I am confused about a claim asserted in the paper ["Higher Order Schrodinger Equations"][1] published in IOP Science. The authors claim that a Fourier multiplier identity $$ \mathcal F(G(-\hbar^2 \Delta)\psi)(p) = G(|p|^2)\hat \psi(p) $$ is valid for any real function $G$. I was initially dubious but decided to believe them on account of the prestige I accredited to IOP Science. Upon asking about this claim in another [question][2], I was kindly informed that the identity only holds for linear $G$. The authors, however, also claim that an exact solution in terms of Fourier multipliers holds for the semirelativistic equation $G(y) = -\sqrt{c^2 y + m^2 c^4}$. They then use this solution later in the paper to prove a bound on the power series approximation $$G(y) = -mc^2 + \sum_{n=1}^N \frac{\alpha(n)}{m^{2n - 1} c^{2n-2}}(-y)^n$$ within its radius of convergence. What am I missing here? Is it really possible for them to treat ***any*** real valued function as a Fourier multiplier? If that were possible, then why would they be using a power series approximation in the first place? I quoted the passage for ease of reading. > **Solving a generalized Schrodinger equation:** Let $G:\mathbb R_+ \to \mathbb R$ be real-valued function, and consider the generalized Schrodinger equation. $$ i\hbar \frac{\partial \psi}{\partial t} + G(-\hbar^2 \Delta) \psi = 0 $$ (3.18) The quantity $G(-\hbar^2 \Delta) \psi$ is defined a s a Fourier multiplier: $$ \mathcal F(G(-\hbar^2 \Delta)\psi)(p) = G(|p|^2)\hat \psi(p) $$ The most important aspect is that we want to encompass the following three cases. >* Schrodinger equation: $G(y) = \frac{-1}{2m} y$. >* Semi-relativistic equation (3.3): $G(y) = -\sqrt{c^2 y + m^2 c^4}$ >* Higher order Schrodinger equation (3.9): $$G(y) = -mc^2 + \sum_{n=1}^N \frac{\alpha(n)}{m^{2n - 1} c^{2n-2}}(-y)^n$$ Applying the Fourier transform to (3.18) with respect to the space variable, we obtain formally a first order ordinary differential equation in time for $\hat \psi$, in view of Proposition (2.1): $$i\hbar \frac{\partial \hat \psi}{\partial t} + G(|p|^2) \hat \psi = 0$$ $$\hat \psi(0, p) = \hat \psi_0(p)$$ It is solved explicitly: $$\hat \psi(p,t) = \hat \psi_0 (p) e^{-it G(|p|^2)}$$ Since $G$ is real-valued, $|\hat \psi(p,t)| = |\hat \psi_0(p)|$, and we obtain, from Proposition 2.1. >**Proposition: 3.4** Let $s\in\mathbb R$ and $\psi_0\in H ^s(\mathbb R^d)$. Then the Cauchy problem (3.18) has a unique solution $\psi \in C(\mathbb R; H^s (\mathbb R^d))$, denoted by $$\psi(t) = e^{-itG(-\hbar^2\Delta)}\psi_0$$ It is given by (3.19). Moreover, we have $$\|\psi(\cdot, t)\|_{H^s} = \|\psi_0\|_{H^s}$$ for all $t \in \mathbb R$. In other words, the propagator $e^{-itG(-\hbar^2 \Delta)}$ is unitary on every Sobolev space $H^s(\mathbb R^d)$. [1]: https://hal.archives-ouvertes.fr/hal-00776157/document [2]: https://mathoverflow.net/questions/437500/find-f-mathbb-r3-to-mathbb-r-s-t-f-1-delta-f-f2-0