\documentclass[fleqn,DIV=11,openright]{scrreprt} 
\usepackage[english]{babel}
\input{expkg}
\input{diffun2}

\begin{document}
\sheettitle{3}{20}{Integral operators and inequalities}\makesheettitle
\begin{ex} For $\phi\in L^{1}(-1,1)$ let
  $(F\phi)(\lambda):=\int\limits_{-1}^{1}e^{i\lambda x}\phi(x)\d x$.
\subex If $\phi\in C_{c}^{\infty}(-1,1)$, show that
$\abs{(F\phi)(\lambda)}\leq C_{k}\left(1+\abs{\lambda}\right)^{-k}\
\forall k\in \N$
\subex If $\phi$ is the restriction of a function in $C^{\infty}(\R)$,
show that
\begin{align*}
  \abs{(F\phi)(\lambda)-\frac{e^{i\lambda}}{i\lambda}\phi(1)+\frac{e^{-i\lambda}}{i\lambda}\phi(-1)}\leq
  C \lambda^{-2}
\end{align*}
\hint{Integration by parts!}
\rem{Related expansions of integrals with oscillating integrands will
  be discussed later in this course.}
\end{ex}
\begin{ex}
  Let $\Omega=\set{(x,y)\in\R^{2}:\Re e^{-ixy}>0}$. Show that the
  integral operator associated to the kernel
  $k(x,y)=e^{-ixy}\mathds{1}_{\Omega}(x,y)$ is a truncation of the
  Fourier transform, which is not bounded on $L^{2}(\R)$.
\hint{Show that
\begin{align*}
  \frac{\Re\langle T_{k}\mathds{1}_{I},\mathds{1}_{I}\rangle_{L^{2}(\R)}}{\langle\mathds{1}_{I},\mathds{1}_{I}\rangle_{L^{2}(\R)}}
\end{align*} is large for large intervals $I\subset\R$.
}
\end{ex}
\begin{ex}[Hardy's inequalities]
  \subex For $\map{w}{(0,\infty)}{(0,\infty)}$ measurable, we denote
  by $L^{p}_{w}$ the $L^{p}$-space associated to the weighted Lebesgue
  measure $\lambda_{w}$, $\lambda_{w}(A)=\int\limits_{A} w(x)\d x$ for
  any measurable $A\subset(0,\infty)$. I.e.
\begin{align*}
  L^{p}_{w}=\set{\map{f}{(0,\infty)}{\C}\text{ measurable:
    }\norm{f}_{p,w}:=\left(\int\limits_{0}^{\infty}\abs{f(x)}^{p}w(x)\d x\right)^{\frac{1}{p}}<\infty}.
  \end{align*}
The triangle inequality in this case says
\begin{align*}
  \norm{\int\limits_{0}^{\infty} f(y,\cdot)\d y}_{p,w}\leq
  \int\limits_{0}^{\infty}\norm{f(y,\cdot)}_{p,w}\d y.
\end{align*}
Let $p\geq 1,\ \alpha\in \R$ and $\map{K}{(0,\infty)\times
  (0,\infty)}{\C}$ a measurable function satisfying $K(\lambda x,\lambda
y)=\lambda^{-1}K(x,y)\ \text{for }\lambda>0$. Assume
\begin{align*}
  C:=\int\limits_{0}^{\infty}\abs{K(x,1)}x^{-\frac{1+\alpha}{p}}\d x <\infty.
\end{align*}
Show that the operator $T_{K}f(y)=\int\limits_{0}^{\infty}K(x,y)f(x)\d
x$ is a bounded operator on $L^{p}_{w}$ for
$w(x)=x^{\alpha}$ and that $\norm{T_{K}}_{L^{p}_{w}\to L^{p}_{w}}\leq
C.$
\hint{Write $T_{K}f(y)=\int\limits_{0}^{\infty}K(x,1)f(x y)\d x$ and
  use the triangle inequality.}
\subex Use a) to show that the "Hilbert integral" given by
$K(x,y)=\frac{1}{x+y}$ defines a continuous operator on
$L^{p}(0,\infty)$ for $p>1$.
\subex Use a) to deduce Hardy's inequalities
\begin{align*} 
  &\left(\int\limits_{0}^{\infty}\left(\int\limits_{0}^{x} f(y) \d
      y\right)^{p}x^{-r-1}\d x\right)^{\frac{1}{p}}
\leq \frac{p}{r}\left(\int\limits_{0}^{\infty}\left(yf(y)\right)^{p}y
  ^{-r-1}\d y\right)^{\frac{1}{p}},\\
 &\left(\int\limits_{0}^{\infty}\left(\int\limits_{x}^{\infty} f(y) \d
      y\right)^{p}x^{r-1}\d x\right)^{\frac{1}{p}}
\leq \frac{p}{r}\left(\int\limits_{x}^{\infty}\left(yf(y)\right)^{p}y
  ^{r-1}\d y\right)^{\frac{1}{p}}
  \end{align*}
where $f\geq 0,\ p\geq 1,\ r>0$.
\end{ex}
\begin{ex}["weak-type Schur test"]
  Slightly more sophisticated than Exercise 11 and Exercise 6 ist the
  weak-type Schur test (use Marcinkiewicz intead of Riesz-Thorin): If
  $\map{k}{X\times Y}{\C}$ is measurable and
  $\norm{k(x,y)}_{L^{q_0,\infty}(Y)}\leq B_{0}$, for almost every
  $x\in X$, $\norm{k(x,y)}_{L^{p_{1}',\infty}(X)}\leq B_{1}$, for
  almost every
  $y\in Y$, $p_{1},\ q_{0}\in (1,\infty)$
\begin{align*}
  \Rightarrow\quad& \forall \theta\in (0,1):\
  \map{T_{k}}{L^{p_{\theta}}(X)}{L^{q_{\theta}}(Y)}\text{ bounded
    for}\\
&\frac{1}{p_{\theta}}=1-\theta+\frac{\theta}{p_{1}},\
\frac{1}{q_{\theta}}=\frac{1-\theta}{q_{0}}\ \text{and}\
\norm{T_{k}}_{L^{p_{\theta}} \to L^{q_{\theta}}}\leq
C_{p_{1},q_{0},\theta} B_{0}^{(1-\theta)}B_{1}^{\theta}.
\end{align*}
Use this to show that $f\mapsto \abs{x}^{-\alpha}*f$ defines a bounded
operator $L^{p}(\Rn)\to L^{r}(\Rn)$ for $1<p,r<\infty,\ 0<\alpha<n$
and $\frac{1}{p}+\frac{\alpha}{n}=\frac{1}{r}+1$.
\end{ex}
\end{document}


