Exercise 4.4.9

Question

[Lipschitz Functions]
  A function $f: A ightarrow \mathrm{R}$ is called Lipschitz if there exists a bound $M>0$ such that
  $$
  \left|\frac{f(x)-f(y)}{x-y}ight| \leq M
  $$
  for all $x 
eq y \in A$. Geometrically speaking, a function $f$ is Lipschitz if there is a uniform bound on the magnitude of the slopes of lines drawn through any two points on the graph of $f$.
  \begin{enumerate}[label=(\alph*)] 
  \item Show that if $f: A ightarrow \mathbf{R}$ is Lipschitz, then it is uniformly continuous on $A$.
  \item Is the converse statement true? Are all uniformly continuous functions necessarily Lipschitz?
   \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] 
  \item Choose $\epsilon > 0$ and set $\delta = \epsilon/M$ to get $|f(x) - f(y)| \le M|x-y| < M\delta < \epsilon$.
  \item No, consider $f(x) = \sqrt x$ over $[0,1]$ which is uniformly continuous by Theorem 4.4.7 however at $x=0$ we have
    $$
    \left|\frac{f(x) - f(y)}{x - y}\right| = \left|\frac{- \sqrt y}{-y}\right| = \frac{1}{\sqrt y}
    $$
    Which is unbounded for small $y$.
   \end{enumerate}

Exercise 4.4.9

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Exercise 4.4.9

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