\documentclass[12pt]{article} \usepackage[fleqn]{amsmath} % This set of parameters are used to distribute the assignment to the class (in paper form) % and for posting on the class web site (for those who missed the class). % With the pdf option the information contained in the keys below are placed % in the document info of the PDF document. If you don't have the AcroTeX Bundle % installed, remove the pdf option. \usepackage[pdf,forpaper,cfg=hw,nopoints,nosolutions]{eqexam} % Note: When using a PDF option like pdf, you need to specify a driver % that is passed to hyperref, web.sty etc. For example, % \usepackage[pdf,pdftex,forpaper,cfg=hw,nopoints,nosolutions]{eqexam} % This set of parameters are used to publish the solutions on the class web site, if % desired. % \usepackage[pdf,forpaper,cfg=hw,pointsonleft,answerkey]{eqexam} % Try compiling the file with vspacewithsolns % % \usepackage[pdf,forpaper,cfg=hw,pointsonleft,vspacewithsolns]{eqexam} % Note the use of the myconfigi parameter. This then inputs eqexami.cfg, there I % have placed some definitions specific to a homework assignment. \subject[AC2]{Advanced Calculus II} \title[HW1]{HW \#1} \author{Dr.\ D. P. Story} \date{Spring 2005} \duedate{01/28/05} \keywords{Homework due \theduedate} \solAtEndFormatting{\eqequesitemsep{3pt}} \begin{document} \maketitle \begin{exam}{HW} \ifanswerkey \begin{instructions}[Solutions] Below, please find a set of solutions to this assignment. \end{instructions} \else \begin{instructions}[] Assignments should be neatly-written, well-organized and concise. If you miss a class and need to get an assignment, see \[ \text{\url{http://www.math.uakron.edu/~dpstory/}} \] All class assignments and other announcements will be posted on this web site. \end{instructions} \fi \begin{eqComments}[]\S4.3, page 155, in the text\end{eqComments} \begin{problem}[4] Problem 15. Use the definition to prove $f(x) = x^2$ is convex on $\mathbb{R}$. \begin{solution} Let $[c,d]$ be any interval and let $t\in[0,1]$, we need to prove \begin{equation} f\bigl( (1-t)c + td \bigr) \le (1-t) f(c) + tf(d)\label{eq0} \end{equation} or, \begin{equation} \bigl( ( 1-t )c + td \bigr)^2 \le (1-t) c^2 + t d^2\label{eq2} \end{equation} We show that the right-side minus the left-side in \eqref{eq2} is nonnegative. Indeed, \begin{align*} (1-t) c^2 + &t d^2 - \left( ( 1-t )c + td \right)^2 \\& = (1-t) c^2 + t d^2 - \left( ( 1-t )^2 c^2 + 2t(1-t)cd + t^2d^2 \right)\\& = (1-t)[1-(1-t)]c^2 - 2t(1-t)cd + t(1-t)d^2\\& = t(1-t)c^2 - 2t(1-t)cd + t(1-t)d^2\\& = t(1-t)( c - d )^2 \ge 0 \end{align*} From the first and last lines we have $(1-t) c^2 + t d^2 - \left( ( 1-t )c + td \right)^2\ge0$. This is equivalent to the desired inequality~\eqref{eq2}. \eqfititin{$\square$} \medskip\noindent\textit{Alternate Solution}: We apply the \textbf{Cauchy-Schwartz Inequality}, page.~16, to the expression on the left side of line~\eqref{eq2}. For convenience, I paraphrase the \textbf{Cauchy-Schwartz Inequality}: \[ \left(\sum_{k=1}^n a_k b_k \right)^2 \le \left(\sum_{k=1}^n a_k^2 \right) \left(\sum_{k=1}^n b_k^2 \right) \] Applying this inequality, with $a_1 = \sqrt{1-t}$, $b_1 = \sqrt{1-t}\,c$, $a_2 = \sqrt{t}$, $b_2 = \sqrt{t}\,d$ (here, $n=2$, two terms), we obtain, \begin{align*} (1-t)^2 c^2 + t^2 d^2 & \le \left( (\sqrt{1-t})^2 + (\sqrt{t})^2\right)\left((\sqrt{1-t}\,c)^2 + (\sqrt{t}\,d)^2\right)\\& = (1-t)c^2 + td^2 \end{align*} Thus, \[ (1-t)^2 c^2 + t^2 d^2 \le (1-t)c^2 + td^2 \] which is line~\eqref{eq2}, what we wanted to prove. \end{solution} \end{problem} \begin{problem}[3] Problem 18. Prove the sum of two convex functions is convex. \begin{solution} Seems simple enough. Suppose $f$ and $g$ be convex on $I$. Let $[\,c,d\,]\subseteq$ and let $t\in[\,0,1\,]$. Then \begin{align*} (f+g)\bigl( (1-t) c + td \bigr) & = f\bigl( (1-t) c + td \bigr) + g\bigl( (1-t) c + td \bigr)\\& \le (1-t) f(c) + tf(d) + (1-t) g(c) + tg(d)\\& = (1-t) (f+g)(c) + t(f+g)(d) \end{align*} Thus, $(f+g)\bigl( (1-t) c + td \bigr) \le (1-t) (f+g)(c) + t(f+g)(d)$, which is what we wanted to prove. \end{solution} \end{problem} \begin{problem}[2] Problem 20. Give an example of a function that is convex and unbounded on $(0,1)$. \begin{solution} Let $ f(x) = 1/x $, $ x \in (0,1) $. This function is clearly unbounded and since $ f''(x) = 1/x^3\ge 0$ on $(0,1)$, it is convex on $(0,1)$. \end{solution} \end{problem} \begin{problem}[4] Problem 21. Define \[ f(x) = \begin{cases} 2, & x = -1;\\ x^2, & -1 < x < 2;\\ 5, & x = 2 \end{cases} \] Show $f$ is convex on $[\,-1,2\,]$ but not continuous on $[\,-1,2\,]$. \begin{solution} Define $g(x) = x^2$, $x\in[\,-1,2\,]$. Then $g$ is twice differentiable on $[\,-1,2\,]$ and $ g''(x) = 2\ge 0$, hence, $g$ is convex on $[\,-1,2\,]$. Note that $ g(x) \le f(x) $ for all $x\in[\,-1,2\,]$. Let $[\,c,d\,]\subseteq [\,-1,2\,]$, we need to show, $\forall t \in [\,0,1\,]$, $$ f\bigl( (1-t)c + td \bigr) \le (1-t) f(c) + tf(d) $$ This inequality is \emph{always true} for $t=0$ and $t=1$, so it suffices to assume $t\in(0,1)$, this implies $(1-t)c \ne -1$ and $ td \ne 2$, hence, $(1-t)c + td\in(-1,2)$ . Thus, $$ f\bigl( (1-t)c + td \bigr) = g\bigl( (1-t)c + td \bigr)) \le (1-t) g(c) + tg(d) = (1-t) f(c) + tf(d) $$ As the assertion about the discontinuity of $f$ (at its endpoints) is obvious, this completes the proof. \end{solution} \end{problem} \begin{problem}[3] Problem 23. Suppose $f$ is convex on $\mathbb R$, prove $f$ is continuous on $\mathbb R$. \begin{solution} This is an application of \textbf{Theorem~4.28}. Let $x\in\mathbb R$, enclose $x$ in a open interval $(a,b)$, where $a$, $b\in\mathbb R$. Then $f$ is convex on $(a,b)$, since it is convex on $\mathbb R$, so by \textbf{Theorem~4.28}, $f$ is continuous on $(a,b)$. Since $f$ is continuous on $(a,b)$, it is, in particular, continuous at $x\in(a,b)$. We have shown that for any $x\in\mathbb R$, $f$ is continuous at $x$, this means that $f$ is continuous on $\mathbb R$. \end{solution} \end{problem} \end{exam} \end{document}