Is the Fourier multiplier $\mathcal F(G(-\hbar^2 \Delta)\psi)(p) = G(|p|^2)\hat \psi(p)$ justified for any real function $G$?

Question

I am confused about a claim asserted in the paper "Higher Order Schrodinger Equations" 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, 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 $H^s(\mathbb R^d)$.

Answer 1

To make sense of your Fourier multiplier, you need only to assume that $p\mapsto G(\vert p\vert^2)$ is a temperate distribution on $\mathbb R^d$. It is true whenever $G$ is a continuous function increasing at most polynomially at infinity: indeed if $\phi$ is a test function in the Schwartz space you may define $$ \langle G(\vert p\vert^2),\phi(p)\rangle_{\mathscr S'(\mathbf R^d), \mathscr S(\mathbf R^d)}=\int_{\mathbb R^d} G(\vert p\vert^2)\phi(p) dp. $$ Once you know that $G(\vert p\vert^2)$ is a temperate distribution on $\mathbb R^d$, you may define the product $G(\vert p\vert^2)\hat \psi(p)$ as a temperate distribution for any $\psi$ in the Schwartz space and thus take the inverse Fourier transform of $G(\vert p\vert^2)\hat \psi(p)$ as a temperate distribution.

Well it is probably too general (even the though the hypotheses on $G$ can be weakened), but my point is that you have to pay attention to the growth rate of $G$ at infinity. A slightly different point of view, a sort of continued abstract nonsense escape course, is to require that $p\mapsto G(\vert p\vert^2)$ belongs to the space $\mathscr O_M(\mathbb R^d)$, coined by Schwartz as the space of multiplier: a function $f$ belongs to that space when it is $C^\infty$ with all derivatives increasing at most polynomially at infinity. Then it is easy to show that $$ \mathscr S(\mathbb R^d)\ni\phi\mapsto f\phi\in \mathscr S(\mathbb R^d) $$ is a continuous mapping and thus by transposition $$ \mathscr S'(\mathbb R^d)\ni T\mapsto f T\in \mathscr S'(\mathbb R^d) $$ is a (weakly) continuous mapping so that $G(\vert p\vert^2) T(p)$ belongs to $ \mathscr S'(\mathbb R^d)$ for any $T\in \mathscr S'(\mathbb R^d)$; thus you can define the inverse Fourier transform of $G(\vert p\vert^2) \hat T(p)$ for any $T\in \mathscr S'(\mathbb R^d)$ .