Partial answer (the compact case for $n=3,p=2/3$ only): Let $M=\mathbb T = \mathbb R/(2\pi\mathbb Z)$ and let $\phi$ be a solution [Footnote 1] of the Klein-Gordon equation at the top of the question with $n=3, p=2/3$ such that $\phi(t_0, \cdot),\frac{\partial\phi}{\partial t}(t_0,\cdot)\in C^{4,\gamma}(\mathbb T^3)$ for some $\gamma>0$ and some $t_0>0$. (We will assume for simplicity that this condition is fulfilled for $t_0=1/27$, but one can check that the argument works no matter what $t_0$ is).
Note: Physically one has $\phi\in C^\infty$, as seen in [2a; Page 27], [2b], so the above conditions are pretty weak.
Note that the argumentation below is valid for all $t\ge 1$, for smaller $t$ one has to be a bit more careful about the estimates, for instance, then we have $t^{-2}\ge t^{-4/3}$ instead of $t^{-2}\le t^{-4/3}$.
Claim. The Fourier modes $c_{\mathbf k}(t) = \int_{\mathbb T^3} \phi(x,t)\exp(-i \langle x, \mathbf k\rangle)\,\mathrm dx$ satisfy $$\tilde d_{\mathbf k}(3 t^{1/3}) = t^{2/3} c_{\mathbf k}(t)$$ where $\tilde d_{\mathbf k}(\tau)$ satisfies $$\tilde d_{\mathbf k}''(\tau)+\left(\lvert\mathbf k\rvert^2-\frac2{\tau^2}\right)\tilde d_{\mathbf k}(\tau)=0$$ for all $\mathbf k\in\mathbb Z^3$ and $\tau>0$.
Proof. For $t\in]0,\infty[$ fixed, the function $x\mapsto \exp(-i \langle x, k\rangle)\phi(t,x)$ is Lebesgue integrable, since $$\phi(t, \cdot)\in C^2(\mathbb T^3)\subset L^\infty(\mathbb T^3)\subset L^1(\mathbb T^3).$$ Furthermore, by assumption, the partial derivative with respect to $t$ of $\exp(-i \langle x, k\rangle)\phi(t,x)$ always exists and it is $C^1$. And finally, since it is in the space $$C^1(]0,\infty[\times\mathbb T^3)\subset C^0(]0,\infty[\times\mathbb T^3),$$ the restriction of the integrand to $K\times\mathbb T^3$ for any compact interval $K\subset]0,\infty[$ is in $L^\infty(K\times\mathbb T^3)$. So it is valid to use the Leibniz rule (the analogous argument applied to $\frac{\partial\phi}{\partial t}\in C^1$ is left to the reader) in order to obtain
$$\dot c_{\mathbf k}(t) = \int_{\mathbb T^3}\exp(-i\langle x,\mathbf k\rangle)\frac{\partial\phi}{\partial t}(x,t)\,\mathrm dx$$ as well as $$\ddot c_{\mathbf k}(t) = \int_{\mathbb T^3}\exp(-
i\langle x,\mathbf k\rangle)\frac{\partial^2\phi}{\partial t^2}(x,t)\,\mathrm dx.$$
From integration by parts (the boundary terms vanish since we are on $\mathbb T^3$; cf. this) we get
$$\lvert\mathbf k\rvert^2 c_{\mathbf k}(t) = -\int_{\mathbb T^3} \exp(-i\langle x,\mathbf k\rangle)\Delta\phi(t,x)\,\mathrm dx.$$ (Where $\lvert\mathbf k\rvert$ is the usual Euclidean norm, i.e. for $\mathbf k = (\mathbf k_1,\mathbf k_2,\mathbf k_3)$, we have $\lvert\mathbf k\rvert^2 = \mathbf k_1^2+ \mathbf k_2^2+ \mathbf k_3^2$.)
Summing those up, since $\phi$ is a solution of the wave equation, we have $$\ddot c_{\mathbf k}(t)+\frac 2t \dot c_{\mathbf k}(t) +\lvert\mathbf k\rvert^2 t^{-4/3} c_{\mathbf k}(t)=0$$ for all $\mathbf k\in\mathbb Z^3$ and $t>0$.
Switching to $d_{\mathbf k}$ now is a straight-forward change of variables, details can be found in my other question. $\square$
Following my argument, this is enough to get (one can usually check that the following solutions are unique using usual uniqueness Theorems for ODEs) $$c_{\mathbf k}(t)=\frac{k \cos \left(k-3 k \sqrt[3]{t}\right) \left(3 A \left(k^2-1\right) \sqrt[3]{t}+A-3 B \sqrt[3]{t}+B\right)-\sin \left(k-3 k \sqrt[3]{t}\right) \left(A k^2 \left(3 \sqrt[3]{t}-1\right)+A+3 B k^2 \sqrt[3]{t}+B\right)}{3 k^3 t}$$ i.e.
$$\dot c_{\mathbf k}(t) = \frac{A \sin \left(k-3 k \sqrt[3]{t}\right)}{3 k^3 t^2}+\frac{A \cos \left(k-3 k \sqrt[3]{t}\right)}{k^2 t^{5/3}}-\frac{A \cos \left(k-3 k \sqrt[3]{t}\right)}{3 k^2 t^2}+\frac{A k \sin \left(k-3 k \sqrt[3]{t}\right)}{t^{4/3}}-\frac{A \sin \left(k-3 k \sqrt[3]{t}\right)}{k t^{4/3}}+\frac{A \sin \left(k-3 k \sqrt[3]{t}\right)}{k t^{5/3}}+\frac{A \cos \left(k-3 k \sqrt[3]{t}\right)}{t^{4/3}}-\frac{A \cos \left(k-3 k \sqrt[3]{t}\right)}{t^{5/3}}-\frac{A \sin \left(k-3 k \sqrt[3]{t}\right)}{3 k t^2}+\frac{B \sin \left(k-3 k \sqrt[3]{t}\right)}{3 k^3 t^2}+\frac{B \cos \left(k-3 k \sqrt[3]{t}\right)}{k^2 t^{5/3}}-\frac{B \cos \left(k-3 k \sqrt[3]{t}\right)}{3 k^2 t^2}-\frac{B \sin \left(k-3 k \sqrt[3]{t}\right)}{k t^{4/3}}+\frac{B \sin \left(k-3 k \sqrt[3]{t}\right)}{k t^{5/3}}+\frac{B \cos \left(k-3 k \sqrt[3]{t}\right)}{t^{4/3}}$$ where $$k\overset{\text{Def.}}=\lvert\mathbf k\rvert\neq 0$$ and $$A=A(\mathbf k)=\tilde d_{\mathbf k}(1)=c_{\mathbf k}(1/27), B=B(\mathbf k)=\tilde d_{\mathbf k}'(1) = \frac19\left(2 c_{\mathbf k}(1/27)+\frac19\dot c_{\mathbf k}(1/27)\right).$$
We also get
$$c_0(t) = \tilde d_{0}(3t^{1/3}) = 3(A+B)+\frac{1}{9t}(2A-B),$$ i.e.
$$\dot c_0(t) = \frac{B-2A}{9 t^2}$$ for
$A=A(0)=\tilde d_0(1)$ and $B=B(0)=\tilde d_0'(1)$.
We note that $\lvert \dot c_{\mathbf k}(t)\rvert\le C t^{-4/3} (\lvert A\rvert+ k \lvert A\rvert +\lvert B\rvert)$ for some real constant $C$ that is independent of $t$ and $\mathbf k$.
By [1; Theorem 3.3.9], we have $k A(\mathbf k), B(\mathbf k)\in O(k^{-3-\gamma})$ as $k\to\infty$.
So there exists a real constant $\tilde C$ independent of $\mathbf k, t$ such that $\lvert \dot c_{\mathbf k}(t)\rvert\le \tilde C t^{-4/3} k^{-3-\gamma}$ for all $k$ large enough. For all other $\mathbf k$, there exists a constant $D$ independent of $\mathbf k, t$ such that $\lvert\dot c_{\mathbf k}(t)\rvert\le D t^{-4/3}$.
Therefore, there exists a constant $\tilde D$ independent of $t$ and $\mathbf k$ such that for all $\mathbf k\neq 0$, \begin{equation}\tag{1}\label{bound-derivative} \lvert\dot c_{\mathbf k}(t)\rvert\le \tilde Dt^{-4/3} k^{-3-\gamma}.\end{equation}
From [3], \begin{equation}\label{main bound}\tag{*}\sum_{\mathbf k\in\mathbb Z^3\setminus\{0\}} \lvert k\rvert^{-3-\gamma}<\infty.\end{equation}
Integrating \eqref{bound-derivative} with respect to $t$ gives \begin{equation}\tag{2}\label{bound-c} \lvert c_{\mathbf k}(t)\rvert \le \lvert c_{\mathbf k}(1)\rvert +H k^{-3-\gamma}\end{equation} for all $\mathbf k\neq 0, t\ge 1$ and a constant $H$ independent of $\mathbf k, t$. Also, $$c_{0}(t) = C_1+C_2/t$$ for some constants $C_1, C_2$. Therefore, by \eqref{main bound} and [1; Proposition 3.2.5], we have the equality $$\phi(x, t) = \sum_{\mathbf k\in\mathbb Z^3} c_{\mathbf k}(t) \exp(i \langle x,\mathbf k\rangle).$$ From the Lebesgue dominated convergence Theorem applied to the counting measure (valid by the estimates from before), we get $$\frac{\partial\phi}{\partial t}(x, t) = \sum_{\mathbf k\in\mathbb Z^3} \dot c_{\mathbf k}(t) \exp(i \langle x,\mathbf k\rangle).$$
So from \eqref{bound-derivative} $$\left\lvert\frac{\partial\phi}{\partial t}(x, t)\right\rvert\le\sum_{\mathbf k\in\mathbb Z^3}\lvert\dot c_{\mathbf k}(t)\rvert\le F t^{-4/3}$$ for some real constant $F$ independent of $t$. Thus, $$\boxed{\left\lVert\frac{\partial\phi}{\partial t}(t, \cdot)\right\rVert_{L^\infty(\mathbb T^3)}\lesssim t^{-4/3},}$$ which is exactly what we wanted to prove.
Literature
- Loukas Grafakos, Classical Fourier Analysis, Third edition. (Springer).
- a. Rafael de la Madrid, The role of the rigged Hilbert space in Quantum
Mechanics. (Eur. J. Phys. 26 (2005) 287-312) https://arxiv.org/abs/quant-ph/0502053
b. Antoine, J.-P., Rigged Hilbert spaces in quantum field theory: a lesson drawn from charge operators (Helvetica Physica Acta 45 (1972)).
- Maximilian Janisch, Reference for convergence of $\sum_{\mathbf k\in\mathbb Z^3\setminus\{0\}} \lvert\mathbf k\rvert^{-p}$, URL (version: 2022-01-10): https://math.stackexchange.com/q/4353143
[Footnote 1]: Meaning that $\phi\in C^2$.
