BarbualatsLemma 2004-12-10 12:26:52 2005-04-09 15:00:55 Theorem % 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 \theoremstyle{theorem} \newtheorem*{thm}{Theorem} \newtheorem*{lemma}{Lemma} \newtheorem*{conj}{Conjecture} \newtheorem*{cor}{Corollary} \theoremstyle{definition} \newtheorem*{defn}{Definition} \begin{lemma}[Barb\u{a}lat] Let $f \colon (0,\infty) \to {\mathbb{R}}$ be Riemann integrable and uniformly continuous then \begin{equation*} \lim_{t \to \infty} f(t) = 0 . \end{equation*} \end{lemma} Note that if $f$ is non-negative, then Riemann integrability is the same as being $L^1$ in the sense of Lebesgue, but if $f$ oscillates then the Lebesgue integral may not exist. Further note that the uniform continuity is required to prevent sharp ``spikes'' that might prevent the limit from existing. For example suppose we add a spike of height 1 and area $2^{-n}$ at every integer. Then the function is continuous and $L^1$ (and thus Riemann integrable), but $f(t)$ would not have a limit at infinity. \begin{thebibliography}{9} \bibitem{LoRy} Hartmut Logemann, Eugene P.\@ Ryan. \PMlinkescapetext{Asymptotic behaviour of nonlinear systems}. \emph{The American Mathematical Monthly}, 111(10):864--889, 2004. \end{thebibliography}