EquivalenceOfConditions2And3 2003-07-10 04:14:00 2007-06-28 12:39:51 Proof distribution 0 % 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}} \newcommand{\cD}[0]{\mathcal{D}} \newcommand{\scomp}[0]{C^\infty_0} Let us show the equivalence of (2) and (3). First, the proof that (3) implies (2) is a direct calculation. Next, let us show that (2) implies (3): Suppose $Tu_i \to 0$ in $\sC$, and 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 $\lVert\cdot\rVert_\infty$ 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.