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\begin{document}


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\bigskip

Prof.\ Dr.\ H.\ Abels, Prof.\ Dr.\ B.\ Ammann \hfill SoSe 2012 \\
Dr.\ C.\ Neira Jimenez  \hfill 8.05.2012 \\
Fakult\"at f\"ur Mathematik\\
Universit{\"a}t Regensburg \\ 

\vskip 15pt
\begin{center}
 \Large{\bf Pseudo-differential operators and applications}\\  \vskip 5pt
 \large{\bf Exercises 3} 
\end{center}

%\vskip 15pt


\bigskip

\begin{enumerate}
\item {\bf (Elementary Composition)}\\
Let $p_1(x,\xi)=\sum_{|\alpha|\leq m_1} c_\alpha(x) \xi^\alpha$ be the
  symbol of a
  differential operator and let $p_2\in S^{m_2}_{1,0}(\Rn\times \Rn)$. In this
  special case it is easy to prove that the composition of the assoziated
  operators is a pseudodifferential operator. Moreover, it is an elementary
  calculation to determine the symbol of the composition. \\
 More precisely, prove  that
  \begin{eqnarray*}
    p_1(x,D_x)p_2(x,D_x)&=& (p_1\# p_2)(x,D_x),\qquad \text{where}\\
    p_1\# p_2(x,\xi)&=&  \sum_{|\beta|\leq m_1} \frac1{\beta!}\partial_\xi^\beta p_1(x,\xi) D_x^\beta p_2(x,\xi).
  \end{eqnarray*}
  In order to prove the statement you may use the identity 
  \[
  \binom{\alpha}{\beta}\xi^{\alpha- \beta} = \frac1{\beta!}\partial_\xi^\beta\xi^\alpha.
  \]
\item {\bf (Properties of Amplitudes) }\\
Let $a_j \in \mathcal{A}_{\tau_j}^{m_j}(\Rn\times \Rn)$, $j=1,2$, $\alpha,\beta \in\N_0^n$. Prove that:
\begin{enumerate} 
\item $a_1 \cdot a_2 \in \mathcal{A}_{\tau_1 + \tau_2}^{m_1 + m_2}(\Rn\times\Rn)$ and for every $k\in \N$ there is a constant $C_k>0$ independent of $a_1,a_2$ such that
  \begin{equation*}
    |a_1\cdot a_2|_{\mathcal{A}_{\tau_1 + \tau_2}^{m_1 + m_2},k}\leq C_k |a_1|_{\mathcal{A}_{\tau_1}^{m_1},k}|a_2|_{\mathcal{A}_{\tau_2}^{m_2},k}.
  \end{equation*}
\item $y^{\alpha} \cdot a_1 \in \mathcal{A}^m_{\tau_1 + |\alpha| }(\Rn\times\Rn)$, $\eta^{\alpha} \cdot a_1 \in \mathcal{A}^{m+|\alpha|}_{\tau_1 }(\Rn\times\Rn)$,
\item $\partial_y^{\alpha}\partial_\eta^\beta a_1 \in \mathcal{A}_{\tau_1}^{m_1}(\Rn\times\Rn)$.
\end{enumerate}


\item {\bf (Simple Properties of the Oscillatory Integrals)}
  \begin{enumerate}
  \item Let $A\in \R^{n\times n}$ such that $\det A\neq 0$ and let $a\in \mathcal{A}^m_\tau(\Rn\times\Rn)$, $m,\tau\in\R$. Prove that
\begin{equation*}
  \Os\iint e^{-iy\cdot A\eta}a(y,A\eta)|\det A|\, dxd\eta = \Os\iint e^{-iy\cdot \eta}a(y,\eta)\, dxd\eta.
\end{equation*}
\item
Let $a_1  \in \mathcal{A}^m_\tau(\Rn\times\Rn)$, $a_2  \in \mathcal{A}^m_\tau(\R^k\times\R^k)$, $m,\tau\in\R$, $n,k\in\N$ and let 
$$
a((y_1,y_2),(\eta_1,\eta_2)):= a_1(y_1,\eta_1)a_2(y_2,\eta_2)\quad \text{for all}\ (y_1,y_2),(\eta_1,\eta_2)\in \Rn\times \R^k.
$$
Prove that
\begin{eqnarray*}
  \lefteqn{\Os\iint_{(\Rn\times \R^k)^2} e^{-iy\cdot \eta} a(y,\eta)\, d y\, d\eta}\\
&=&\Os\iint_{(\Rn)^2} e^{-iy_1\cdot \eta_1} a_1(y_1,\eta_1)\, d y_1\, d\eta_1 \Os\iint_{(\R^k)^2} e^{-iy_2\cdot \eta_2} a_2(y_2,\eta_2)\, d y_2\, d\eta_2
\end{eqnarray*}
where $y=(y_1,y_2), \eta=(\eta_1,\eta_2)$.
 
  \end{enumerate}
\end{enumerate}


% \newpage
%  \begin{eqnarray*}
%    p_1(x,D_x)p_2(x,D_x)f &=& \sum_{|\alpha|\leq m_1} c_\alpha(x) D_x^\alpha \int e^{ix\cdot \xi}
%    p_2(x,\xi) \hat{f}(\xi)\dd \xi \\
%    &=& \sum_{|\alpha|\leq m_1} c_\alpha(x) \int  D_x^\alpha \left( e^{ix\cdot \xi}p_2(x,\xi)\right) \hat{f}(\xi)\dd \xi \\
%    &=& \sum_{|\alpha|\leq m_1} c_\alpha(x)\int e^{ix\cdot \xi}\left(\sum_{\beta\leq
%    \alpha}  \binom{\alpha}{\beta}\xi^{\alpha- \beta} D_x^\beta p_2(x,\xi)\right)
%    \hat{f}(\xi)\dd \xi \\
%    &=& \sum_{|\beta|\leq m_1} \int e^{ix\cdot \xi}\left(\sum_{\alpha\geq
%    \beta}  c_\alpha(x)\frac1{\beta!}\partial_\xi^\beta \xi^\alpha D_x^\beta p_2(x,\xi)\right)
%    \hat{f}(\xi)\dd \xi  \\
%    &=& \sum_{|\beta|\leq m_1} \frac1{\beta!} \int e^{ix\cdot
%    \xi}\partial_\xi^\beta \left(\sum_{|\alpha|\leq m_1}  c_\alpha(x)\xi^\alpha\right) D_x^\beta p_2(x,\xi)
%    \hat{f}(\xi)\dd \xi \\
%  &=& \sum_{|\beta|\leq m_1} \frac1{\beta!} \int e^{ix\cdot
%    \xi}\partial_\xi^\beta p_1(x,\xi) D_x^\beta p_2(x,\xi)
%    \hat{f}(\xi)\dd \xi
%  \end{eqnarray*}
%  where we have used
%  \[
%   \binom{\alpha}{\beta}\xi^{\alpha- \beta} = \frac1{\beta!}\partial_\xi^\beta\xi^\alpha.
%  \]
%  Hence $p_1(x,D_x)p_2(x,D_x)$ is again a pseudodifferential operator with symbol
%  \begin{equation}\label{eq:simpleProduct}
%    (p_1\# p_2)(x,\xi):= \sum_{|\beta|\leq m_1} \frac1{\beta!}\partial_\xi^\beta p_1(x,\xi) D_x^\beta p_2(x,\xi)
%  \end{equation}


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