Exercise 4.4.1

Question

\begin{enumerate}[label=(\alph*)] 
  \item Show that $f(x)=x^{3}$ is continuous on all of $\mathbf{R}$.
  \item Argue, using Theorem 4.4.5, that $f$ is not uniformly continuous on $\mathbf{R}$.
  \item Show that $f$ is uniformly continuous on any bounded subset of $\mathbf{R}$.
   \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] 
  \item True since the product of continuous functions is continuous
  \item Take $x_n = n$ and $y_n = n + 1/n$ has $|x_n - y_n| 	o 0$ but
    $$
    \left|f(y_n) - f(x_n)ight|
    = \left|(n + 1/n)^3 - n^3ight|
    = \left|3n^2 \cdot \frac{1}{n} + 3n \cdot \frac 1{n^2} + \frac{1}{n^3}ight| 	o \infty
    $$
    Which shows $x^3$ is not uniformly continuous by Theorem 4.4.5
  \item Let $A$ be a bounded subset of $\mathbf R$ with $A \subset (-M, M)$. Let $\epsilon > 0$ and note that
    $$
    \left|\frac{x^3 - y^3}{x - y}ight|
    = \left|\frac{(x-y)(x^2 + xy + y^2)}{(x-y)}ight|
    = \left|x^2 + xy + y^2ight|
    $$
    Is clearly bounded on $(-M, M)$. Thus the Lipschitz condition allows us to conclude $f$ is uniformly continuous on $A$.
   \end{enumerate}

Exercise 4.4.1

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Exercise 4.4.1

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