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The derivative hypothesis implies that $1/Q(\xi )$ is a symbol of order $-\delta N$ at high frequency. Choose cutoffs $\chi \prec \psi$ in $X$ and a high-frequency cutoff in $\xi$. The corresponding Fourier multiplier is a parametrix for $Q(D)$, and the symbol calculus, together with the product formula from part (b), gives the localized estimatefor some sufficiently negative $t$. The commutator terms contain derivatives $Q^{(\alpha )}$; the assumed factor $|\xi {|}^{-\delta |\alpha |}$ lowers their order and lets them be absorbed inductively. Therefore

If $Q(D)u$ is smooth, it belongs locally to $H^s$ for every $s$. Starting from the fact that every compactly supported distribution has some negative Sobolev order and repeatedly applying the gain $\delta N>0$ places $u$ in every local Sobolev space. The Sobolev embedding theorem then gives $u\in C^{\infty }(X)$. Thus $Q(D)$ is a hypoelliptic differential operator.

The assumed lower bound gives $|a|=|Q{|}^{-1}{\lesssim }_K⟨\lambda {⟩}^{-M}$ at large frequency. Differentiating $1/Q$ repeatedly expresses every derivative as a finite sum of products of derivatives of $Q$ divided by powers of $Q$. Since $D_{\lambda }^{\beta }Q$ has polynomial order at most $M-|\beta |$, induction and symbol calculus giveThe interpolation region $R\le |\lambda |\le R+1$ is compact in frequency and causes no problem. Hence

The symbol class $\mathrm{Sym}(X,{\mathbb{R}}^n;N)=S_{1,0}^N$ consists of $c\in C^{\infty }(X\times {\mathbb{R}}^n)$ such that, for every compact $K\subset X$ and all multi-indices $\alpha ,\beta$,The defining estimate givesThe Leibniz rule givesand the triangle inequality givesThese are the basic rules of symbol calculus.

TakeBy symbol calculus, $a_2\in S^{-M-1}=S^{N_Q-1}$. Its zeroth-order contribution is $Q{a}_2=b_1$ outside the compact transition region and therefore cancels the order $-1$ remainder. Every term incontains at least one $\lambda$ derivative of $Q$ and has order at most $(M-1)+(-M-1)=-2$. Absorbing these terms and the smoothing cutoff contribution into $b_2$ gives