|
|
|
Viewing Version
1
of
'equivalence of conditions (2) and (3)'
|
[ view 'equivalence of conditions (2) and (3)'
|
back to history
]
| Title of object: |
equivalence of conditions (2) and (3) |
| Canonical Name: |
EquivalenceOfConditions2And3 |
| Type: |
Definition |
| Created on: |
2003-07-10 04:14:00 |
| Modified on: |
2003-07-10 04:14:00 |
| Classification: |
msc:46F05, msc:46-00 |
Preamble:
% this is the default PlanetMath preamble. as your knowledge
% of TeX increases, you will probably want to edit this, but
% it should be fine as is for beginners.
% almost certainly you want these
\usepackage{amssymb}
\usepackage{amsmath}
\usepackage{amsfonts}
% used for TeXing text within eps files
%\usepackage{psfrag}
% need this for including graphics (\includegraphics)
%\usepackage{graphicx}
% for neatly defining theorems and propositions
%\usepackage{amsthm}
% making logically defined graphics
%\usepackage{xypic}
% there are many more packages, add them here as you need them
% define commands here
\newcommand{\sR}[0]{\mathbb{R}}
\newcommand{\sC}[0]{\mathbb{C}}
\newcommand{\sN}[0]{\mathbb{N}}
\newcommand{\sZ}[0]{\mathbb{Z}}
\renewcommand{\bibname}{References} |
Content:
\newcommand{\cD}[0]{\mathcal{D}}
\newcommand{\scomp}[0]{C^\infty_0}
(The below proof follows \cite{hormander}, pp. 35.)
The proof that (3) implies (2) is a direct calculation.
Next, let us show that (2) implies (3):
Suppose $Tu_i \to 0$ in $\sC$,
if $K$ is a compact set in $U$, and
$\{u_i\}_{i=1}^\infty$ is a sequence in $\cD_K$ such that
for any multi-index $\alpha$, we have
$$ D^\alpha u_i \to 0$$
in the supremum norm $||\cdot||_\infty$ as $i\to \infty$.
For a contradiction, suppose there is a compact set $K$ in $U$
such that for all constants $C>0$ and $k\in\{0, 1,2,\ldots\}$ there exists
a function $u\in \cD_K$ such that
$$|T(u)|> C\sum_{|\alpha|\le k} ||D^\alpha u||_\infty.$$
Then, for $C=k=1,2,\ldots$ we obtain functions $u_1,u_2,\ldots$ in $\cD(K)$
such that
$ |T(u_i)| > i\sum_{|\alpha|\le i} ||D^\alpha u_i||_\infty.$
Thus $|T(u_i)|>0$ for all $i$, so for $v_i=u_i/|T(u_i)|$, we have
$$ 1> i\sum_{|\alpha|\le i} ||D^\alpha v_i||_\infty.$$
It follows that $||D^\alpha u_i||_\infty< 1/i$
for any multi-index $\alpha$ with $|\alpha|\le i$.
Thus $\{v_i\}_{i=1}^\infty$ satisfies our assumption, whence $T(v_i)$ should tend to $0$.
However, for all $i$, we have $T(v_i)= 1$. This contradiction
completes the proof. $\Box$
\begin{thebibliography}{9}
\bibitem{hormander}
L. H\"ormander, \emph{The Analysis of Linear Partial Differential Operators I,
(Distribution theory and Fourier Analysis)}, 2nd ed, Springer-Verlag, 1990.
\end{thebibliography} |
|
|
|
|
|
