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\begin{document}
\noindent
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\begin{center}
{\bf{\Large\bf Section 3.7:} 2, 5 (optional)}
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\begin{enumerate}
\item[2.] Let $(X,d)$ be a complete metric space, and let $f:X\to X$
  be a mapping such that
  \begin{align*}f^{(p)}\doteq \underbrace{f\circ f\circ\cdots\circ
    f}_{p}
  \end{align*}
  is a contraction for some integer $p\geq 2$, with contraction constant
  $\gamma\in[0,1)$.  In other words, $d(f^{(p)}(x), f^{(p)}(y))\leq
  \gamma d(x,y)$ for all $x,y\in X$.  Note that the fact that $f^{(p)}$ is a
  contraction does not imply that $f$ is continuous.  Show that there
  is one and only one fixed point $x$ of $f$, and that, for any
  $x_0\in X$, the sequence generated by $x_{n+1}=f(x_n)$ for $n\geq
  0$ converges to $x$.
 

\item[5.] Let $(X,d)$ be a metric space, $A\subset X$, and  $L\geq 0$.
  Suppose that $f:A\to\RR$ is Lipschitz continuous (on $A$), with
  Lipschitz constant $L$.  Show that there is a $g:X\to\RR$ such that
  $g(x)=f(x)$ for all $x\in A$, and that $g$ is Lipschitz continuous (on $X$), with
  Lipschitz constant $L$.  This result is often called the MacShane Lemma.
 
\end{enumerate}


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