We show how the spectral relation \(\partial_z \widehat{\varphi}(k, 0)=k \tanh (k h) \widehat{\varphi}(k, 0)\) implies the physical-space operator identity \(\varphi_z(\cdot, \cdot, 0)=(|D| \tanh (h|D|)) \varphi(\cdot, \cdot, 0)\), by inverse Fourier transform and the commutation of \(\partial_z\) with the horizontal Fourier transform.
Let \(\varphi(x, y, z)\) be a sufficiently regular function in \((x, y)\) for each fixed \(z\), and denote by \(\mathcal{F}_{x, y}\) the two-dimensional Fourier transform in the horizontal variables only. Write
\[ \widehat{\varphi}\left(k_x, k_y, z\right):=\mathcal{F}_{x, y}[\varphi(\cdot, \cdot, z)]\left(k_x, k_y\right), \quad k:=\sqrt{k_x^2+k_y^2} . \]
Fourier-multiplier definition of \(|D| \tanh (h|D|)\)
Define the pseudodifferential operator \(|D| \tanh (h|D|)\) on functions \(\psi(x, y)\) by its Fourier symbol \(k \tanh (k h)\) :
\[ (|D| \tanh (h|D|) \psi)(x, y):=\mathcal{F}_{x, y}^{-1}\left[k \tanh (k h) \hat{\psi}\left(k_x, k_y\right)\right](x, y) . \]
Equivalently,
\[ \mathcal{F}_{x, y}[(|D| \tanh (h|D|) \psi)]=k \tanh (k h) \widehat{\psi} . \]
Recovering \(\varphi_z(\cdot, \cdot, 0)\) from its Fourier transform
For any horizontal function \(g(x, y)\), one has the identity \(g=\mathcal{F}_{x, y}^{-1}[\widehat{g}]\). Applying this to \(g=\varphi_z(\cdot, \cdot, 0)\) yields
\[ \varphi_z(\cdot, \cdot, 0)=\mathcal{F}_{x, y}^{-1}\left[\widehat{\varphi_z}\left(k_x, k_y, 0\right)\right] . \]
Since \(\mathcal{F}_{x, y}\) acts only on \((x, y)\), it commutes with \(\partial_z\) (under standard regularity/decay assumptions), hence
\[ \widehat{\varphi_z}\left(k_x, k_y, 0\right)=\partial_z \widehat{\varphi}\left(k_x, k_y, 0\right) \]
and therefore
\[ \varphi_z(\cdot, \cdot, 0)=\mathcal{F}_{x, y}^{-1}\left[\partial_z \widehat{\varphi}\left(k_x, k_y, 0\right)\right] \]
Substitution of the spectral Dirichlet-Neumann relation
Assume the (previously derived) spectral Dirichlet-Neumann relation at the free surface,
\[ \partial_z \widehat{\varphi}\left(k_x, k_y, 0\right)=k \tanh (k h) \widehat{\varphi}\left(k_x, k_y, 0\right) . \]
Substituting into the preceding identity gives
\[ \varphi_z(\cdot, \cdot, 0)=\mathcal{F}_{x, y}^{-1}\left[k \tanh (k h) \widehat{\varphi}\left(k_x, k_y, 0\right)\right] . \]
Finally, set \(\psi(x, y):=\varphi(x, y, 0)\), so that \(\widehat{\psi}\left(k_x, k_y\right)=\widehat{\varphi}\left(k_x, k_y, 0\right)\). By the Fouriermultiplier definition in Step 1,
\[ \mathcal{F}_{x, y}^{-1}\left[k \tanh (k h) \hat{\varphi}\left(k_x, k_y, 0\right)\right]=(|D| \tanh (h|D|) \psi)(x, y) . \]
Hence the physical-space operator identity follows:
\[ \varphi_z(\cdot, \cdot, 0)=(|D| \tanh (h|D|)) \varphi(\cdot, \cdot, 0) . \]