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\title{Distributions in One Variable}
\author{V.A.Malyshev}
\date{}

\newcommand{\Fact}{\mbox{Fact}}
\newcommand{\Int}{\mbox{Int}}
\newcommand{\Cl}{\mbox{Cl}}
\newcommand{\Fr}{\mbox{Fr}}
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\begin{document}
\maketitle


\begin{center}
ø{\bf Contents}
\end{center}

\noindent {Introduction}

\medskip\noindent {$\;\;1$. Test functions}

\noindent {$\;\;2$. Distributions}

\noindent {$\;\;3$. Distributions with compact supports}

\noindent {$\;\;4$. Positive distributions}

\noindent {$\;\;5$. The derivative}

\noindent {$\;\;6$. The integral}

\noindent {$\;\;7$. The convolution of functions}

\noindent {$\;\;8$. The convolution of functions and distributions }

\noindent {$\;\;9$. The convolution of distributions }

\noindent {10. Differential operators}

\noindent {11. Sobolev spaces}

\noindent {12. The Green function}

\medskip\noindent {Bibliography}


%%%%%%%%%%%%%%%%%%%%%%%
\section*{Introduction}

The little book is a self-contained introduction to the theory of distributions. All
sections are suitable for students who have had one or two years of calculus. The manual
is based on the book \cite{Gel}. Necessary facts about measurable and absolutely
continuous functions may be taken from the book \cite{Nat}. In three last sections we
need standard facts about linear differential equations with constant coefficients.
\bigskip
\bigskip
\bigskip

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\newpage


%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section*{1. Test functions}

Let $\varphi$ be a real function. The set
$$
\supp\,\varphi = \Cl\{x\in \mathbb{R}:\; \varphi(x)\ne 0\}
$$
is called the support. Here Cl is the closure in $\mathbb{R}$. The function $\varphi$ is
called finite if the suppurt $\supp\,\varphi$ is compact. The space of all finite
$\varphi\in C^\infty(\mathbb{R})$ is denoted by $C_0^\infty(\mathbb{R})$. Analogously the
space of all finite $\varphi\in C(\mathbb{R})$ is denoted by $C_0(\mathbb{R})$. Functions
$\varphi\in C^\infty_{0}(\mathbb{R})$ are called test functions.



Let $\Omega=(a,b)$ be a open interval. Usually we assume that $-\infty<a<b<\infty$.
Consider the space
$$
C_0^\infty(\Omega) = \{\varphi\in C_0^\infty(\mathbb{R}):\; \supp\varphi\subset\Omega \}.
$$
We say that $\varphi_1,\varphi_2,\ldots \in C_0^\infty(\Omega)$ converges to $\varphi\in
C_0^\infty(\Omega)$ if there exists $[\alpha,\beta]\subset\Omega$ such that
$
\supp\varphi_k\subset [\alpha,\beta]
$
for all $k=1,2,\ldots$ and
$$
\lim_{k\to\infty}\|\varphi^{(i)}-\varphi_k^{(i)}\|_{C[\alpha,\beta]}=0
$$
for all $i=0,1,\ldots\,$.

Let us construct a nonzero test function. Put
$$
\lambda=\int\limits_{-1}^1\exp\left(\frac{1}{x^2-1}\right)\,dx
$$
and define
$$
h(x)=\left\{\begin{array}{lll}
 {\displaystyle\frac{1}{\lambda}\exp\left(\frac{1}{x^2-1}\right)} &{\rm for} &|x|<1\medskip\\
 0 &{\rm for} &|x|\ge1\end{array}\right.
$$
By direct calculation, we see that $h\in C^\infty(\mathbb{R})$ and
$
\supp\,h = [-1,1].
$
Hence $h\in C^\infty_{0}(\mathbb{R})$. Obviously
$$
\int\limits_{-1}^1h(x)\,dx=1.
$$
The test function $h$ is called the hat.

Let $\varepsilon>0$. The test function
$$
h_\varepsilon(x)=\frac{1}{\varepsilon}h\left(\frac{x}{\varepsilon}\right)
$$
is called the $\varepsilon$-hat. We see that
$
\supp\,h_\varepsilon = [-\varepsilon,\varepsilon]
$
and
$$
\int\limits_{-\varepsilon}^\varepsilon h_\varepsilon(x)\,dx=1.
$$

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Let $\alpha<\beta$. Then
$$
H^{\alpha,\beta}_{\varepsilon}(x)=\int\limits_{\alpha}^{\beta}h_{\varepsilon}(x-y)\,dy
$$
is a test function. Obviously,
$$
H^{\alpha,\beta}_\varepsilon(x)=\left\{\begin{array}{lll}
 1 &{\rm for} &x\in[\alpha+\varepsilon,\beta-\varepsilon] \medskip\\
 0 &{\rm for} &x\notin[\alpha-\varepsilon,\beta+\varepsilon]
\end{array}\right.
$$
This means that the test function $H^{\alpha,\beta}_{\varepsilon}$ has the form
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If $[\alpha,\beta]\subset\Omega$, then $H^{\alpha,\beta}_{\varepsilon}\in
C_0^\infty(\Omega)$. In particular, the space $C_0^\infty(\Omega)$ has nonzero elements.

Consider the space
$$
C_0(\Omega) = \{\varphi\in C_0(\mathbb{R}):\; \supp\varphi\subset\Omega \}.
$$
We say that $\varphi_1,\varphi_2,\ldots\in C_0(\Omega)$ converges to $\varphi\in
C_0(\Omega)$ if there exists $[\alpha,\beta]\subset\Omega$ such that
$
\supp\varphi_k\subset [\alpha,\beta]
$
for all $k=1,2,\ldots$ and
$$
\lim_{k\to\infty}\|\varphi-\varphi_k\|_{C[\alpha,\beta]}=0.
$$
Let $*$ be the convolution
$$
f*g(x)=\int\limits_{-\infty}^{\infty}f(x-y)g(y)\,dy.
$$
If $\varphi\in C_0(\Omega)$, then $h_\varepsilon*\varphi\in C_0^\infty(\Omega)$ for all
sufficiently small $\varepsilon>0$.

\begin{theorem}
\label{convden} If $\varphi\in C_0(\Omega)$, then
$$
\varphi=\lim_{\varepsilon\to0}\varphi*h_\varepsilon
$$
in the space $C_0(\Omega)$.
\end{theorem}
{\bf Proof.} There exists $[\alpha,\beta]\subset\Omega$ such that
$\supp\,\varphi\subset[\alpha,\beta]$ and
$\supp(\varphi*h_\varepsilon)\subset[\alpha,\beta]$ for all sufficiently small
$\varepsilon$. We see
$$
\mid\varphi(x)-(\varphi*h_\varepsilon)(x)\mid
=\left|\int\limits_{-\infty}^{\infty}[\varphi(x)-\varphi(x-y)]h_\varepsilon(y)\,dy\right|\le
 \sup_{x\in{\rm R}}\sup_{\mid y\mid\le\varepsilon} \mid\varphi(x)-\varphi(x-y)\mid
$$
Since $\varphi$ is uniformly continuous, we have
$$
\lim_{\varepsilon\to 0}\|\varphi-\varphi*h_\varepsilon\|_{C[\alpha,\beta]}=0.
$$
As was to be proved. $\bullet$
\bigskip

\noindent {\bf Corollary.}  The space $C_0^\infty(\Omega)$ is dence in the space
$C_0(\Omega)$.




%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section*{2. Distributions}

A linear continuous functional on the space $C_0^\infty(\Omega)$ is called  a
distribution. The space of all distributions is denoted by ${\cal D}'(\Omega)$. The value
of the distribution $\Lambda\in {\cal D}'(\Omega)$ at the test function $\varphi\in
C_0^\infty(\Omega)$ is denoted by $(\Lambda,\varphi)$. The continuity of $\Lambda\in
{\cal D}'(\Omega)$ means that
$$
(\Lambda,\varphi)=\lim_{k\to\infty}(\Lambda,\varphi_k)
$$
whenever
$$
\varphi=\lim_{k\to\infty}\varphi_k.
$$
In other words $\Lambda$ is sequentially continuous.

We say that $\Lambda_1,\Lambda_2,\ldots\in {\cal D}'(\Omega)$ converges to $\Lambda\in
{\cal D}'(\Omega)$ if
$$
(\Lambda,\varphi)=\lim_{k\to\infty}(\Lambda_k,\varphi)
$$
for any $\varphi\in C_0^\infty(\Omega)$.

Let $f$ be a measurable function on the interval $\Omega$. The function $f$ is called
locally integrable if
$$
\int\limits_\alpha^\beta|f(x)|\,dx<\infty
$$
for any $[\alpha,\beta]\subset\Omega$. Let $L_1(\Omega)^{\rm loc}$ be the space of all
locally integrable functions. We say that $f_1,f_2,\ldots\in L_1(\Omega)^{\rm loc}$
converges to $f\in L_1(\Omega)^{\rm loc}$ if
$$
\lim_{k\to\infty}\|f-f_k\|_{L_1[\alpha,\beta]}=0
$$
for any $[\alpha,\beta]\subset\Omega$.

Let $f\in L_1(\Omega)^{\rm loc}$. Define $\Lambda_f\in {\cal D}'(\Omega)$ by the rule
$$
(\Lambda_f,\varphi)=\int\limits_\Omega f(x)\varphi(x)\,dx,
$$
where $\varphi\in C_0^\infty(\Omega)$. A distribution $\Lambda\in {\cal D}'(\Omega)$ is
called regular if $\Lambda = \Lambda_f$ for some $f\in L_1(\Omega)^{\rm loc}$.

\begin{theorem}
\label{l1indis} Let $f\in L_1(\Omega)^{\rm loc}$. If $f\ne 0$, then $\Lambda_{f}\ne 0$.
\end{theorem}
{\bf Proof.} Assume that $f\ne 0$. Put
\begin{eqnarray*}
S^+_\delta&=&\{x\in\Omega:\;f(x)\ge+\delta\},\medskip\\
S^-_\delta&=&\{x\in\Omega:\;f(x)\le-\delta\}.
\end{eqnarray*}
We see that $\mes\{S^+_\delta\}>0$ or $\mes\{S^-_\delta\}>0$ for some $\delta>0$. Suppose
that
$$
\mes\{S^+_\delta\}=2\mu>0.
$$
Then there exists a compact set $A\subset S^+_\delta$ such that
$
\mes\{A\}\ge\mu.
$
The compact set $A$ is representable in the form
$$
A=\bigcap_{n=1}^{\infty}U^n,
$$
where open sets $\Omega\supset U^{1}\supset U^{2}\supset\ldots$ are such that
$$
\mes\{U^n\setminus A\}\le\frac{1}{n}.
$$
We may assume that  $U^n$ consists of a finite number of open intervals
$$
U^n=\bigcup_{k=1}^{m_n}(\alpha^n_k,\beta^n_k)
$$
where
$
\beta^n_k<\alpha^n_{k+1}
$
for all $k=1,\ldots,m_n-1$. Let
$$
\varphi^{n}_\varepsilon=\sum_{k=1}^{m_n}
H^{\alpha^n_k-\varepsilon,\beta^n_k+\varepsilon}_\varepsilon.
$$
We see that
$$
ø\int\limits_\Omega f(x)\varphi_\varepsilon^n(x)\,dx
\ge\int\limits_Af(x)\,dx-\int\limits_{U^n\setminus A}\mid f(x)\mid\,dx-
\left(\sum_{k=1}^{m_n} \int\limits_{\alpha^n_k-2\varepsilon}^{\alpha^n_k}\mid
f(x)\mid\,dx +\sum_{k=1}^{m_n} \int\limits_{\beta^n_k}^{\beta^n_k+2\varepsilon}\mid
f(x)\mid\,dx\right).
$$
We note that
$$
\int\limits_Af(x)\,dx\ge\delta\mu>0.
$$
Obviously,
$$
\lim_{n\to\infty}\int\limits_{U^n\setminus A}\mid f(x)\mid\,dx=0
$$
and
$$
\lim_{\varepsilon\to 0}\left(\sum_{k=1}^{m_n}
\int\limits_{\alpha^n_k-2\varepsilon}^{\alpha^n_k}\mid f(x)\mid\,dx +\sum_{k=1}^{m_n}
\int\limits_{\beta^n_k}^{\beta^n_k+2\varepsilon}\mid f(x)\mid\,dx\right)=0.
$$
Hence
$$
\int\limits_\Omega f(x)\varphi_\varepsilon^n(x)\,dx>0
$$
for some $n$ and $\varepsilon$. This means that $\Lambda_{f}\ne0$. $\bullet$
\bigskip

By Theorem~\ref{l1indis}, the mapping
$$
L_1(\Omega)^{\rm loc}\longrightarrow {\cal D}'(\Omega)
$$
such that
$$
f\longmapsto \Lambda_f
$$
is a sequentially continuous embedding.

Let $x_0\in\Omega$. The distribution $\delta_{x_0}\in {\cal D}'(\Omega)$ such that
$$
(\delta_{x_0},\varphi)=\varphi(x_0)
$$
for all $\varphi\in C_0^\infty(\Omega)$ is called the delta-function.

\begin{theorem}
\label{nonreg} The delta-function $\delta_0\in {\cal D}'(\mathbb{R})$ is not regular.
\end{theorem}
{\bf Proof.} Let $f\in L_1(\mathbb{R})^{\rm loc}$ be such that
$$
(\delta_0,\varphi)=\int\limits_{-\infty}^{\infty}f(x)\varphi(x)\,dx
$$
for all $\varphi\in C_0^\infty(\mathbb{R})$. Put
$$
\varphi_{\varepsilon}(x)=h(x/\varepsilon).
$$
On the one hand, we have
$$
(\delta_0,\varphi_{\varepsilon})=\varphi_{\varepsilon}(0)=h(0)>0.
$$
On the other hand, we have
$$
\left|\int\limits_{-\infty}^{\infty}f(x)\varphi_{\varepsilon}(x)\,dx\right|=
\left|\int\limits_{-\varepsilon}^\varepsilon f(x)h(x/\varepsilon)\,dx\right| \le
h(0)\int\limits_{-\varepsilon}^\varepsilon \mid f(x)\mid\,dx.
$$
Since
$$
\lim_{\varepsilon\to 0} \int\limits_{-\varepsilon}^\varepsilon \mid f(x)\mid\,dx=0,
$$
we have the contradiction. $\bullet$
\bigskip



\begin{theorem}
\label{delta} We have
$$
\delta_0=\lim_{\varepsilon\to 0}h_\varepsilon
$$
in the space ${\cal D}'(\mathbb{R})$.
\end{theorem}
{\bf Proof.} Let $\varphi\in C_0^\infty(\mathbb{R})$. Then
\begin{eqnarray*}
\mid(\delta_0,\varphi)-(h_\varepsilon,\varphi)\mid
&=&\left|\varphi(0)-\int\limits_{\infty}^{\infty}h_\varepsilon(x)\varphi(x)\,dx\right|\medskip\\
&=&\left|\int\limits_{\infty}^{\infty}h_\varepsilon(x)(\varphi(0)-\varphi(x))\,dx\right|\medskip\\
&\le&\int\limits_{\infty}^{\infty}h_\varepsilon(x)\mid\varphi(0)-\varphi(x)\mid\,dx\medskip\\
&=&\int\limits_{\infty}^{\infty}h_\varepsilon(x)\left|\int\limits_0^x\varphi'(t)\,dt\right|\,dx\medskip\\
&\le&\varepsilon\|\varphi'\|_{C({\rm R})}.
\end{eqnarray*}
As was to be proved. $\bullet$
\bigskip

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section*{3. Distributions with compact supports}


We say that $\Lambda\in {\cal D}'(\Omega)$ is equal to zero at $x\in\Omega$ if there
exists $\varepsilon>0$ such that $(\Lambda,\varphi)=0$ for all  $\varphi\in
C_0^\infty(\Omega)$ such that $\supp\varphi\subset [x-\varepsilon,x+\varepsilon]$. For
example, the delta-function $\delta_{x_0}$ is equal to zero outside of $x_0$.

Let ${\cal O}(\Lambda)$ be the set of all points $x\in\Omega$ at which $\Lambda$ is equal
to zero. Then ${\cal O}(\Lambda)$ is open in $\Omega$. Hence $\Omega\setminus{\cal
O}(\Lambda)$ is closed in $\Omega$. The set $\supp\,\Lambda=\Omega\setminus{\cal
O}(\Lambda)$ is called the support of $\Lambda$. The space of distributions $\Lambda\in
{\cal D}'(\Omega)$ with compact supports is denoted by ${\cal E}'(\Omega)$. We see that
$\supp\delta_{x_0}=\{x_0\}$. Hence $\delta_{x_0}\in {\cal E}'(\Omega)$ for all $x_{0}\in
\Omega$.

\begin{theorem}
\label{supdis} Let $\Lambda\in {\cal D}'(\Omega)$ and $\varphi\in C_0^\infty(\Omega)$. If
$
{\rm supp}\Lambda\cap{\rm supp}\varphi=\emptyset,
$
then $(\Lambda,\varphi)=0$.
\end{theorem}
{\bf Proof.} Consider the test function
$$
\psi_\varepsilon^k(x)=\left(\sum_{i=-\infty}^\infty h_\varepsilon(x-\varepsilon
i)\right)^{-1}h_\varepsilon(x-\varepsilon k).
$$
Note that
$
\supp\,\psi_\varepsilon^k=[\varepsilon(k-1),\varepsilon(k+1)].
$
Since
$$
\sum_{k=-\infty}^\infty \psi_\varepsilon^k(x)=1,
$$
we have
$$
\varphi=\sum_{k=-\infty}^\infty \psi_\varepsilon^k\varphi.
$$
Note that the right-hand sum consists of a finite number of nonzero terms. Since
$\supp\varphi$ is compact, since $\supp\Lambda$ is closed in $\Omega$, and since
$\supp\Lambda\cap\supp\varphi=\emptyset$, we have
$$
(\Lambda,\varphi)= \sum_{k=-\infty}^\infty (\Lambda,\psi_\varepsilon^k\varphi)=0,
$$
where $\varepsilon>0$ is sufficiently small. $\bullet$
\bigskip



Consider the space $C^\infty(\Omega)$. We say that $\varphi_1,\varphi_2,\ldots\in
C^\infty(\Omega)$ converges to $\varphi\in C^\infty(\Omega)$ if
$$
\lim_{k\to\infty}\|\varphi^{(i)}-\varphi_k^{(i)}\|_{C[\alpha,\beta]}=0
$$
for any $[\alpha,\beta]\subset\Omega$ and for any $i=0,1,\ldots\,$. We see that
$C_0^\infty(\Omega)$ is dense in $C^\infty(\Omega)$.


\begin{theorem}
\label{comsup} A distribution $\Lambda\in {\cal D}'(\Omega)$ has a compact support if and
only if $\Lambda$ may be continuously extended from $C_0^\infty(\Omega)$ onto
$C^\infty(\Omega)$.
\end{theorem}
{\bf Proof. Necessity.} Let $\Lambda\in {\cal D}'(\Omega)$ has a compact support. Then
there exist an open set $U$ such that $\supp\Lambda\subset U\subset\Omega$ and a test
function $\psi\in C_0^\infty(\Omega)$ such that $\psi(x)=1$ for all $x\in U$. Define a
functional $\tilde\Lambda$ on the space $C^\infty(\Omega)$ by the rule
$$
(\tilde\Lambda,\varphi)=(\Lambda,\psi\varphi),
$$
where $\varphi\in C^\infty(\Omega)$. Let $\varphi\in C_0^\infty(\Omega)$. Since
$\supp\Lambda\cap\supp[(1-\psi)\varphi]=\emptyset$, we have
$$
(\tilde\Lambda,\varphi)=(\Lambda,\psi\varphi)
=(\Lambda,\psi\varphi)+(\Lambda,(1-\psi)\varphi)=(\Lambda,\varphi).
$$
{\bf Sufficiency.} Let $\Lambda$ be continuously extended from $C_0^\infty(\Omega)$ onto
$C^\infty(\Omega)$, and let $\tilde\Lambda$ be the extension. Assume that $\supp\Lambda$
is not compact. Then there exists a sequence $x_1,x_2,\ldots\in\supp\Lambda$ such that
$$
\lim_{k\to\infty}x_k=a
$$
or
$$
\lim_{k\to\infty}x_k=b.
$$
Let $0<\varepsilon_k<1$ be such that
$[x_k-\varepsilon_k,x_k+\varepsilon_k]\subset\Omega$. Then let $\psi_k\in
C_0^\infty(\Omega)$ be such that
$\supp\psi_k\subset[x_k-\varepsilon_k,x_k+\varepsilon_k]$ and $(\Lambda,\psi_k)=1$. We
see that
$$
\lim_{k\to\infty}\psi_k=0
$$
in in the space $C^\infty(\Omega)$. Hence,
$$
\lim_{k\to\infty}(\tilde\Lambda,\psi_k)=0.
$$
Since
$$
(\tilde\Lambda,\psi_k)=(\Lambda,\psi_k)=1
$$
for all $k=1,2,\ldots$, we have the contradiction. Thus $\supp\Lambda$ is compact.
$\bullet$
\bigskip


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section*{4. Positive distributions}

Let $f\in L_1(\Omega)^{\rm loc}$. By Theorem~\ref{l1indis}, we see that
$
f(x)\ge 0
$
for almost all $x\in\Omega$ if and only if
$$
\int\limits_\Omega f(x)\varphi(x)\,dx\ge 0
$$
for all positive $\varphi\in C_0^\infty(\Omega)$. This leads to the following definition.

A distribution $\Lambda\in {\cal D}'(\Omega)$ is called positive if
$$
(\Lambda,\varphi)\ge 0
$$
for all positive $\varphi\in C_0^\infty(\Omega)$. To denote the fact that $\Lambda\in
{\cal D}'(\Omega)$ is positive, we write $\Lambda\ge 0$.

The delta-function is positive. Indeed, if  $x_0\in\Omega$, then
$$
(\delta_{x_0},\varphi)=\varphi(x_0)\ge 0
$$
for all positive $\varphi\in C_0^\infty(\Omega)$. Hence $\delta_{x_0}\ge 0$.

Let $x_1,\ldots,x_k\in\Omega$ and $\lambda_1\ge 0,\ldots,\lambda_k\ge 0$. Then
$$
\Lambda=\lambda_1\delta_{x_1}+\ldots+\lambda_k\delta_{x_k}
$$
is positive.

\begin{theorem}
\label{posn1} Let $\Lambda\in {\cal D}'(\Omega)$ be positive. Given
$[\alpha,\beta]\subset\Omega$, there exists $C>0$ such that
$$
\mid(\Lambda,\varphi)\mid\le C\|\varphi\|_{C[\alpha,\beta]}
$$
for all $\varphi\in C_0^\infty(\alpha,\beta)$.
\end{theorem}
{\bf Proof.} Let $\psi\in C_0^\infty(\Omega)$ be positive, and let $\psi(x)=1$ for all
$x\in [\alpha,\beta]$. If $\varphi\in C_0^\infty(\alpha,\beta)$, then
$$
-\|\varphi\|_{C[\alpha,\beta]}\psi\le \varphi\le\|\varphi\|_{C[\alpha,\beta]}\psi.
$$
Hence,
$$
-\|\varphi\|_{C[\alpha,\beta]}(\Lambda,\psi)\le
(\Lambda,\varphi)\le\|\varphi\|_{C[\alpha,\beta]}(\Lambda,\psi).
$$
This means that
$$
\mid(\Lambda,\varphi)\mid\le C\|\varphi\|_{C[\alpha,\beta]}
$$
for all $\varphi\in C_0^\infty(\alpha,\beta)$ with $C=(\Lambda,\psi)$. $\bullet$
\bigskip

We say that $\Lambda\in {\cal D}'(\Omega)$ is a zero order distribution if given
$[\alpha,\beta]\subset\Omega$, there exists $C\ge 0$ such that
$$
\mid(\Lambda,\varphi)\mid\le C\|\varphi\|_{C[\alpha,\beta]}
$$
for all $\varphi\in C_0^\infty(\alpha,\beta)$. By Theorem~\ref{posn1}, if $\Lambda\ge 0$,
then $\Lambda$ is a zero order distribution.

\begin{theorem}
\label{zerordn1} Let $\Lambda$ be a linear continuous functional on $C_0(\Omega)$, then
$\Lambda$ is a zero order distribution.
\end{theorem}
{\bf Proof.}  Since $C_0^\infty(\Omega)\subset C_0(\Omega)$, we have $\Lambda\in {\cal
D}'(\Omega)$. Assume that $\Lambda$ is not a zero order distribution. Hence there exists
$[\alpha,\beta]\subset\Omega$ such that for each $k=1,2,\ldots$ we can find $\varphi_k\in
C_0^\infty(\alpha,\beta)$ satisfying
$$
\mid(\Lambda,\varphi_k)\mid > k\|\varphi_k\|_{C[\alpha,\beta]}.
$$
Let
$$
\psi_k=\frac{1}{k}\|\varphi_k\|_{C[\alpha,\beta]}^{-1}\varphi_k.
$$
On the one hand, we see that
$$
\lim_{k\to\infty}\psi_k=0
$$
in the space $C_0(\Omega)$. Hence
$$
\lim_{k\to\infty}(\Lambda,\psi_k)=0.
$$
On the other hand, we see that
$$
\mid(\Lambda,\psi_k)\mid= \frac{1}{k}\|\varphi_k\|_{C[\alpha,\beta]}^{-1}
\mid(\Lambda,\varphi_k)\mid>1.
$$
The contradiction shows that $\Lambda$ is a zero order distribution. $\bullet$
\bigskip


\begin{theorem}
\label{ext} Let $\Lambda\in {\cal D}'(\Omega)$ be a zero order distribution. Then
$\Lambda$ can be continuously extended from $C_0^\infty(\Omega)$ onto $C_0(\Omega)$.
\end{theorem}
{\bf Proof.} The unique continuous extension $\tilde\Lambda$ is defined by the rule
$$
(\tilde\Lambda,\varphi)= \lim_{\varepsilon\to 0}(\Lambda,\varphi*h_\varepsilon),
$$
where $\varphi\in C_0(\Omega)$. $\bullet$
\bigskip

In the following statement we need the Stieltjes integral and the Riesz theorem about
linear continuous functionals on the space of continuous functions. Assume that any
increasing function $f(x)$ is continuous from the right. This means, that
$$
øf(x_{0})=f(x_{0}+0)=\lim_{\varepsilon\searrow0}f(x_{0}+\varepsilon)
$$
for all $x_{0}$.


\begin{theorem}
\label{riepon1} Let $\Lambda\in {\cal D}'(\Omega)$ be positive. Given $x_0\in\Omega$,
there exists a unique increasing function $f_\Lambda$ such that $f_\Lambda(x_0)=0$ and
$$
(\Lambda,\varphi)=\int\limits_\Omega\varphi(x)\,df_\Lambda(x)
$$
for all $\varphi\in C_0^\infty(\Omega)$.
\end{theorem}
{\bf Proof.} We may assume that $\Lambda$ is extended from $C_0^\infty(\Omega)$ onto
$C_0(\Omega)$. Let $x_{0}\in [\alpha,\beta]\subset\Omega$. Consider the space
$$
S[\alpha,\beta]=\{\varphi\in C[\alpha,\beta]:\;\varphi(\alpha) = \varphi(\beta) = 0\}.
$$
We may assume that $\varphi\in S[\alpha,\beta]$ is equal to zero outside of
$[\alpha,\beta]$. Since $\Lambda$ is positive on $S[\alpha,\beta]$, there exists a unique
increasing function $f_\Lambda^{\alpha,\beta}$ such that
$f_\Lambda^{\alpha,\beta}(x_0)=0$ and
$$
(\Lambda,\varphi)=\int\limits_{\alpha}^{\beta}\varphi(x)\,df_\Lambda^{\alpha,\beta}(x)
$$
for all $\varphi\in S[\alpha,\beta]$. If
$[\alpha,\beta]\subset[\gamma,\tau]\subset\Omega$, then
$
\left.f_\Lambda^{\gamma,\tau}\right|_{[\alpha,\beta]}=f_\Lambda^{\alpha,\beta}.
$
Therefore there exists a unique increasing function $f_\Lambda$ such that
$$
f_\Lambda(x_0)=0
$$
and
$$
(\Lambda,\varphi)=\int\limits_\Omega\varphi(x)\,df_\Lambda(x)
$$
for all $\varphi\in C_0(\Omega)$. $\bullet$
\bigskip

\noindent {\bf Corollary.} Let $\Lambda\in {\cal D}'(\Omega)$ be positive. There exists a
unique Borel measure $\mu_\Lambda$ such that
$$
(\Lambda,\varphi)=\int\limits_\Omega\varphi(x)\,d\mu_\Lambda(x)
$$
for all $\varphi\in C_0^\infty(\Omega)$. Indeed, we put
$$
\mu_\Lambda\{(\alpha,\beta]\}=f_\Lambda(\beta)-f_\Lambda(\alpha),
$$
where $[\alpha,\beta]\subset\Omega$. Then we define the measure $\mu_\Lambda$ on the
$\sigma$-algebra of Borel sets by the standard extension. In particular, we have
$$
ø\mu_\Lambda\{[\alpha,\beta]\}=f_\Lambda(\beta)-f_\Lambda(\alpha-0)\;\;\;\;{\rm
and}\;\;\;\; \mu_\Lambda\{(\alpha,\beta)\}=f_\Lambda(\beta-0)-f_\Lambda(\alpha)
$$



%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section*{5. The derivative}

Consider $f\in C^1(\Omega)$ and $\varphi\in C_0^\infty(\Omega)$. Integrating by parts, we
have
$$
\int\limits_\Omega f'(x)\varphi(x)\,dx = -\int\limits_\Omega f(x)\varphi'(x)\,dx.
$$
The right-hand side integral has sense for any $f\in L_1(\Omega)^{\rm loc}$. This leads
to the following definition.

Let $\Lambda\in {\cal D}'(\Omega)$. Define the distribution $\Lambda'\in {\cal
D}'(\Omega)$ by the rule
$$
(\Lambda',\varphi) = -(\Lambda,\varphi'),
$$
where $\varphi\in C_0^\infty(\Omega)$. The distribution $\Lambda'$ is called the
derivative of the distribution $\Lambda$.


\begin{theorem}
\label{divc1} If $f\in C^1(\Omega)$, then
$
(\Lambda_f)'=\Lambda_{f'}.
$
\end{theorem}
{\bf Proof.} Indeed, let $\varphi\in C_0^\infty(\Omega)$. Then
$$
((\Lambda_f)',\varphi)=-(\Lambda_f,\varphi')=-\int\limits_\Omega
f(x)\varphi'(x)\,dx=\int\limits_\Omega f'(x)\varphi(x)\,dx=(\Lambda_{f'},\varphi).
$$
As was to be proved. $\bullet$
\bigskip

A function $f$ is called absolutly continuous on $[\alpha,\beta]$ if given
$\varepsilon>0$, there exists $\delta>0$ such that
$$
\sum_{k=1}^n \mid f(\beta_k)-f(\alpha_k)\mid<\varepsilon
$$
whenever
$
\alpha\le\alpha_1<\beta_1<\ldots<\alpha_n<\beta_n\le\beta
$
satisfy
$$
\sum_{k=1}^n (\beta_k-\alpha_k)<\delta.
$$
A function $f$ is called locally absolutly continuous on $\Omega$ if $f$ is absolutly
continuous on any $[\alpha,\beta]\subset\Omega$. If $f$ is locally absolutly continuous
on $\Omega$, then the classic derivative $f'(x)$ exists for almost all $x\in\Omega$ and
$f'\in L_1(\Omega)^{\rm loc}$.

\begin{theorem}
\label{divabs} If $f$ is locally absolutly continuous on $\Omega$, then
$
(\Lambda_f)'=\Lambda_{f'}.
$
\end{theorem}
{\bf Proof.} Indeed, let $\varphi\in C_0^\infty(\Omega)$ and
$\supp\varphi\subset[\alpha,\beta]\subset\Omega$. Then
$$
ø-\int\limits_\alpha^\beta f(x)\varphi'(x)\,dx=-\int\limits_\alpha^\beta
f(x)\,d\varphi(x)=\int\limits_\alpha^\beta \varphi(x)\,df(x)=\int\limits_\alpha^\beta
f'(x)\varphi(x)\,dx.
$$
Hence,
$$
ø((\Lambda_f)',\varphi) =-\int\limits_\alpha^\beta
f(x)\varphi'(x)\,dx=\int\limits_\alpha^\beta f'(x)\varphi(x)\,dx=(\Lambda_{f'},\varphi).
$$
As was to be proved. $\bullet$
\bigskip

Consider the Heaviside function
$$
\Theta(x)=\left\{\begin{array}{lll} 1&{\rm if}&x\ge 0\\ 0&{\rm if}&x < 0
\end{array}\right..
$$

\begin{theorem}
\label{divhev} We have
$
\Theta'=\delta_0.
$
\end{theorem}
{\bf Proof.} Indeed, let $\varphi\in C_0^\infty(\mathbb{R})$. Then
$$
(\Theta',\varphi)=-(\Theta,\varphi')=-\int\limits_{-\infty}^{\infty}\Theta(x)\varphi'(x)\,dx=
 -\int\limits_0^\infty\varphi'(x)\,dx=\varphi(0)=(\delta_0,\varphi).
$$
As was to be proved. $\bullet$
\bigskip

\begin{theorem}
\label{divc1hev} If $f\in C^1(\mathbb{R})$, then
$
(\Theta f)'=f(0)\delta_0+\Theta f'.
$
\end{theorem}
{\bf Proof.} Indeed, let $\varphi\in C_0^\infty(\mathbb{R})$. Then
\begin{eqnarray*}
((\Theta f)',\varphi)&=&-(\Theta f,\varphi')\medskip\\
&=&-\int\limits_{-\infty}^{\infty}\Theta(x)f(x)\varphi'(x)\,dx=-\int\limits_0^\infty
f(x)\varphi'(x)\,dx=-\int\limits_0^\infty f(x)\,d\varphi(x)\medskip\\
&=&f(0)\varphi(0)+\int\limits_0^\infty
f'(x)\varphi(x)\,dx=f(0)\varphi(0)+\int\limits_{-\infty}^{\infty}\Theta(x)f'(x)\varphi(x)\,dx
\medskip\\
&=&(f(0)\delta_0+\Theta f',\varphi).
\end{eqnarray*}
As was to be proved. $\bullet$
\bigskip

Let
$$
x_+^k=\left\{\begin{array}{lll} x^k&{\rm if}&x\ge 0\\ 0&{\rm if}&x < 0
\end{array}\right..
$$
We see that $x_+^k=\Theta x^k$. Hence
$$
\frac{d^l}{dx^l}\frac{x_+^{l-1}}{(l-1)!}=\delta_0.
$$


Let $f\in C^\infty(\Omega)$ and $\Lambda\in {\cal D}'(\Omega)$. The product $f\Lambda\in
{\cal D}'(\Omega)$ is defined by the rule
$$
(f\Lambda,\varphi) = (\Lambda,f\varphi),
$$
where $\varphi\in C_0^\infty(\Omega)$.

\begin{theorem}
\label{divprod} Let $f\in C^\infty(\Omega)$ and $\Lambda\in {\cal D}'(\Omega)$. Then
$
(f\Lambda)'=f\Lambda'+f'\Lambda.
$
\end{theorem}
{\bf Proof.} Indeed, let $\varphi\in C_0^\infty(\mathbb{R})$. Then
\begin{eqnarray*}
((f\Lambda)',\varphi)&=&-(f\Lambda,\varphi')=-(\Lambda,f\varphi')\medskip\\
  &=&-(\Lambda,(f\varphi)'-f'\varphi)=-(\Lambda,(f\varphi)')+(\Lambda,f'\varphi)\medskip
  \\&=&
  (\Lambda',f\varphi)+(\Lambda,f'\varphi)=(f\Lambda',\varphi)+(f'\Lambda,\varphi)\medskip\\
  &=& (f\Lambda'+f'\Lambda,\varphi)
\end{eqnarray*}
As was to be proved. $\bullet$
\newpage

%%%%%%%%%%%%%%%%%%%%%%%%%%
\section*{6. The Integral}

\begin{theorem}
\label{difzer} Let $u\in {\cal D}'(\Omega)$ be such that
$
u'=0.
$
Then $u$ is a constant function.
\end{theorem}
{\bf Proof.} Let $\psi\in C_0^\infty(\Omega)$ be such that
$$
\int\limits_a^{b}\psi(x)\,dx=1.
$$
Define the mapping
$$
ø\Phi_\psi:\;C_0^\infty(\Omega)\longrightarrow C_0^\infty(\Omega)
$$
by the rule
$$
\Phi_\psi(\varphi)(x)=\int\limits_a^x\varphi(y)\,dy-
\int\limits_a^x\psi(y)\,dy\int\limits_a^{b}\varphi(y)\,dy.
$$
Put $C=(u,\psi)$. Let $\varphi\in C_0^\infty(\Omega)$. Then
\begin{eqnarray*}
(u,\varphi)
&=&\left(u,\Phi_\psi(\varphi)'+\psi\int\limits_a^{b}\varphi(y)\,dy\right)\medskip\\
&=&(u,\Phi_\psi(\varphi)')+(u,\psi)\int\limits_a^{b}\varphi(y)\,dy\medskip\\
&=&-(u',\Phi_\psi(\varphi))+C\int\limits_a^{b}\varphi(y)\,dy\medskip\\
&=&\int\limits_a^{b} C\varphi(y)\,dy\medskip\\ &=&(C,\varphi).
\end{eqnarray*}
Therefore $u=C$. $\bullet$
\bigskip

\begin{theorem}
\label{difequ} Let $\psi\in C_0^\infty(\Omega)$ be such that
$$
\int\limits_a^{b}\psi(x)\,dx=1.
$$
Given $\Lambda\in {\cal D}'(\Omega)$, there exists a unique $u\in {\cal D}'(\Omega)$ such
that
$
u'=\Lambda
$
and
$
(u,\psi)=0.
$
\end{theorem}
{\bf Proof.} Define the distribution $u\in {\cal D}'(\Omega)$ by the rule
$$
(u,\varphi)=-(\Lambda,\Phi_\psi(\varphi)),
$$
where again
$$
\Phi_\psi(\varphi)(x)=\int\limits_a^x\varphi(y)\,dy-
\int\limits_a^x\psi(y)\,dy\int\limits_a^{b}\varphi(y)\,dy.
$$
Since
$$
\int\limits_a^{b}\varphi'(y)\,dy=0,
$$
we have
$
\Phi_\psi(\varphi')=\varphi.
$
Therefore
$$
(u',\varphi)=-(u,\varphi')=(\Lambda,\Phi_\psi(\varphi'))=(\Lambda,\varphi).
$$
This means that
$
u'=\Lambda.
$
Since
$
\Phi_\psi(\psi)=0,
$
we have
$
(u,\psi)=0.
$
By Theorem~\ref{difzer}, we see that $u$ is a unique distribution satisfying the
conditions. $\bullet$
\bigskip

\noindent {\bf Corollary.} There exists a linear continuous operator
$$
I:\;{\cal D}'(\Omega)\longrightarrow {\cal D}'(\Omega)
$$
such that
$$
\frac{d}{dx}I(\Lambda)=\Lambda
$$
for all $\Lambda\in{\cal D}'(\Omega)$.



%%%%%%%%%%%%%$%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section*{7. The convolution of functions}

We say that the family of $k$ functions and distributions is admissible if at least $k-1$
of them have compact supports. Usually we have $k=2$ or $k=3$.

Let $\varphi_1,\varphi_2\in C^\infty(\mathbb{R})$ be admissible. The convolution of
$\varphi_1$ and $\varphi_2$ is define by the rule
$$
\varphi_1*\varphi_2(x)= \int\limits_{-\infty}^{\infty}\varphi_1(x-y)\varphi_2(y)\,dy.
$$
By the identity
$$
\int\limits_{-\infty}^{\infty}\varphi_1(x-y)\varphi_2(y)\,dy=
\int\limits_{-\infty}^{\infty}\varphi_1(y)\varphi_2(x-y)\,dy,
$$
we have
$$
\varphi_1*\varphi_2=\varphi_2*\varphi_1.
$$
This means that the convolution is commutative.

By differentiation under the integral, we have
$$
(\varphi_1*\varphi_2)'=\varphi_1'*\varphi_2=\varphi_1*\varphi_2'.
$$
In particular, we have
$$
\varphi_1*\varphi_2\in C^\infty(\mathbb{R}).
$$

Let us check that
$$
\supp(\varphi_1*\varphi_2)\subset\supp\varphi_1+\supp\varphi_2.
$$
Indeed, let $z_0\in \mathbb{R}$ be such that $\varphi_1*\varphi_2(z_0)\ne 0$. Since
$$
\int\limits_{-\infty}^{\infty}\varphi_1(z_0-y)\varphi_2(y)\,dy\ne 0,
$$
there exists $y_0\in \mathbb{R}$ such that
$
\varphi_1(z_0-y_0)\ne 0
$
and
$
\varphi_2(y_0)\ne 0.
$
Let $x_0=z_0-y_0$. We see that $x_0\in\supp\varphi_1$ and $y_0\in\supp\varphi_2$. Since
$\supp\varphi_1$ or $\supp\varphi_2$ is compact, the set $\supp\varphi_1+\supp\varphi_2$
is closed. Therefore $\supp(\varphi_1*\varphi_2)\subset\supp\varphi_1+\supp\varphi_2$.

In particular, if $\varphi_1,\varphi_2\in C_0^\infty(\mathbb{R})$, then
$
\varphi_1*\varphi_2\in C_0^\infty(\mathbb{R}).
$
Moreover, by the identity
$$
\int\limits_{-\infty}^{\infty}\varphi_1*\varphi_2(x)\varphi(x)\,dx=
\int\limits_{-\infty}^{\infty}\int\limits_{-\infty}^{\infty}
\varphi_1(x)\varphi_2(y)\varphi(x+y)\,dxdy,
$$
where $\varphi\in C_0^\infty(\mathbb{R})$, we have
$$
\begin{array}{lll}
\min\,\supp(\varphi_1*\varphi_2) &=&\min\,\supp\varphi_1+\min\,\supp\varphi_2,\\
\max\,\supp(\varphi_1*\varphi_2) &=&\max\,\supp\varphi_1+\max\,\supp\varphi_2.
\end{array}
$$

Let $\varphi_1,\varphi_2,\varphi_3\in C^\infty(\mathbb{R})$ be admissible, then
$$
\varphi_1*(\varphi_2*\varphi_3)=(\varphi_1*\varphi_2)*\varphi_3.
$$
This means that the convolution is associative.

\begin{theorem}
\label{limconv} Let $\varphi\in C^\infty(\mathbb{R})$, then
$$
\varphi=\lim_{\varepsilon\to0}\varphi*h_\varepsilon
$$
in the space $C^\infty(\mathbb{R})$.
\end{theorem}
{\bf Proof.} Indeed, let $\alpha<\beta$, and let $x\in [\alpha,\beta]$, then
\begin{eqnarray*}
\left|\varphi^{(k)}(x)-(\varphi*h_\varepsilon)^{(k)}(x)\right|
&=&\left|\varphi^{(k)}(x)-\varphi^{(k)}*h_\varepsilon(x)\right|\\
&=&\left|\varphi^{(k)}(x)-
\int\limits_{-\infty}^{\infty}\varphi^{(k)}(x-y)h_\varepsilon(y)\,dy\right|\\ &=&\left|
\int\limits_{-\varepsilon}^\varepsilon
\left(\varphi^{(k)}(x)-\varphi^{(k)}(x-y)\right)h_\varepsilon(y)\,dy\right|\\
&\le&\varepsilon\|\varphi^{(k+1)} \|_{C[\alpha-\varepsilon,\beta+\varepsilon]}\|.
\end{eqnarray*}
Therefore
$$
\|\varphi^{(k)}-(\varphi*h_\varepsilon)^{(k)}\|_{C[\alpha,\beta]}\le
\varepsilon\|\varphi^{(k+1)}\|_{C[\alpha-\varepsilon,\beta+\varepsilon]}
$$
for all $k=0,1,\ldots$. $\bullet$
\bigskip

\noindent {\bf Corollary.} Let $\varphi\in C_0^\infty(\mathbb{R})$, then
$$
\varphi=\lim_{\varepsilon\to 0}\varphi*h_\varepsilon
$$
in the space $C_0^\infty(\mathbb{R})$.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section*{8. The convolution of distributions and functions}

Let $\Lambda\in {\cal D}'(\mathbb{R})$ and $\varphi\in C^\infty(\mathbb{R})$ be
admissible. By definition, this means that $\Lambda\in {\cal D}'(\mathbb{R})$ and
$\varphi\in C^\infty_{0}(\mathbb{R})$ or $\Lambda\in {\cal E}'(\mathbb{R})$ and
$\varphi\in C^\infty(\mathbb{R})$. In the second case we assume that $\Lambda$ is
extended from $C^\infty_{0}(\mathbb{R})$ onto $C^\infty(\mathbb{R})$.

The convolution of $\Lambda$ and $\varphi$ is defined by the rule
$$
\Lambda*\varphi(x)=\left(\Lambda,\varphi(x-\,\cdot\,)\right),
$$
where $x\in \mathbb{R}$. In particular, the convolution $\Lambda*\varphi$ is a real
function.

We shall use the notation
$$
\stackrel{\vee}{\varphi}(x)=\varphi(-x).
$$
Note that
$$
(\Lambda,\varphi)=\left(\Lambda*\stackrel{\vee}{\varphi}\right)(0).
$$


\begin{theorem}
\label{condf} Let $\Lambda\in {\cal D}'(\mathbb{R})$ and $\varphi\in
C^\infty(\mathbb{R})$ be admissible. Then $(\Lambda*\varphi)'=\Lambda*\varphi'$.
\end{theorem}
{\bf Proof.} Let $x_{0}\in \mathbb{R}$. If $\varphi\in C^\infty_{0}(\mathbb{R})$, then
$$
\varphi(x_{0}-\,\cdot\,)=\lim_{\Delta x\to 0} \frac{\varphi(x_{0}+\Delta
x-\,\cdot\,)-\varphi(x_{0}-\,\cdot\,)}{\Delta x}
$$
in the space $C^\infty_{0}(\mathbb{R})$. If $\varphi\in C^\infty(\mathbb{R})$, then
analogously
$$
\varphi(x_{0}-\,\cdot\,)=\lim_{\Delta x\to 0} \frac{\varphi(x_{0}+\Delta
x-\,\cdot\,)-\varphi(x_{0}-\,\cdot\,)}{\Delta x}
$$
in the space $C^\infty(\mathbb{R})$. Since
$$
\frac{\Lambda*\varphi(x_{0}+\Delta x)-\Lambda*\varphi(x_{0})}{\Delta x}= \left(\Lambda,
\frac{\varphi(x_{0}+\Delta x-\,\cdot\,)-\varphi(x_{0}-\,\cdot\,)}{\Delta x}\right),
$$
we have
$$
(\Lambda*\varphi)'(x_{0})=\Lambda*\varphi'(x_{0}).
$$
As was to be proved. $\bullet$
\bigskip

\noindent {\bf Corollary.} We have $\Lambda*\varphi\in C^\infty(\mathbb{R})$.
\bigskip

\begin{theorem}
Let $\Lambda\in {\cal D}'(\mathbb{R})$ and $\varphi\in C^\infty(\mathbb{R})$ be
admissible. Then ${\rm supp}(\Lambda*\varphi)\subset{\rm supp}\Lambda+{\rm supp}\varphi$.
\end{theorem}
{\bf Proof.} Let $\psi(x)=1$ on the whole $\mathbb{R}$ if $\supp\Lambda$ is unbounded,
and let $\psi(x)=1$ on a bounded neighborhood of $\supp\Lambda$ if $\supp\Lambda$ is
bounded. Consider $z_0\in \mathbb{R}$ such that $\Lambda*\varphi(z_0)\ne 0$. Since
$$
\Lambda*\varphi(z_0)=(\Lambda,\varphi(z_0-\,\cdot\,))=
(\Lambda,\psi(\cdot)\varphi(z_0-\,\cdot\,))
$$
and
$
\psi(\cdot)\varphi(z_0-\,\cdot\,)\in C_0^\infty(\mathbb{R}),
$
we have
$$
\supp\Lambda\cap\supp[\psi(\cdot)\varphi(z_0-\,\cdot\,)]\ne\emptyset.
$$
Obviously,
$$
\supp[\psi(\cdot)\varphi(z_0-\,\cdot\,)]\subset\supp\varphi(z_0-\,\cdot\,)
$$
Let
$
x_0\in\supp\Lambda\cap\supp\varphi(z_0-\,\cdot\,)
$
and
$
y_0=z_0-x_0.
$
We see that $x_0\in\supp\Lambda$ and $y_0\in\supp\varphi$. Therefore
$
\supp(\Lambda*\varphi)\subset\supp\Lambda+\supp\varphi.
$
As was to be proved. $\bullet$
\bigskip

%Let $\Lambda*$ be the operator, defined by the rule
%$$
%\Lambda*:\;\varphi\longmapsto \Lambda*\varphi.
%$$
%If $\Lambda\in {\cal D}'(\mathbb{R})$, then
%$$
%\Lambda*:\;C_0^\infty(\mathbb{R})\longrightarrow C^\infty(\mathbb{R}).
%$$
%If $\Lambda\in {\cal E}'(\mathbb{R})$, then
%$$
%\Lambda*:\;C_0^\infty(\mathbb{R})\longrightarrow C_0^\infty(\mathbb{R})
%$$
%and
%$$
%\Lambda*:\;C^\infty(\mathbb{R})\longrightarrow C^\infty(\mathbb{R}).
%$$
%In addition, all these operators are continuous.

\begin{theorem}
\label{ass} Let $\Lambda\in {\cal D}'(\mathbb{R})$ and $\varphi,\psi\in
C^\infty(\mathbb{R})$ be admissible. Then
$
(\Lambda*\varphi)*\psi=\Lambda*(\varphi*\psi).
$
\end{theorem}
{\bf Proof.} Let us prove the statement in the case $\Lambda\in {\cal D}'(\mathbb{R})$
and $\varphi,\psi\in C_0^\infty(\mathbb{R})$. We see that
$$
\int\limits_{-\infty}^{\infty}\varphi(\,\cdot\,-y)\psi(y)\,dy= \lim_{\varepsilon\to
0}\sum_{k=-\infty}^{\infty} \varphi(\,\cdot\,-\varepsilon k)\psi(\varepsilon
k)\varepsilon
$$
in the space $C_0^\infty(\mathbb{R})$. The sum consists of a finite number of nonzero
terms. Therefore,
\begin{eqnarray*}
[\Lambda*(\varphi*\psi)](x)&=&\left(\Lambda,\varphi*\psi(x-\,\cdot\,)\right)\\
&=&\lim_{\varepsilon\to 0}\sum_{k=-\infty}^{\infty}
\left(\Lambda,\varphi(x-\,\cdot\,-\varepsilon k)\right) \psi(\varepsilon k)\varepsilon\\
&=&\lim_{\varepsilon\to 0}\sum_{k=-\infty}^{\infty} \Lambda*\varphi(x-\varepsilon
k)\psi(\varepsilon k)\varepsilon\\
&=&\int\limits_{-\infty}^{\infty}\Lambda*\varphi(x-y)\psi(y)\,dy\\
&=&[(\Lambda*\varphi)*\psi](x)
\end{eqnarray*}
for all $x\in R$. $\bullet$
\bigskip

\begin{theorem}
\label{appdist} Let $\Lambda\in {\cal D}'(\mathbb{R})$. Then
$$
\Lambda=\lim_{\varepsilon\to 0}\Lambda*h_\varepsilon
$$
in the space ${\cal D}'(\mathbb{R})$.
\end{theorem}
{\bf Proof.} Let $\varphi\in C_0^\infty(\mathbb{R})$. Then
$$
ø\lim_{\varepsilon\to 0}(\Lambda*h_\varepsilon,\varphi) =\lim_{\varepsilon\to
0}[(\Lambda*h_\varepsilon)* \stackrel{\vee}{\varphi}](0)=\lim_{\varepsilon\to
0}[\Lambda*(h_\varepsilon*
\stackrel{\vee}{\varphi})](0)=\Lambda*\stackrel{\vee}{\varphi}(0)=(\Lambda,\varphi).
$$
As was to be proved. $\bullet$
\bigskip

\noindent {\bf Corollary.} The space $C_0^\infty(\mathbb{R})$ is dense in the space
${\cal D}'(\mathbb{R})$.

\begin{theorem}
\label{commdis} Let $\Lambda_1,\Lambda_2\in {\cal D}'(\mathbb{R})$ and $\varphi\in
C^\infty(\mathbb{R})$ be admissible. Then
$$
\Lambda_1*(\Lambda_2*\varphi)=\Lambda_2*(\Lambda_1*\varphi).
$$
\end{theorem}
{\bf Proof.} On the one hand, we see that
\begin{eqnarray*}
\Lambda_1*[\Lambda_2*(\varphi*h_\varepsilon)]
&=&\Lambda_1*[(\Lambda_2*\varphi)*h_\varepsilon]\\
&=&\Lambda_1*[h_\varepsilon*(\Lambda_2*\varphi)]\\
&=&(\Lambda_1*h_\varepsilon)*(\Lambda_2*\varphi).
\end{eqnarray*}
On the other hand, we see that
\begin{eqnarray*}
\Lambda_2*[\Lambda_1*(\varphi*h_\varepsilon)]
&=&\Lambda_2*[\Lambda_1*(h_\varepsilon*\varphi)]\\
&=&\Lambda_2*[(\Lambda_1*h_\varepsilon)*\varphi]\\
&=&\Lambda_2*[\varphi*(\Lambda_1*h_\varepsilon)]\\
&=&(\Lambda_2*\varphi)*(\Lambda_1*h_\varepsilon)\\
&=&(\Lambda_1*h\varepsilon)*(\Lambda_2*\varphi).
\end{eqnarray*}
Therefore
$$
\Lambda_1*[\Lambda_2*(\varphi*h_\varepsilon)]=
\Lambda_2*[\Lambda_1*(\varphi*h_\varepsilon)].
$$
We let $\varepsilon\to 0$, and obtain
$$
\Lambda_1*(\Lambda_2*\varphi)=\Lambda_2*(\Lambda_1*\varphi).
$$
As was to be proved. $\bullet$

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section*{9. The convolution of distributions}

Let $\Lambda_1,\Lambda_2\in {\cal D}'(\mathbb{R})$ be admissible. By definition, this
means that $\Lambda_{1}\in {\cal D}'(\mathbb{R})$ and $\Lambda_{2}\in {\cal
E}'(\mathbb{R})$ or $\Lambda_{1}\in {\cal E}'(\mathbb{R})$ and $\Lambda_{2}\in {\cal
D}'(\mathbb{R})$. If $\Lambda\in {\cal E}'(\mathbb{R})$, then we assume that $\Lambda$ is
extended from $C^\infty_{0}(\mathbb{R})$ onto $C^\infty(\mathbb{R})$.


The convolution $\Lambda_1*\Lambda_2$ of the distributions $\Lambda_1$ and $\Lambda_2$ is
called a distribution $\Lambda\in {\cal D}'(\mathbb{R})$ such that
$$
\Lambda*\varphi=\Lambda_1*(\Lambda_2*\varphi)
$$
for all $\varphi\in C_0^\infty(\mathbb{R})$. We have to prove the existence and the
uniqueness of this convolution.
\bigskip

\noindent {\bf Existence.} Define the distribution $\Lambda\in {\cal D}'(\mathbb{R})$ by
the rule
$$
(\Lambda,\varphi)=[\Lambda_1*(\Lambda_2*\stackrel{\vee}{\varphi})](0),
$$
where $\varphi\in C_0^\infty(\mathbb{R})$ and $\stackrel{\vee}{\varphi}(x)=\varphi(-x)$.
We claim that
$$
\Lambda*\varphi=\Lambda_1*(\Lambda_2*\varphi)
$$
for all $\varphi\in C_0^\infty(\mathbb{R})$. Indeed, let $x\in R$, then
\begin{eqnarray*}
\Lambda*\varphi(x)&=&(\Lambda,\varphi(x-\,\cdot\,))\\
&=&\left(\Lambda_1*\{\Lambda_2*[\varphi(x+\,\cdot\,)]\}\right)(0)\\
&=&\left(\Lambda_1*\{(\Lambda_2*\varphi)(x+\,\cdot\,)\}\right)(0)\\
&=&\left([\Lambda_1*(\Lambda_2*\varphi)](x+\,\cdot\,)\}\right)(0)\\
&=&[\Lambda_1*(\Lambda_2*\varphi)](x)
\end{eqnarray*}
\bigskip

\noindent {\bf Uniqueness.} Let $\Lambda\in {\cal D}'(\mathbb{R})$ as above, and let
$\tilde\Lambda\in {\cal D}'(\mathbb{R})$ satisfy the identity
$$
\tilde\Lambda*\varphi=\Lambda_1*(\Lambda_2*\varphi)
$$
for all $\varphi\in C_0^\infty(\mathbb{R})$. Then
$$
\Lambda*\varphi=\tilde\Lambda*\varphi
$$
for all $\varphi\in C_0^\infty(\mathbb{R})$. Therefore
$$
(\Lambda,\varphi)= \Lambda*\stackrel{\vee}{\varphi}(0)=
\tilde\Lambda*\stackrel{\vee}{\varphi}(0)= (\tilde\Lambda,\varphi)
$$
for all $\varphi\in C_0^\infty(\mathbb{R})$.

\begin{theorem}
\label{procon} Let $\Lambda_1,\Lambda_2,\Lambda_3\in {\cal D}'(\mathbb{R})$ be
admissible. Then
$$
\begin{array}{lll}
\Lambda_1*\Lambda_2&=&\Lambda_2*\Lambda_1,\\
\Lambda_1*(\Lambda_2*\Lambda_3)&=&(\Lambda_1*\Lambda_2)*\Lambda_3,\\ {\rm
supp}(\Lambda_1*\Lambda_2)&\subset&{\rm supp}\Lambda_1+{\rm supp}\Lambda_2.
\end{array}
$$
\end{theorem}
{\bf Proof.} The first follows from Theorem~\ref{commdis}. The second follows from the
definition. Let us check the third. Let $z_0\in \supp(\Lambda_1*\Lambda_2)$. Then for
each $k=1,2,\ldots$ there exists $\varphi_k\in C_0^\infty(\mathbb{R})$ such that
$$
(\Lambda_1*\Lambda_2,\varphi_k)= [\Lambda_1*(\Lambda_2*\stackrel{\vee}{\varphi}_k)](0)\ne
0
$$
and
$$
\supp\varphi_k\subset[z_0-1/k,z_0+1/k].
$$
By Theorem~\ref{condf}, we see that
$$
0\in \supp\Lambda_1+\supp\Lambda_2-\supp\varphi_k.
$$
Therefore there exist $x_k\in \supp\Lambda_1$, $y_k\in \supp\Lambda_2$ and $\mid
t_k\mid\le 1/k$ such that
$$
z_0=x_k+y_k+t_k.
$$
Since
$$
\lim_{k\to\infty}t_{k}=0,
$$
and since $\supp\Lambda_1$ or $\supp\Lambda_2$ is compact, there exist $x_0\in
\supp\Lambda_1$ and $y_0\in \supp\Lambda_2$ such that
$$
z_0=x_0+y_0.
$$
As was to be proved. $\bullet$
\bigskip

\noindent {\bf Corollary.} If $\Lambda_1,\Lambda_2\in {\cal E}'(\mathbb{R})$, then
$\Lambda_1*\Lambda_2\in {\cal E}'(\mathbb{R})$.

\begin{theorem}
\label{condel} Let $\Lambda\in {\cal D}'(\mathbb{R})$, then $\Lambda*\delta_0=\Lambda$.
\end{theorem}
{\bf Proof.} If $\varphi\in C_0^\infty(\mathbb{R})$, then
$$
\delta_0*\varphi(x)=(\delta_0,\varphi(x-\,\cdot\,))=\varphi(x).
$$
Therefore
$$
(\Lambda*\delta_0)*\varphi=\Lambda*(\delta_0*\varphi)=\Lambda*\varphi.
$$
As was to be proved. $\bullet$
\bigskip

\noindent {\bf Remark.} Let $\Lambda_1,\Lambda_2\in {\cal E}'(\mathbb{R})$. It can be
proved that
\begin{eqnarray*}
\min\,\supp(\Lambda_1*\Lambda_2)&=& \min\,\supp\Lambda_1+\min\,\supp\Lambda_2,\\
\max\,\supp(\Lambda_1*\Lambda_2)&=& \max\,\supp\Lambda_1+\max\,\supp\Lambda_2.
\end{eqnarray*}
%This means that
%$$
%\begin{array}{l}
%\min\,\supp:\;{\cal E}'(\mathbb{R})\longrightarrow R,\\ \max\,\supp:\;{\cal
%E}'(\mathbb{R})\longrightarrow R
%\end{array}
%$$
%are homomorphisms from the Abelian group ${\cal E}'(\mathbb{R})$ with the operation $*$
%onto the Abelian group $R$ with the operation $+$.

\begin{theorem}
\label{difcon} Let $\Lambda_1,\Lambda_2\in {\cal D}'(\mathbb{R})$ be admissible. Then
$$
(\Lambda_1*\Lambda_2)'=\Lambda_1'*\Lambda_2=\Lambda_1*\Lambda_2'.
$$
\end{theorem}
{\bf Proof.} Let $\Lambda\in {\cal D}'(\mathbb{R})$ and $\varphi\in C^\infty(\mathbb{R})$
be admissible. We claim that
$$
(\Lambda*\varphi)'=\Lambda'*\varphi.
$$
Indeed, we have
$$
ø(\Lambda*\varphi)'(x)=(\Lambda*\varphi')(x)=(\Lambda,\varphi'(x-\,\cdot\,)=
 -(\Lambda,-\varphi'(x-\,\cdot\,)=(\Lambda',\varphi(x-\,\cdot\,)=(\Lambda'*\varphi)(x).
$$
Let $\varphi\in C_0^\infty(\mathbb{R})$. Then
\begin{eqnarray*}
\left((\Lambda_1*\Lambda_2)',\varphi\right)
&=&-\left(\Lambda_1*\Lambda_2,\varphi'\right)\\
&=&-\left((\Lambda_1*\Lambda_2)*(\varphi')^\vee\right)(0)\\
&=&\left((\Lambda_1*\Lambda_2)*\stackrel{\vee}{\varphi}'\right)(0)\\
&=&\left(\Lambda_1*(\Lambda_2*\stackrel{\vee}{\varphi}'\right)(0)\\
&=&\left(\Lambda_1*(\Lambda_2'*\stackrel{\vee}{\varphi}\right)(0)\\
&=&\left((\Lambda_1*\Lambda_2')*\stackrel{\vee}{\varphi}\right)(0)\\
&=&(\Lambda_1*\Lambda_2',\varphi)
\end{eqnarray*}
As was to be proved. $\bullet$


\begin{theorem}
\label{conl1} Let $f,g\in L_1(\mathbb{R})$ be admissible. Then
$
\Lambda_{f*g}=\Lambda_{f}*\Lambda_{g}.
$
\end{theorem}
{\bf Proof.} We recall that
$$
f*g(x)=\int\limits_{-\infty}^{\infty}f(x-y)g(y)\,dy=\int\limits_{-\infty}^{\infty}f(y)g(x-y)\,dy.
$$
Let $\varphi\in C_0^\infty(\mathbb{R})$. Then
$$
\Lambda_g*\varphi(x)=(\Lambda_g,\varphi(x-\,\cdot\,))=
\int\limits_{-\infty}^{\infty}g(y)\varphi(x-y)\,dy.
$$
Therefore
\begin{eqnarray*}
(\Lambda_f*(\Lambda_g*\varphi))(z)
&=&\int\limits_{-\infty}^{\infty}f(x)\left\{\int\limits_{-\infty}^{\infty}g(y)\varphi(z-y-x)\,dy\right\}\,dx\medskip\\
&=&\int\limits_{-\infty}^{\infty}\left\{\int\limits_{-\infty}^{\infty}f(t-y)g(y)\,dy\right\}\varphi(z-t)\,dt\medskip\\
&=&\int\limits_{-\infty}^{\infty}f*g(t)\varphi(z-t)\,dt\medskip\\
&=&(\Lambda_{f*g},\varphi(z-\,\cdot\,))\medskip\\ &=&\Lambda_{f*g}*\varphi(z)
\end{eqnarray*}
for all $z\in R$. Therefore
$$
\Lambda_{f*g}*\varphi=\Lambda_f*(\Lambda_g*\varphi).
$$
As was to be proved. $\bullet$

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section*{10. Differential operators}

Consider the differential operator
$$
{\cal L}=\sum_{k=0}^la_k\frac{d^k}{dx^k}
$$
and the dual differential operator
$$
\stackrel{\vee}{{\cal L}}=\sum_{k=0}^l(-1)^ka_k\frac{d^k}{dx^k}.
$$
We assume that $a_l=1$.

We see that
$$
({\cal L}\Lambda,\varphi)=(\Lambda,\stackrel{\vee}{{\cal L}}\varphi)
$$
for all $\Lambda\in{\cal D}'(\Omega)$ and $\varphi\in C_0^\infty(\Omega)$. Besides, the
linear operators
$$
{\cal L}:\;{\cal D}'(\Omega)\longrightarrow {\cal D}'(\Omega)
$$
and
$$
\stackrel{\vee}{{\cal L}}:\;{\cal D}'(\Omega)\longrightarrow {\cal D}'(\Omega)
$$
are continuous.

Let $Z$ satisfy the differential equation
$$
{\cal L}Z=0
$$
with initial conditions
$$
\begin{array}{lcl}
\left.\begin{array}{l}
      Z^{(0)}(0)\,\;\;=0\\
      \cdots\\
      Z^{(l-2)}(0)=0\\
      \end{array}
\right\}&{\rm if}& l\ge 2\medskip\\ \,\;Z^{(l-1)}(0)=1&&
\end{array}
$$

\begin{theorem}
\label{lequ} There exists a linear continuous operator
$$
I_{\cal L}:\;{\cal D}'(\Omega)\longrightarrow {\cal D}'(\Omega)
$$
such that
$
{\cal L}I_{\cal L}(\Lambda)=\Lambda
$
for all $\Lambda\in{\cal D}'(\Omega)$.
\end{theorem}
{\bf Proof.} By Corollary to Theorem 17, there exists a linear continuous operator
$$
I:\;{\cal D}'(\Omega)\longrightarrow {\cal D}'(\Omega)
$$
such that
$$
\frac{d}{dx}I(\Lambda)=\Lambda
$$
for all $\Lambda\in {\cal D}'(\Omega)$.

Let
$$
\Phi=\left(\begin{array}{llcl} Z^{(0)}&Z^{(1)}&\cdots&Z^{(l-1)}\\
Z^{(1)}&Z^{(2)}&\cdots&Z^{(l)}\\ \cdots&\cdots&\cdots&\cdots\\
Z^{(l-1)}&Z^{(l)}&\cdots&Z^{(2l-2)}\end{array}\right).
$$
It is not hard to check that $ \det\Phi(x)\ne 0
$
for all $x\in \mathbb{R}$.

We claim that
$$
I_{\cal L}(\Lambda)=\sum_{k=0}^{l-1}Z^{(k)}I(\varphi_k\Lambda),
$$
where $\varphi_0,\ldots,\varphi_{l-1}\in C^\infty(\Omega)$ are elements of the last
column of the matrix $\Phi^{-1}$.

Put
$$
A=\left(\begin{array}{ccccc}
 0&1&0&\cdots&0\\
 0&0&1&\cdots&0\\
\cdots&\cdots&\cdots&\cdots&\cdots\\
 0&0&0&\cdots&1\\
 -a_0\;\,&-a_1\;&-a_2\;&\cdots&-a_{l-1}\end{array}\right)
 \quad\quad F=\left(\begin{array}{l} 0\\ 0\\ \;\vdots\\ 0\\
 \Lambda\end{array}\right)\quad\quad
 U=\left(\begin{array}{l} u_0\\ u_1\\ \;\vdots\\
 u_{l-2}\\ u_{l-1}\end{array}\right),
$$
where $\Lambda\in {\cal D}'(\Omega)$ and $u_0,\ldots,u_{l-1}\in {\cal D}'(\Omega)$. Note
that
$$
\Phi'=A\Phi
$$
Consider the differential equation
$$
{\cal L}u=\Lambda \eqno (1)
$$
and the system of differential equations
$$
U'=AU+F. \eqno (2)
$$
We see that (1) and (2) are equivalent in the following sense. If $u$ is a solution of
(1), then $U$ with $u_k=u^{(k)}$ is a solution of (2). If $U$ is a solution of (2), then
$u=u_0$ is a solution of (1). Let
$$
U=\Phi V.
$$
Since
$$
U'=\Phi'V+\Phi V'=A\Phi V+\Phi V'=AU+\Phi V',
$$
we see that (2) is equivalent to
$$
V'=\Phi^{-1}F. \eqno (3)
$$
From (3) we have
$$
\begin{array}{lll}
v_0'&=&\varphi_0\Lambda,\\ &\cdots&\\ v_{l-1}'&=&\varphi_{l-1}\Lambda.
\end{array}
$$
Therefore
$$
\begin{array}{lll}
v_0&=&I(\varphi_0\Lambda)+c_0,\\ &\cdots&\\ v_{l-1}&=&I(\varphi_{l-1}\Lambda)+c_{l-1}
\end{array}
$$
with some constants $c_0,\ldots,c_{l-1}$. Since $U=\Phi V$, we see that $u=u_0$ has the
representation
$$
u=\sum_{k=0}^{l-1}Z^{(k)}I(\varphi_k\Lambda)+\sum_{k=0}^{l-1}c_kZ^{(k)}.
$$
We rewrite the last formula in the form
$$
u=\sum_{k=0}^{l-1}c_kZ^{(k)}+I_{\cal L}(\Lambda).
$$
and obtain
$$
{\cal L}I_{\cal L}(\Lambda)=\Lambda.
$$
As was to be proved. $\bullet$
\bigskip

A distribution $E\in{\cal D}'(R)$ such that
$$
{\cal L}E=\delta_0
$$
is called a fundamental solution of ${\cal L}$.

\begin{theorem}
\label{fundsol} The function $E=Z\Theta$ is the fundamental solution of ${\cal L}$.
\end{theorem}
{\bf Proof.} By Theorem~\ref{divc1hev}, we have
$$
\begin{array}{lcl}
E^{(0)}&=&Z^{(0)}\Theta\\ E^{(1)}&=&Z^{(1)}\Theta\\ &\cdots&\\
E^{(l-1)}&=&Z^{(l-1)}\Theta\\ E^{(l)}&=&Z^{(l)}\Theta+\delta_0
\end{array}
$$
Therefore
$$
{\cal L}E=({\cal L}Z)\Theta+\delta_0=\delta_0.
$$
As was to be proved. $\bullet$
\bigskip

Let us put
$$
E=Z\Theta.
$$
We see that
$$
\stackrel{\vee}{E}(x)=E(-x)
$$
is a fundamental solution of  $\stackrel{\vee}{{\cal L}}$.

\begin{theorem}
\label{difcom} Let $\Lambda\in{\cal E}'(\mathbb{R})$. Then
$$
{\cal L}(E*\Lambda)=\Lambda.
$$
\end{theorem}
{\bf Proof.} Indeed, we see that
$$
{\cal L}(E*\Lambda)=({\cal L}E)*\Lambda=\delta*\Lambda=\Lambda.
$$
As was to be proved. $\bullet$
\bigskip

\noindent {\bf Remark.} Let
$$
{\cal L}(\lambda)=\lambda^l+a_{l-1}\lambda^{l-1}+\cdots+a_0
=(\lambda-\lambda_1)^{q_1}\ldots(\lambda-\lambda_m)^{q_m}
$$
where $q_i$ is the multiplicity of the root $\lambda_i$. The function $Z$ is
representable in the form
$$
Z(x)=\sum_{i=1}^m \left(\sum_{k=1}^{q_i}C_{ik}\frac{x^k}{k!}\right)e^{\lambda_ix},
$$
where
$$
C_{ik}=\left.\frac{1}{(q_i-k)!} \left[\frac{(\lambda-\lambda_i)^{q_i}}{{\cal
L}(\lambda)}\right]^{(q_i-k)} \right|_{\lambda=\lambda_i}.
$$






%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section*{11. Sobolev spaces}

Let $l\ge 0$ be an integer number. The space of all distributions $U\in {\cal
D}'(\Omega)$ such that
$
U^{(k)}\in L_2(\Omega)
$
for all $k=0,1,\ldots,l$ is denoted by $W^l_2(\Omega)$ and is called the Sobolev space.
The Sobolev space is a Hilbert space with the scalar product
$$
(U,V)_{W^l_2(\Omega)}=\sum_{k=0}^l(U^{(k)},V^{(k)})_{L_2(\Omega)}
$$
and the norm
$$
\|U\|^2_{W^l_2(\Omega)}=\sum_{k=0}^l\|U^{(k)}\|^2_{L_2(\Omega)}.
$$
Let $C^m[a,b]$ be the Banach space of $C^m$-functions  with the norm
$$
\| U\|_{C^k[a,b]}=\sum_{k=0}^m\| U^{(k)}\|_{C[a,b]}.
$$
\newpage

\begin{theorem}
\label{sobemb} Let $l\ge 1$. Then
$
W^l_2(\Omega)\subset C^{l-1}[a,b]
$
is an embedding.
\end{theorem}
{\bf Proof.} Let $\alpha\in\Omega$ and $U\in W^l_2(\Omega)$. We see that
$$
U^{(k)}(x)=U^{(k)}(\alpha) + \int\limits_\alpha^xU^{(k+1)}(t)\,dt
$$
for all $k=0,1,\ldots,l-1$. This means that $U\in C^{l-1}[a,b]$.

From
$$
U^{(k)}(\alpha)=U^{(k)}(x) - \int\limits_\alpha^xU^{(k+1)}(t)\,dt.
$$
we obtain the inequality
$$
(b-a)^{1/2}\mid U^{(k)}(\alpha)\mid \le\| U^{(k)}\|_{L_2(\Omega)} +(b-a)^{1/2}\|
U^{(k+1)}\|_{L_2(\Omega)}.
$$
Therefore,
$$
\| U^{(k)}\|_{C[a,b]} \le(b-a)^{-1/2}\| U^{(k)}\|_{L_2(\Omega)}+\|
U^{(k+1)}\|_{L_2(\Omega)}.
$$
This means that
$$
\| U\|_{C^{l-1}[a,b]}\le C\| U\|_{W^l_2(\Omega)}
$$
with some constant $C$. Hence
$
W^l_2(\Omega)\subset C^{l-1}[a,b]
$
is an embedding. $\bullet$
\bigskip

\begin{theorem}
øLet $U\in W^l_2(\Omega)$ and $\alpha\in\Omega$. Then
$$
U(x)=\sum_{k=0}^{l-1}U^{(k)}(\alpha)\frac{(x-\alpha)^k}{k!}+
\int\limits_\alpha^x\frac{(x-y)^{l-1}}{(l-1)!}U^{(l)}(y)\,dy.
$$
\end{theorem}
{\bf Proof.} We see that
$$
\int\limits_\alpha^x\frac{(x-y)^k}{k!}f(y)\,dy= \int\limits_\alpha^x
\left(\int\limits_\alpha^y\frac{(y-t)^{k-1}}{(k-1)!}f(t)\,dt\right)\,dy.
$$
Indeed,
\begin{eqnarray*}
\int\limits_\alpha^x
\left(\int\limits_\alpha^y\frac{(y-t)^{k-1}}{(k-1)!}f(t)\,dt\right)\,dy
&=&\int\limits_\alpha^x
\left(\int\limits_\alpha^x\frac{(y-t)_+^{k-1}}{(k-1)!}f(t)\,dt\right)\,dy\medskip\\
&=&\int\limits_\alpha^x
\left(\int\limits_\alpha^x\frac{(y-t)_+^{k-1}}{(k-1)!}\,dy\right)f(t)\,dt\medskip\\
&=&\int\limits_\alpha^x
\left(\int\limits_t^x\frac{(y-t)^{k-1}}{(k-1)!}\,dy\right)f(t)\,dt\medskip\\
&=&\int\limits_\alpha^x \frac{(x-t)^k}{k!}f(t)\,dt.
\end{eqnarray*}
Since
$$
U^{(k)}(x)=U^{(k)}(\alpha) + \int\limits_\alpha^xU^{(k+1)}(t)\,dt,
$$
where $k=0,1,\ldots,l-1$, we have
$$
U(x)=U(\alpha)+\int\limits_\alpha^xU'(y)\,dy=U(\alpha)+U'(\alpha)\frac{(x-\alpha)}{1!}+
 \int\limits_\alpha^x\frac{x-y}{1!}U''(y)\,dy=\cdots
$$
As was to be proved. $\bullet$
\bigskip

%An embedding is usually denoted by $\hookrightarrow$. We see that
%$$
%C^l[a,b]\hookrightarrow W^l_2(\Omega)\hookrightarrow C^{l-1}[a,b].
%$$

\begin{theorem}
\label{densob} The space $C^l[a,b]$ is dense in the space $W^l_2(\Omega)$.
\end{theorem}
{\bf Proof.} Let $U\in W^l_2(\Omega)$ and $\alpha\in\Omega$. Then
$$
U(x)=\sum_{k=0}^{l-1}U^{(k)}(\alpha)\frac{(x-\alpha)^k}{k!}+
\int\limits_\alpha^x\frac{(x-y)^{l-1}}{(l-1)!}U^{(l)}(y)\,dy.
$$
Since $U^{(l)}\in L_2(\Omega)$, given $\varepsilon>0$, there exists $f\in C[a,b]$ such
that
$$
\| U^{(l)}-f\|_{L_2(\Omega)}\le\varepsilon.
$$
Let
$$
F(x)=\sum_{k=0}^{l-1}U^{(k)}(\alpha)\frac{(x-\alpha)^k}{k!}+
\int\limits_\alpha^x\frac{(x-y)^{l-1}}{(l-1)!}f(y)\,dy.
$$
We see that $F\in C^l[a,b]$ and
$$
\| U-F\|_{W^l_2(\Omega)}\le C\varepsilon
$$
with some constant $C>0$. $\bullet$
\bigskip

Let us define the space
$$
\stackrel{0}{W^l_2}\!\!(\Omega)={\rm Cl}C^\infty_0(\Omega)
$$
where {\rm Cl} is the closure in the space $W^l_2(\Omega)$.

\begin{theorem}
\label{sobequ} Let $U\in W^l_2(\Omega)$. Then $U\in\stackrel{0}{W^l_2}\!\!(\Omega)$ if
and only if
$$
U^{(k)}(a)=U^{(k)}(b)=0
$$
for all $0\le k\le l-1$.
\end{theorem}
{\bf Proof. Sufficiency.} Let
$
U^{(k)}(a)=U^{(k)}(b)=0
$
for all $0\le k\le l-1$. Define the function
$$
V_\varepsilon(x)=\int\limits_{a+2\varepsilon}^{b-2\varepsilon}h_\varepsilon(x-y)\,dy.
$$
We claim that
$$
\lim_{\varepsilon\to 0}\| U-V_\varepsilon U\|_{W^l_2(\Omega)}=0.
$$
We see that
$$
(V_\varepsilon U)^{(k)}=V_\varepsilon U^{(k)}+\sum_{p=1}^k C^p_kV_\varepsilon^{(p)}
U^{(k-p)}
$$
and
$$
\lim_{\varepsilon\to 0}\| U^{(k)}-V_\varepsilon U^{(k)}\|_{L_2(\Omega)}=0
$$
for all $0\le k\le l$. Let us prove that
$$
\lim_{\varepsilon\to 0}\| V_\varepsilon^{(p)}U^{(k-p)}\|_{L_2(\Omega)}=0
$$
for all $1\le p\le k\le l$.

Since
$$
V_\varepsilon^{(p)}(x)=\int\limits_{a+2\varepsilon}^{b-2\varepsilon}h^{(p)}_\varepsilon(x-y)\,dy
,
$$
we have
$$
\| V_\varepsilon^{(p)}\|_{C[a,b]}\le \frac{C}{\varepsilon^p}
$$
where $C>0$ does not depend on $\varepsilon>0$ and $0\le p\le l$. Since
$$
U^{(k)}(a)=U^{(k)}(b)=0,
$$
where $0\le k\le l-1$, we have
$$
U^{(k-p)}(x)=\int\limits_a^x\frac{(x-y)^{l-k+p-1}}{(l-k+p-1)!}U^{(l)}(y)\,dy
$$
and
$$
U^{(k-p)}(x)=\int\limits_b^x\frac{(x-y)^{l-k+p-1}}{(l-k+p-1)!}U^{(l)}(y)\,dy,
$$
where $1\le p\le k\le l-1$. Therefore,
\begin{eqnarray*}
\| V_\varepsilon^{(p)}U^{(k-p)}\|_{L_2(\Omega)}^2 &=&\int\limits_a^b\mid
V_\varepsilon^{(p)}(x)U^{(k-p)}(x)\mid^2\,dx\medskip\\
&=&\int\limits_a^{a+\varepsilon}\mid V_\varepsilon^{(p)}(x)\mid^2
\left|\int\limits_a^x\frac{(x-y)^{l-k+p-1}}{(l-k+p-1)!}U^{(l)}(y)\,dy\right|^2
\,dx\medskip\\ &&{}+\int\limits_{b-\varepsilon}^b\mid V_\varepsilon^{(p)}(x)\mid^2
\left|\int\limits_b^x\frac{(x-y)^{l-k+p-1}}{(l-k+p-1)!}U^{(l)}(y)\,dy\right|^2
\,dx\medskip\\ &\le&\frac{C}{\varepsilon^{2p}}\left\{ \int\limits_a^{a+\varepsilon}
\left|\int\limits_a^x\frac{(x-y)^{l-k+p-1}}{(l-k+p-1)!}U^{(l)}(y)\,dy\right|^2
\,dx\right.\medskip\\ &&{}\left.+\int\limits_{b-\varepsilon}^b
\left|\int\limits_b^x\frac{(x-y)^{l-k+p-1}}{(l-k+p-1)!}U^{(l)}(y)\,dy\right|^2
\,dx\right\}\medskip\\ &\le&\frac{C}{\varepsilon^{2p}}\left\{
\int\limits_a^{a+\varepsilon}\mid U^{(l)}(y)\mid^2\,dy
\int\limits_a^{a+\varepsilon}\int\limits_a^x\left[
\frac{(x-y)^{l-k+p-1}}{(l-k+p-1)!}\right]^2\,dydx\right.\medskip\\
&&\left.+\int\limits_{b-\varepsilon}^b\mid U^{(l)}(y)\mid^2\,dy
\int\limits_{b-\varepsilon}^b\int\limits_x^b\left[
\frac{(x-y)^{l-k+p-1}}{(l-k+p-1)!}\right]^2\,dydx\right\}\medskip\\
&\le&\frac{C}{\varepsilon^{2p}}\varepsilon^{2(m-k+p-1)+2}
\left\{\int\limits_a^{a+\varepsilon}\mid U^{(l)}(y)\mid^2\,dy
+\int\limits_{b-\varepsilon}^b\mid U^{(l)}(y)\mid^2\,dy\right\}\medskip\\
&=&C\varepsilon^{2(m-k)} \left\{\int\limits_a^{a+\varepsilon}\mid U^{(l)}(y)\mid^2\,dy
+\int\limits_{b-\varepsilon}^b\mid U^{(l)}(y)\mid^2\,dy\right\}
\end{eqnarray*}
This means that
$$
\lim_{\varepsilon\to 0}\| V_\varepsilon^{(p)}U^{(k-p)}\|_{L_2(\Omega)}=0
$$
for all $1\le p\le k\le l$. Thus
$$
\lim_{\varepsilon\to 0}\| U-V_\varepsilon U\|_{W^l_2(\Omega)}=0.
$$
Note that
$
\supp(V_\varepsilon U)\subset [a+\varepsilon,b-\varepsilon].
$

Let $u\in L_2(\Omega)$. We assume that $u$ is extended by zero outside of $\Omega$. Let
us check the inequality
$$
\| u*h_\varepsilon\|_{L_2(\Omega)}\le \| u\|_{L_2(\Omega)}.
$$
Indeed,
\begin{eqnarray*}
\| u*h_\varepsilon\|_{L_2(\Omega)}
&=&\int\limits_a^b\left\{\int\limits_{-\varepsilon}^\varepsilon
[u(x-y)h_\varepsilon(y)^{1/2}]h_\varepsilon(y)^{1/2}\,dy\right\}\,dx
\medskip\\
&\le&\int\limits_a^b\left\{ \int\limits_{-\varepsilon}^\varepsilon\mid
u(x-y)\mid^2h_\varepsilon(y)\,dy \int\limits_{-\varepsilon}^\varepsilon
h_\varepsilon(y)\,dy\right\}\,dx
\medskip\\
&\le&\int\limits_{-\varepsilon}^\varepsilon h_\varepsilon(y)
\int\limits_{-\infty}^\infty\mid u(x-y)\mid^2h_\varepsilon(y)\,dxdy
\medskip\\
&=&\| u\|^2_{L_2(\Omega)}.
\end{eqnarray*}

Thus for any $\varepsilon>0$ there exists $\varepsilon_1>0$ such that
$$
\| U-V_{\varepsilon_1} U\|_{W^l_2(\Omega)}\le\frac{\varepsilon}{2}.
$$
Then there exists $\varepsilon_2>0$ such that
$$
\| V_{\varepsilon_1} U- (V_{\varepsilon_1} U)*h_{\varepsilon_2}\|_{W^l_2(\Omega)} \le
\frac{\varepsilon}{2}.
$$
We see that
$$
\| U-(V_{\varepsilon_1} U)*h_{\varepsilon_2}\|_{W^l_2(\Omega)}\le\varepsilon
$$
and
$$
(V_{\varepsilon_1} U)*h_{\varepsilon_2}\in C^\infty_0(\Omega).
$$
This proves the inclusion
$
U\in \stackrel{0}{W^l_2}\!\!(\Omega).
$

{\bf Necessity.} Let
$
U\in\stackrel{0}{W^l_2}\!\!(\Omega).
$
Then there exists a sequence $U_1,U_2,\ldots\in C^\infty_0(\Omega)$ such that
$$
\lim_{k\to\infty}U_k=U
$$
in the space $W_2^l(\Omega)$. Since
$$
U_k^{(p)}(a)=U_k^{(p)}(b)=0
$$
for all $0\le p\le l-1$, we have
$$
U^{(p)}(a)=U^{(p)}(b)=0
$$
for all $0\le p\le l-1$. $\bullet$
\bigskip

Consider the cone
$$
\stackrel{+}{C^\infty_0}\!\!(\Omega)= \{\varphi\in C^\infty_0(\Omega):\; \varphi\ge 0 \}
$$
of positive functions in the space $C^\infty_0(\Omega)$ and  the cone
$$
\stackrel{+}{W^l_2}\!\!(\Omega)= \{U\in\stackrel{0}{W^l_2}\!\!(\Omega):\; U\ge 0 \}
$$
of positive functions in the space $\stackrel{0}{W^l_2}\!\!(\Omega)$.



\noindent {\bf Corollary 1.} Let $U\in\stackrel{0}{W^l_2}\!\!(\Omega)$. Then given
$\varepsilon>0$, there exists $V\in\stackrel{+}{C^\infty_0}\!\!(\Omega)$ such that
$$
\| U-VU\|_{W_2^l(\Omega)}\le\varepsilon.
$$
\bigskip

\noindent {\bf Corollary 2.} We have
$
\stackrel{+}{W^l_2}\!\!(\Omega)=\Cl\stackrel{+}{C^\infty_0}\!\!(\Omega)
$
where Cl is the closure in the space $W^l_2(\Omega)$.
\bigskip

\noindent {\bf Corollary 3.} Let $U\in W^l_2(\Omega)$ and
$
U^{(k)}(a)=U^{(k)}(b)=0
$
for all $0\le k\le l-1$. If $U$ is extended by zero outside of $\Omega$, then $U\in
\stackrel{0}{W^l_2}\!\!(\mathbb{R})$.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section*{12. The Green function}

Consider the differential operators
$$
{\cal L}=\sum_{k=0}^la_k\frac{d^k}{dx^k}
$$
and
$$
\stackrel{\vee}{{\cal L}}=\sum_{k=0}^l(-1)^ka_k\frac{d^k}{dx^k}.
$$
Let $f\in L_2(\Omega)$. We assume that $f$ is extended by zero outside of $\Omega$. Hence
$f\in L_2(\mathbb{R})$ and $\supp f\subset [a,b]$. In particular, $f\in {\cal E}'(R)$. By
Theorem~\ref{difcom} and Theorem~\ref{conl1}, the differential equation
$$
{\cal L}u=f
$$
has a solution $u=E*f$. We see that
$$
u(x)=\int\limits_a^bE(x-y)f(y)\,dy=\int\limits_a^xZ(x-y)f(y)\,dy.
$$
Define the operator
$$
\stackrel{\vee}{E}*:\;L_2(\Omega)\longrightarrow W^l_2(\Omega)
$$
by the rule
$$
\stackrel{\vee}{E}*:\;u\longmapsto (\stackrel{\vee}{E}*u)\mid_\Omega.
$$
We see that
$$
\stackrel{\vee}{E}*u(x)=\int\limits_x^bZ(y-x)u(y)\,dy.
$$


\begin{theorem}
\label{L2Wl2} The operator
$$
\stackrel{\vee}{E}*:\;L_2(\Omega)\longrightarrow W^l_2(\Omega)
$$
is continuous.
\end{theorem}
{\bf Proof.} Let $u\in L_2(\Omega)$. Since
$$
\stackrel{\vee}{E}{}^{(k)}=(-1)^{k}(Z^{(k)}\Theta)^\vee
$$
for $0\le k \le l-1$ and
$$
\stackrel{\vee}{E}{}^{(l)}=(-1)^{l}(Z^{(l)}\Theta)^\vee+(-1)^{l}\delta_0,
$$
we have
$$
(\stackrel{\vee}{E}*u)^{(k)}(x)= (\stackrel{\vee}{E}{}^{(k)}*u)(x)=
(-1)^k\int\limits_x^bZ^{(k)}(y-x)u(y)\,dy
$$
for $0\le k \le l-1$ and
$$
(\stackrel{\vee}{E}*u){}^{(l)}(x)= (\stackrel{\vee}{E}{}^{(l)}*u)(x)=
(-1)^l\int\limits_x^bZ^{(l)}(y-x)u(y)\,dy+(-1)^lu(x).
$$
Therefore
$$
\|\stackrel{\vee}{E}*u\|_{W^l_2(\Omega)}\le C\|u\|_{L^2(\Omega)}
$$
where $C>0$ does not depend on $u\in L_2(\Omega)$. $\bullet$


Define the space
$$
P_{\cal L}(\Omega)=\{u\in L_2(\Omega):\; {\cal L}u=0\}.
$$
Obviously
$$
\stackrel{\vee}{\cal L}:\;W_2^l(\Omega)\longrightarrow L_2(\Omega)
$$
is a linear continuous operator.

\begin{theorem}
\label{n1linis} The operators
$$
\begin{array}{llll}
\stackrel{\vee}{E}*:&P_{\cal L}(\Omega)^\perp
&\longrightarrow&\stackrel{0}{W^l_2}\!\!(\Omega)\medskip\\ \stackrel{\vee}{{\cal
L}}:&\stackrel{0}{W^1_2}\!\!(\Omega) &\longrightarrow&P_{\cal L}(\Omega)^\perp
\end{array}
$$
are inverse continuous isomorphisms.
\end{theorem}
{\bf Proof. 1.} Let $u\in P_{\cal L}(\Omega)^\perp$ and $U=\stackrel{\vee}{E}*u$. We note
that $Z^{(k)}(\,\cdot\,-x)\in P_{\cal L}(\Omega)$ for all $k=0,1,\ldots,l-1$. Since
$$
(\stackrel{\vee}{E}*u)^{(k)}(x)=(-1)^k\int\limits_x^bZ^{(k)}(y-x)u(y)\,dy
$$
for all $k=0,1,\ldots,l-1$, we have
$$
U(a)=(\stackrel{\vee}{E}*u)^{(k)}(a)=(-1)^k\int\limits_a^bZ^{(k)}(y-a)u(y)\,dy=0
$$
and
$$
U(b)=(\stackrel{\vee}{E}*u)^{(k)}(b)=(-1)^k\int\limits_b^bZ^{(k)}(y-b)u(y)\,dy=0
$$
all $k=0,1,\ldots,l-1$. By Theorem~\ref{sobequ}, we conclude that
$U\in\stackrel{0}{W^l_2}\!\!(\Omega)$.

{\bf 2.} Let $U\in \stackrel{0}{W^l_2}\!\!(\Omega)$ and $v\in P_{\cal L}(\Omega)$. Let
$\varphi_1,\varphi_2,\ldots\in C^\infty_0(\Omega)$ be such that
$$
\lim_{k\to\infty}\varphi_k=U
$$
in the space $W_2^l(\Omega)$. Then
$$ (\stackrel{\vee}{{\cal
L}}U,v)_{L_2(\Omega)} =\lim_{k\to\infty}(\stackrel{\vee}{{\cal
L}}\varphi_k,v)_{L_2(\Omega)} =\lim_{k\to\infty}(\varphi_k,{\cal L}v)_{L_2(\Omega)}=0
$$
Therefore $\stackrel{\vee}{{\cal L}}U\in P_{\cal L}(\Omega)^\perp$.

{\bf 3.} By Theorem~\ref{difcom}, we have
$$
\stackrel{\vee}{{\cal L}}(\stackrel{\vee}{E}*u)=u
$$
for all $u\in L_2(\Omega)$. Let $U\in \stackrel{0}{W^l_2}\!\!(\Omega)$ and
$
\tilde U=\stackrel{\vee}{E}*(\stackrel{\vee}{{\cal L}}U).
$
Obviously
$
\stackrel{\vee}{{\cal L}}\tilde U=\stackrel{\vee}{{\cal L}}U
$
Since
$
\stackrel{\vee}{{\cal L}}(\tilde U-U)=0,
$
and since
$
\tilde U-U\in\stackrel{0}{W^l_2}\!\!(\Omega),
$
we have
$
\tilde U=U.
$
Therefore
$$
\stackrel{\vee}{E}*(\stackrel{\vee}{{\cal L}}U)=U
$$
for all $U\in \stackrel{0}{W^l_2}\!\!(\Omega)$. $\bullet$
\bigskip





\begin{theorem}
\label{scapro} The space $\stackrel{0}{W^l_2}\!\!(\Omega)$ is a Hilbert space with the
scalar product
$$
(U,V)_{\cal L}= (\stackrel{\vee}{{\cal L}}U,\stackrel{\vee}{{\cal L}}V)_{L_2(\Omega)}
$$
and the norm
$$
\|U\|_{\cal L}=\|\stackrel{\vee}{{\cal L}}U\|_{L_2(\Omega)}.
$$
\end{theorem}
{\bf Proof.} We claim that the norms $\|\,\cdot\,\|_{\cal L}$ and
$\|\,\cdot\,\|_{W_2^l(\Omega)}$ are equivalent in the space
$\stackrel{0}{W^l_2}\!\!(\Omega)$. This means that
$$
C_0\|U\|_{W^l_2(\Omega)}\le \|U\|_{\cal L}\le C_1\|U\|_{W^l_2(\Omega)}
$$
for some constants $C_0,C_1>0$. Let $U\in\stackrel{0}{W^l_2}\!\!(\Omega)$ and
$u=\stackrel{\vee}{{\cal L}}U$.

On the one hand,
$$
(\stackrel{\vee}{E}*u)^{(k)}(x)= (-1)^k\int\limits_x^bZ^{(k)}(y-x)u(y)\,dy
$$
for $0\le k \le l-1$ and
$$
(\stackrel{\vee}{E}*u)^{(l)}(x)= (-1)^l\int\limits_x^bZ^{(l)}(y-x)u(y)\,dy+(-1)^lu(x).
$$
Hence
$$
U^{(k)}(x)= (-1)^k\int\limits_x^bZ^{(k)}(y-x)\stackrel{\vee}{{\cal L}}U(y)\,dy
$$
for $0\le k \le l-1$ and
$$
U^{(l)}(x)=(-1)^l\int\limits_x^bZ^{(l)}(y-x)\stackrel{\vee}{{\cal L}}U(y)\,dy
+(-1)^l\stackrel{\vee}{{\cal L}}U(x).
$$
Therefore
$$
\|U\|_{W^l_2(\Omega)}\le C\|U\|_{\cal L}
$$
with a constant $C>0$.

On the other hand,
\begin{eqnarray*}
\|U\|_{\cal L}^2 &=&(\stackrel{\vee}{{\cal L}}U,\stackrel{\vee}{{\cal L}}U)_{L_2(\Omega)}
\medskip\\
&=&\sum_{i=0}^l\sum_{j=0}^l(-1)^{i+j}a_ia_j(U^{(i)},U^{(j)})_{L_2(\Omega)}
\medskip\\
&\le&\sum_{i=0}^l\sum_{j=0}^l\mid a_i\mid\mid a_j\mid
\|U^{(i)}\|_{L_2(\Omega)}\|U^{(j)})\|_{L_2(\Omega)}\medskip\\ &=&\left(\sum_{i=0}^l\mid
a_i\mid\|U^{(i)}\|_{L_2(\Omega)}\right)^2
\medskip\\
&\le&C\|U\|_{W^l_2(\Omega)}^2
\end{eqnarray*}
with a constant $C>0$. $\bullet$
\bigskip

The function $G(x,y)$ is called the Green function of the differential operator ${\cal
L}\stackrel{\vee}{{\cal L}}$ if
$$
{\cal L}\stackrel{\vee}{{\cal L}}G(\,\cdot\,,y)=\delta_y
$$
and
$$
G(\,\cdot\,,y)\in\stackrel{0}{W^l_2}\!\!(\Omega)
$$
for any $y\in\Omega$.



Let ${\cal E}$ be a fundamental solution of the differential operator ${\cal
L}\stackrel{\vee}{{\cal L}}$, and let the function $R(x,y)$ satisfy the differential
equation
$$
{\cal L}_x\stackrel{\vee}{{\cal L}}_xR(x,y)=0
$$
with boundary conditions
\begin{eqnarray*}
 R^{(k)}_x(a,y)&=&{\cal E}^{(k)}(a-y),\\
 R^{(k)}_x(b,y)&=&{\cal E}^{(k)}(b-y),\\
\end{eqnarray*}
where $k=0,1,\ldots,l-1$. Then obviously
$$
G(x,y)={\cal E}(x-y)-R(x,y)
$$
the Green function of the differential operator ${\cal L}\stackrel{\vee}{{\cal L}}$.

Define the function
$$
\gamma_y={\cal P}_{P_L(\Omega)^\perp} \left\{E(\,\cdot\,-y)\mid_\Omega\right\},
$$
where
$$
{\cal P}_{P_L(\Omega)^\perp}:\;L_2(\Omega)\longrightarrow P_L(\Omega)^\perp
$$
is the projection.

\begin{theorem}
\label{n1grpre} The Green function of the operator ${\cal L}\stackrel{\vee}{{\cal L}}$
has the representation
$$
G(x,y)=(\gamma_x,\gamma_y)_{L_2(\Omega)}.
$$
\end{theorem}
{\bf Proof.} We see that
$$
(\gamma_x,\gamma_y)_{L_2(\Omega)}=(\stackrel{\vee}{E}*\gamma_y)(x).
$$
Since $\gamma_y\in P_L(\Omega)$, by Theorem~\ref{n1linis}, we have
$$
G(\,\cdot\,,y)\in\stackrel{0}{W^l_2}\!\!(\Omega).
$$
Then
$$
{\cal L}\stackrel{\vee}{{\cal L}}(\stackrel{\vee}{E}**\gamma_y)= {\cal
L}\gamma_y=\delta_y.
$$
This proves the representation. $\bullet$
\bigskip


For example, calculate the Green function $G(x,y)$ for the differential operator
$$
{\cal L}=\frac{d^l}{dx^l}
$$
in the interval
$
\Omega=(-1,1).
$

\begin{theorem}
\label{n1green} The Green function of the differential operator
$$
(-1)^l\frac{d^{2l}}{dx^{2l}}
$$
has the representation
$$
G(x,y)=\frac{(-1)^l}{2(2l-1)!}\mid x-y\mid^{2l-1}-R(x,y)
$$
with
$$
R(x,y)=\frac{(-1)^l}{2(2l-1)!}
(1-xy)\sum_{k=0}^{l-1}C_k[(x^2-1)(y^2-1)]^k(x-y)^{2(l-1-k)},
$$
where
$$
C_k=\frac{\bigl((1+\lambda)^{-1/2}\bigr)^{(k)}}{k!}(0) =\frac{(-1)^k}{2^{k-1}}C^k_{2k-1}.
$$
\end{theorem}
{\bf Proof.} Let $y\in\Omega$. Since
$$
(-1)^l\frac{d^{2l}}{dx^{2l}}G(x,y)=\delta_y,
$$
we have only check the boundary conditions. Let $x\ne y$, then
$$
\frac{x^2y^2-2xy+1}{\mid x-y\mid^2}=1+\frac{(x^2-1)(y^2-1)}{\mid x-y\mid^2}.
$$
Therefore
$$
G(x,y)=\frac{(-1)^l}{2(2l-1)!}\mid x-y\mid^{2l-1} g\left(\frac{(x^2-1)(y^2-1)}{\mid
x-y\mid^2}\right)
$$
where
$$
g(\lambda)=(1+\lambda)^{1/2}
\Bigl\{(1+\lambda)^{-1/2}-\sum_{k=0}^{l-1}C_k\lambda^k\Bigr\}.
$$
Since
$$
g(0)=g'(0)=\ldots=g^{(l-1)}(0)=0,
$$
we have
$$
\frac{d^k}{dx^k}G(-1,y)=\frac{d^k}{dx^k}G(1,y)=0
$$
for all $k=0,1,\ldots,l-1$. $\bullet$
\bigskip

\noindent {\bf Corollary.} We have
$
G(x,y)>0
$
for all $x,y\in \Omega$.

\noindent {\bf Proof.} Obviously
$
G(x,x)>0
$
for all $x\in \Omega$. Let $x\ne y$. By Theorem~\ref{n1green}, we have
$$
G(x,y)=\frac{(-1)^l}{2(2l-1)!}\mid x-y\mid^{2l-1} (1+\lambda)^{1/2}f(\lambda)
$$
where
$$
\lambda=\frac{(x^2-1)(y^2-1)}{\mid x-y\mid^2}
$$
and
$$
f(\lambda)=(1+\lambda)^{-1/2}-\sum_{k=0}^{l-1}C_k\lambda^k.
$$
We note that
$$
f(0)=f'(0)=\ldots=f^{(l-1)}(0)=0
$$
and
$$
f^{(l)}(\lambda)=(-1)^l\frac{1}{2}\frac{3}{2}\cdots\frac{2l-1}{2} (1+\lambda)^{-1/2-l}.
$$
Therefore
$$
{\rm sign}f(\lambda)=(-1)^l
$$
for all $\lambda>0$. $\bullet$
\bigskip


\begin{thebibliography}{9}
\bibitem{Gel} I.M. Gelfond, G.E.Shilov. Genegalized functions. Vol. 1, Moscow, 1958.

\bibitem{Nat} I.P.Natanson. The theory of functions of the real variable. Moscow, "Nauka", 1974.

\end{thebibliography}








\end{document}
