ProofofBarbualatsLemma 2005-04-06 03:55:47 2005-04-14 03:28:57 Proof Barb\u{a}lat's lemma 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 We suppose that $y(t) \not\rightarrow 0$ as $t \rightarrow \infty$. There exists a sequence $(t_{n})$ in $\mathbb{R}_{+}$ such that $t_{n} \rightarrow \infty$ as $n \rightarrow \infty$ and $|y(t_{n})| \geq \varepsilon$ for all $n \in \mathbb{N}$. By the uniform continuity of $y$, there exists a $\delta > 0$ such that, for all $n \in \mathbb{N}$ and all $t \in \mathbb{R}$, \[ |t_{n} - t| \leq \delta \; \Rightarrow \; |y(t_{n}) - y(t)| \leq \frac{\varepsilon}{2}. \] So, for all $t \in [t_{n}, t_{n}+\delta]$, and for all $n \in \mathbb{N}$ we have \begin{eqnarray*} |y(t)| & = & |y(t_{n}) - (y(t_{n})- y(t))| \geq |y(t_{n})| - |y(t_{n})- y(t)| \geq \\ & \geq & \varepsilon - \frac{\varepsilon}{2} = \frac{\varepsilon}{2}. \end{eqnarray*} Therefore, \[ \left| \int_{0}^{t_{n}+\delta} y(t) dt - \int_{0}^{t_{n}} y(t) dt \right| = \left| \int_{t_{n}}^{t_{n}+\delta} y(t) dt \right| = \int_{t_{n}}^{t_{n}+\delta} |y(t)| dt \geq \frac{\varepsilon \delta}{2} > 0 \] for each $n \in \mathbb{N}$. By the hypothesis, the improprer Riemann integral $\int_{0}^{\infty} y(t) dt$ exists, and thus the left hand side of the inequality converges to 0 as $n \rightarrow \infty$, yielding a contradiction.