= Solution
Assume the pinning coefficient $t>0$. Here $t$ belongs to an actual energy, so its Boltzmann coefficient is $\beta t$. The quadratic kernel and <covariance> become
$$
H_2+V=\frac12\int_{\mathbf q}(\sigma q^2+t)|h_{\mathbf q}|^2,\qquad
\langle h_{\mathbf q}h_{\mathbf q'}\rangle=\frac{k_BT}{\sigma q^2+t}(2\pi)^D\delta^{(D)}(\mathbf q+\mathbf q').
$$
Therefore the <pinned capillary-wave correlation> is
$$
\boxed{B_t(r)=\frac{2k_BT}{\sigma}\int_{\mathbf q}\frac{1-\cos(\mathbf q\cdot\mathbf r)}{q^2+\xi^{-2}},\qquad \xi=\sqrt{\sigma/t}.}
$$
The normal-translation <symmetry (physics)> is now explicitly broken, and the formerly massless <Goldstone mode> has a restoring term. For $r\ll\xi$, but beyond the microscopic cutoff, the unpinned behavior is recovered. For $r\gg\xi$, the connected height <covariance> decays and $B_t$ approaches $2\langle h^2\rangle$, so roughness no longer grows without bound even for $d=2,3$.
In internal dimension one the exact continuum answer is
$$
B_t(r)=\frac{k_BT\xi}{\sigma}(1-e^{-|r|/\xi}).
$$
In internal dimension two the connected <covariance> at $r>0$ is $k_BT K_0(r/\xi)/(2\pi\sigma)$, where $K_0$ is a <modified Bessel function>; its zero-separation value needs the cutoff and the saturation value grows as $(k_BT/\pi\sigma)\log(\xi/a)$ for $\xi\gg a$. In internal dimension three the connected <covariance> is approximately $k_BT e^{-r/\xi}/(4\pi\sigma r)$ beyond the cutoff. Thus pinning replaces the long-distance massless tails by exponential decay. A negative $t$ would destabilize the quadratic <statistical Hamiltonian> rather than bind the membrane.
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