Let $n \in \mathbb{N}$, let $U \subset \mathbb{R}^n$ be a nonempty open convex set, let $m \in \mathbb{R}$, and let $t \in [0,1]$. For a symbol $a \in S^m(U \times \mathbb{R}^n)$, define the $t$-quantisation operator
for every $u \in C_c^\infty(U)$ and $x \in U$. The convexity of $U$ ensures that $(1-t)x+ty \in U$ for all $x,y \in U$. Here $D_x^\alpha := (1/i)^{|\alpha|}\partial_x^\alpha$ for every multi-index $\alpha \in \mathbb{N}_0^n$, and $\operatorname{Op}_L$ denotes the left quantisation $\operatorname{Op}_0$.
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Then there exists a left symbol $a_L \in S^m(U \times \mathbb{R}^n)$ such that $\operatorname{Op}_t(a)-\operatorname{Op}_L(a_L)$ is smoothing, and
in the sense of asymptotic expansions of symbols. In particular, for each fixed $a \in S^m(U \times \mathbb{R}^n)$ and any $s,t \in [0,1]$, the operators $\operatorname{Op}_t(a)$ and $\operatorname{Op}_s(a)$ agree modulo $\Psi^{m-1}(U)$, up to smoothing operators.