= Solution
Use the <Fourier transform> convention
$$
h(\mathbf x)=\int_{\mathbf q}h_{\mathbf q}e^{i\mathbf q\cdot\mathbf x},\qquad
\int_{\mathbf q}=\int\frac{d^Dq}{(2\pi)^D},\qquad h_{-\mathbf q}=h_{\mathbf q}^*.
$$
<Parseval's identity> and $\nabla\mapsto i\mathbf q$ give the diagonal quadratic form
$$
\boxed{\beta H_2=\frac{\beta\sigma}{2}\int_{\mathbf q}q^2h_{\mathbf q}h_{-\mathbf q}.}
$$
In a finite periodic box one can equivalently use orthonormal discrete <Fourier modes>, for which each independent real amplitude has a positive stiffness $\sigma q^2$. Fourier orthogonality eliminates all cross terms between different modes. The zero mode is removed as explained in part (a).
The stiffness vanishes as $q\to0$, so <capillary waves> at long wavelength have arbitrarily low energy. The <statistical Hamiltonian> is invariant under $h(\mathbf x)\mapsto h(\mathbf x)+a$ for any constant $a$. Selecting a particular mean membrane height breaks this continuous normal-translation <symmetry (physics)>. A varying local translation therefore has a <gradient> cost but no restoring mass: these are <Goldstone modes>. The diagonal static kernel is $q^2$; it should not be identified with a dynamical frequency without specifying the surrounding fluid or membrane inertia.
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