$G \in L^1(\mathbb{R}^d)$ such that $\left|\mathcal{F}_2a(x,k)\right| \leq G(k)$ for a.e. $x,k \in \mathbb{R}^d$. Hence we find for a.e. $y \in \mathbb{R}^d$ $$\begin{aligned}
\left|(a(Q,P)f)(y)\right| &= \frac{1}{(2\pi)^{d/2}c^d} \left|\int_{\mathbb{R}^{d}}\mathcal{F}_2a\left(\frac{x+y}{2c}, \frac{x-y}{c}\right) \exp\left(\frac{1}{2}(\phi(y)-\phi(x))\right)f(x) dx\right| \\
&\leq \frac{1}{(2\pi)^{d/2}c^d} \int_{\text{supp}(f)}G\left(\frac{x-y}{c}\right) \exp\left(\frac{1}{2}(\phi(y)-\phi(x))\right)|f(x)| dx\end{aligned}$$ Since $f \in \mathcal{C}_\phi$, $\text{supp}(f)$ is compact and $||f||_{L^\infty(\lambda)}$ is bounded. Since $\phi$ is continuous, it will be bounded on $\text{supp}(f)$, by $C>0$ say. So $$\begin{aligned}
\left|(a(Q,P)f)(y)\right| &\leq \frac{1}{(2\pi)^{d/2}c^d} \exp(C/2)||f||_{L^\infty(\lambda)}\exp\left(\phi(y)/2\right)\int_{\text{supp}(f)}G\left(\frac{x-y}{c}\right)  dx \\
&\leq \frac{1}{(2\pi)^{d/2}c^d} \exp(C/2)||f||_{L^\infty(\lambda)}||G||_{L^1(\lambda)}\exp\left(\phi(y)/2\right).\\\end{aligned}$$ Since $\phi$ is continuous, $\left(y \mapsto \exp\left(\phi(y)/2\right) \right) \in L^1_{loc}(\lambda)$, and thus so is $a(Q,P)f$.

We now show our main theorem: that for $a \in HS(M)$ for the correct $M$, $a(Q,P)$ extends to a bounded operator on $L^p(\mu)$. The correct $M$ is as follows.

A pair $(M,\epsilon)$ consisting of a measurable function $M:\mathbb{R}^d \to \mathbb{R}^d$ and a number $\epsilon \geq 0$ is a called a valid growth pair for $\phi \in C^2(\mathbb{R}^d)$ and $p \in [1, \infty]$ if for all $x,y \in \mathbb{R}^d$, $$\left|\left( \left|\frac{1}{2}-\frac{1}{p}\right|\left|\phi(x)-\phi(y)\right| - \left|\left(\frac{x-y}{c}\right) M\left(\frac{x+y}{2c}\right)\right|\right)\right| \leq \epsilon.$$

\[thmBoundedness0\] Fix a potential $\phi \in C^2$, let $(Q, P)$ be the associated generalised Weyl pair, and fix $p \in [1,\infty]$. Suppose there exists a valid growth pair $(M,\epsilon)$ for $\phi$ and $p$. Then for $a \in HS_0(M)$ the operator 