Hamiltonian as printed has no stable ordered phase and no literal spontaneous-magnetization exponent. The displayed $\beta_m$ is its Gaussian scaling index, obtained by assigning the field its Gaussian scaling dimension; it is not a construction of a stable ordered equilibrium. For $D<4$, a stabilizing quartic interaction is relevant and generally changes the actual critical fixed point. For $D\leq2$, massless Gaussian infrared fluctuations also obstruct interpreting the formal zero or negative index as an ordinary bounded-spin ordering law. These qualifications do not alter the requested thinning calculation, but are necessary for its physical interpretation.
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