Exercise 4.4.7

Question

Prove that $f(x)=\sqrt{x}$ is uniformly continuous on $[0, \infty)$.

Ulisse Mini · Accepted Answer

We will show $f$ is uniformly continuous on $[0,1]$ and $[1, \infty)$ then combine them similar to Exercise 4.4.5.

\begin{enumerate}[label=(\roman*)] 
  \item Since $f$ is continuous over $[0,1]$ Theorem 4.4.7 implies $f$ is uniformly continuous on $[0,1]$.
  \item $f$ is Lipschitz on $[1, \infty)$ since $\sqrt x$ is sublinear over $[1, \infty)$
    $$
    \left|\frac{\sqrt x - \sqrt y}{x - y}\right|
    = \left|\frac{(\sqrt x - \sqrt y)}{(\sqrt x - \sqrt y)(\sqrt x + \sqrt y)}\right|
    = \left|\frac{1}{\sqrt x + \sqrt y}\right| \le 1
    $$
   \end{enumerate}

Let $\delta = \min\{\delta_1, \delta_2\}$ where $\delta_1$ is for $[0,1]$ and $\delta_2$ is for $[1, \infty)$. If $x,y$ are both in one of $[0,1]$ or $[1,\infty)$ we have $|f(x)-f(y)|<\epsilon/2$ and are done. If $x \in [0,1]$ and $y \in [1, \infty)$ then
  $$
  |f(x) - f(y)| \le |f(x) - f(1)| + |f(1) - f(y)| < \epsilon/2 + \epsilon/2 = \epsilon
  $$
  Thus $f(x) = \sqrt x$ is uniformly continuous on $[0,\infty)$.

Exercise 4.4.7

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Exercise 4.4.7

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