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Denote by $S^m$ the set of functions $p: \mathbb{R}^n \times \mathbb{R}^n \to \mathbb{C}$ such that $p \in C^{\infty}(\mathbb{R}^n \times \mathbb{R}^n)$ and $$\left| \frac{\partial^{\alpha+\beta}}{\partial \xi^{\alpha} \partial x^{\beta}} p(x,\xi) \right| \leq c_{\alpha,\beta} \cdot (1+|\xi|)^{m-|\alpha|} \tag{1}$$ for $\alpha,\beta \in \mathbb{N}_0^n$. $p$ is called the symbol of the pseudo-differential operator $$p(x,D)u(x) := (2\pi)^{-n} \cdot \int e^{\imath \, x \cdot \xi} \cdot p(x,\xi) \cdot \hat{u}(\xi) \, d\xi$$ where $u \in \mathcal{S}$ is a Schwartz function, $\hat{u}$ denotes the Fourier-transform of $u$.

Theorem For $p \in S^{m_1}, q \in S^{m_2}$ we have that the composition $(p(\cdot,D) \circ q(\cdot,D))$ is again a pseudo-differential operator, with symbol in $S^{m_1+m_2}$.

Using the definition of the composition of operators, one can show that $$\begin{align} t(x,\xi) &:=e^{-\imath \, x \cdot \xi} \cdot p(x,D)[q(x,\xi) \cdot e^{\imath \, x \cdot \xi}] \\ &=\int k(x,x-y) \cdot q(y,\xi) \cdot e^{\imath \, (y-x) \cdot \xi} \, dy \end{align}$$ is the symbol of the operator $(p(\cdot,D) \circ q(\cdot,D))$ where $k(x,\cdot)$ denotes the inverse Fourier-transform (in the sense of distributions) of the mapping $\xi \mapsto p(x,\xi)$. So, I have to show that $t$ satisfies $(1)$ for $m:=m_1+m_2$. But I don't see how to proceed...

Any hints would be appreciated.

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2 Answers 2

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I realized that the following theorem is quite helpful to prove the statement:

Theorem For all $N>0$ there exists $N_1 \in \mathbb{N}_0$ such that $T_{N_1} \in S^{-N}$,$$p(\cdot,D) \circ q(\cdot,D)- \sum_{|\alpha| \leq N_1} \tau_{\alpha}(\cdot,D) = T_{N_1}(\cdot,D)$$ where $$\tau_{\alpha}(x,\xi) := \frac{1}{\alpha!} \cdot \partial_\xi^{\alpha} p(x,\xi) \cdot D_x^{\alpha}q(x,\xi)$$

In particular, by choosing $N=\max\{|m_1+m_2|,1\}$, I obtain $$p(\cdot,D) \circ q(\cdot,D) = T_{N_1}(\cdot,D)+\sum_{|\alpha| \leq N_1} \tau_{\alpha}(\cdot,D)$$ where $T_{N_1} \in S^{-\max\{|m_1+m_2|,1\}} \subseteq S^{m_1+m_2}$ (note that $S^{r_1} \subseteq S^{r_2}$ for $r_1 \leq r_2$). Moreover, using the definition of $\tau_{\alpha}$, one can easily show that $\tau_{\alpha} \in S^{m_1+m_2-|\alpha|}$ for $\alpha \in \mathbb{N}_0^n$: $$\begin{align} \partial_\xi^{\gamma} \tau_{\alpha}(x,\xi) &= \frac{1}{\alpha!} \sum_{\gamma_1+\gamma_2=\gamma} c_{\gamma_1,\gamma_2} \partial_\xi^{\alpha+\gamma_1} p(x,\xi) \cdot D_x^{\alpha} \partial_\xi^{\gamma_2} q(x,\xi) \\ \Rightarrow |\partial_\xi^{\gamma} \tau_{\alpha}(x,\xi)| &\leq \frac{1}{\alpha!} \cdot \sum_{\gamma_1+\gamma_2=\gamma} c_{\gamma_1,\gamma_2} (1+|\xi|)^{m_1-|\alpha|-|\gamma_1|} \cdot (1+|\xi|)^{m_2-|\gamma_2|} \\ &\leq c \cdot (1+|\xi|)^{(m_1+m_2-|\alpha|)-|\gamma|} \end{align}$$ for all $\gamma \in \mathbb{N}_0^n$. Similar proof works for derivative with respect to $x$.

Thus $p(\cdot,D) \circ q(\cdot,D)$ has a symbol in $S^{m_1+m_2}$ since the operator can be represented as finite sum of operators with symbols in $S^{m_1+m_2}$.

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I did the calculation from the start and got (ignoring multiples of $2\pi$) $$t(x, \xi) = \int\int p(x, \eta)q(y, \xi)e^{i(y - x).(\xi - \eta)}dyd\eta$$ This form of $t$ seems to me easier to work with in this case. Now one can differentiate straightaway under the integral sign. The details of the calculation are extremely nasty, but one can profitably try out the first few steps. If one differentiates with respect to $x_1$, say, then, by Leibniz rule, $\partial_{x_1}p(x, \xi)$ will give an estimate from the fact that $p \in S^{m_1}$ and $p(x, \xi)$ will balance out the factor $(\xi_1 - \eta_1)$ coming from the differentiation of the exponential term (by appealing again to the fact that $p \in S^{m_1}$). Similarly you can do for the $\xi$.

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  • $\begingroup$ I don't see your point. $p \in S^{m_1}$ implies $|p(x,\eta)| \leq c \cdot (1+|\xi|)^{m_1}$ ... thus $$|p(x,\eta) \cdot \partial_1 e^{\ldots}| \leq |\xi_1-\eta_1| \cdot (1+|\xi|)^{m_1}$$ i.e. the factor $(\xi_1-\eta_1)$ does not vanish, as far as I can see. $\endgroup$
    – saz
    Apr 4, 2013 at 7:22

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