Exercise 5.3.1

Question

Recall from Exercise $4.4 .9$ that a function $f: A ightarrow \mathbf{R}$ is Lipschitz on $A$ if there exists an $M>0$ such that
  $$
  \left|\frac{f(x)-f(y)}{x-y}ight| \leq M
  $$
  for all $x 
eq y$ in $A$
  \begin{enumerate}[label=(\alph*)] 
  \item Show that if $f$ is differentiable on a closed interval $[a, b]$ and if $f^{\prime}$ is continuous on $[a, b]$, then $f$ is Lipschitz on $[a, b]$.
  \item Review the definition of a contractive function in Exercise 4.3.11. If we add the assumption that $\left|f^{\prime}(x)ight|<1$ on $[a, b]$, does it follow that $f$ is contractive on this set?
   \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] 
  \item Since $f'$ is continuous on the compact set $[a,b]$ we can set $M$ such that $|f'(x)| \le M$ over $[a,b]$, then pick $x,y \in [a,b]$ with $x<y$. Apply MVT on $[x,y]$ to get a $c \in (x,y)$ with
    $$
    \frac{f(x)-f(y)}{x-y} = f'(c)
    $$
    Which implies
    $$
    \left|\frac{f(x)-f(y)}{x-y}\right| = |f'(c)| \le M
    $$
    Since $x,y$ were arbitrary this shows $f$ is Lipschitz.

\item For $f$ to be contractive we need some $c \in (0, 1)$ with $|f(x)-f(y)| \le c|x-y|$.
Let $x$ and $y$ be arbitrary, and consider
    $$c = \left|\frac{f(x) - f(y)}{x -y} \right|$$
By the mean value theorem, there must be some $d \in (0, 1)$ where $\left|f'(d)\right| = c$. But since $x$ and $y$ are arbitrary and $\left|f'(d)\right| < 1$, we must have $c < 1$ always, and therefore $f$ is contractive.
   \end{enumerate}

Exercise 5.3.1

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Exercise 5.3.1

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