Answer to Question #310448 in Real Analysis for Nikhil Singh

Question

Check whether the function, ,f defined below, is uniformly continuous or not:

f(x)=x^(1/2), x∈[1,2]

Answer 1

ANSWER: The function $f(x)=x^ \frac{1}{2}$ is uniformly continuous on [1,2]

EXPLANATION 1.

Let $\epsilon>0$ . We need to show that exists $\delta>0$ such that

$x,y \in[1,2]$ and $\left | x-y \right |<\delta$ imply $|f(x)-f(y)|<\epsilon$ .

We have

$f(x)-f(y) =\sqrt{x}-\sqrt{y}=\frac{\left ( \sqrt{x} -\sqrt{y}\right )\cdot\left ( \sqrt{x}+\sqrt{y} \right )}{\sqrt{x}+\sqrt{y}}=\frac{x-y}{\sqrt{x}+\sqrt{y}}$ .

Because $\sqrt{x}\geq1, \sqrt{y}\geq1$ , then $\frac{\left | x-y \right |}{\sqrt{x}+\sqrt{y}}\leq\frac{\left | x-y \right |}{2}$ . If $\delta=2\epsilon$ and $|x-y| \leq\delta$ , then $\left | f(x)-f(y) \right |\leq\frac{\left | x-y \right |}{2}\leq\frac{\delta}{2}=\frac{2\epsilon}{2}=\epsilon$ . So, we have shown, by the definition, that $f$ is uniformly continuous on [1,2].

EXPLANATION 2.

The function $f$ is continuous on a closed interval [1,2] , so by Cantor's theorem the function $f$ is uniformly continuous on [1,2].