\documentclass[fleqn,DIV=11,openright]{scrreprt} 
\usepackage[english]{babel}
\input{expkg}
\input{diffun2}
\newlength{\tlength}
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\begin{document}
\sheettitle{5}{21}{}\makesheettitle
\addtocounter{exnum}{-1}
\begin{ex}
Let $\phi_{1}(x):=c_{1}\left(1+\abs{x}^{2}\right)^{-\frac{n+1}{2}},\ 
\phi_{2}(x):=c_{2}e^{-\frac{\abs{x}^{2}}{4}}$ with $c_{1}, c_{2}$ such that
$\int\phi_{1/2}=1.$ As in Exercise 13, write $\phi_{t}(x):=t^{-n}\phi(\tfrac{x}{t}).$
Given $f\in L^{p}(\Rn)$, let $u_{1}(t,x):= f*(\phi_{1})_{t}(x)$ and
$u_{2}(t,x):= f*(\phi_{2})_{\sqrt{t}}(x)$. Then
$\partial^{2}_{t}u_{1}(t,x)+\Delta_{x} u_{1}(t,x)=0$ and $(\partial_{t}-\Delta)u_{2}(t,x)=0$ for
$t>0,\ x\in\Rn$, and $\lim\limits_{t\to 0^{+}}u_{1/2}(t,x)=f(x)$ for
almost every $x\in\Rn$.
  \hint{This is a continuation of Exercise 13 on Sheet 4.}
\end{ex}
\begin{ex}
  Let
  \begin{align*}
    S^{m}:=\left\{a\in C^{\infty}(\Rn\times\Rn):\
      \forall\alpha,\beta\in \N_{0}^{n}\exists C_{\alpha\beta}:\
      \abs{\partial^{\beta}_{x}\partial^{\alpha}_{\xi}a(x,\xi)}\leq
      C_{\alpha\beta}\left(1+\abs{\xi}\right)^{m-\abs{\alpha}} \right\}
    \end{align*}
    The operator
    $L_{\xi}:=\left(1+\abs{x}^{2}\right)^{-1}\!(1-\Delta_{\xi})$
    satisfies $L^{N}_{\xi}e^{ix\xi}=e^{ix\xi}\ \forall N.$
\subex Check that for $a\in S^{m}$
\begin{align*}
  \operatorname{op}(a)f(x):=&\int\d \xi a(x,\xi)\hat{f}(\xi)e^{ix\xi}\\
  =&\int\d \xi L^{N}_{\xi}\left[a(x,\xi)\hat{f}(\xi)\right]e^{ix\xi}
\end{align*}
defines an operator $\map{\operatorname{op(a)}}{S(\Rn)}{S(\Rn)}$. If $a$
is a polynomial in $\xi$, $\operatorname{op}(a)$ is a differential
operator.
\subex
Let $A := \sum\limits_{\alpha} a_{\alpha} \partial^{\alpha}$ be a differential operator on
$\Rn$. Show that $A f(x) = \sum\limits_{\alpha} a_{\alpha}(x) ((\partial)^{\alpha} \delta *
f)(x)$ and conclude that $A$ is a singular integral operator. 
What is its kernel? Find an explicit
 distribution $K$ on $\Rn \times \Rn$ such that $\langle g, A f
\rangle = \langle K(x,y), f(x) g(y) \rangle$ for all test functions
$f,g$.
\rem{Operators of the form $\operatorname{op}(a),\ a\in S^{m}$, are
  called "pseudodifferential operators". In DifFun 1, only
  $x$-independent symbols were considered.}
\end{ex}
\begin{ex}
  Let $A$ be a singular integral operator on $\Rn$ with kernel $k(x,y) > c
\abs{x-y}^{-n}$. Show that $A$ does not extend to a bounded operator on
$L^{2}(\Rn)$ and, in particular, is not a Calderon-Zygmund operator.

\hint{Let $Q=(-\tfrac{1}{4},\tfrac{ 1}{4})^{n}$, $Q_{y} = Q+y$, $S_{R} = \bigcup\limits_{y \in \mathds{Z}^{n},\abs{y}<R}
Q_y$ and $f = \mathds{1}_{S_{R}}$.
Note that both of $S_{R}$ and $\Rn\setminus S_{R}$ have a
volume of at least $c_{n} R^n$ and show that $\frac{\norm
{A f}_{2} }{\norm{f}_{2}}$
is unbounded as $R$ goes to infinity, using that $\norm{A f}_{L^{2}(\Rn)}\geq
\norm{A f}_{L^{2}(\Rn\backslash S_{R})}$.}
\end{ex}
\end{document}


