 Solution

Use the Fourier transform conventionParseval's identity and $\nabla \mapsto i\mathbf{q}$ give the diagonal quadratic formIn 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 $q\to 0$, 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. 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.