# Real Analysis - Uniform Continuity Question

So I was given this question on a worksheet and I'm pretty sure I have it but the solution seems a little too simple. Can someone let me know if I'm missing something?

"Why do you know, without any computation, that $f(x) = \sqrt x$ is uniformly continuous on the interval $[0.01,100]$. Then, find a $\delta$ that corresponds to an arbitrary choice of $\epsilon \gt 0$ that satisfies the definition of uniform continuity."

I think the answer is that since $[0.01,100]$ is a compact interval it is then uniformly continuous. Then for part 2 of the question I'm thinking that $\delta = \epsilon$ which seems a little trivial. Can someone help me confirm this?

• In mathematics, the boundaries is often crucial. $\epsilon$ can be important however small it is. – fantasie Jun 25 '18 at 5:12

You are correct on part. Since $f(x)=\sqrt{x}$ is continuous on $[0.01,100]$, and $[0.01,100]$ is compact, $f$ is necessarily uniformly continuous on $[0.01,100]$. As for the $\epsilon$ we have that following. Assume $x\ge y$, then we have $$\sqrt{|x-y|}\ge |\sqrt{x}-\sqrt{y}|$$ because by squaring both sides we see this is equivalent to $$x-y\ge x+y-2\sqrt{xy}\Leftrightarrow \sqrt{xy}\ge y$$ which is clearly true by $x\ge y$. Hence we see that one such valid $\epsilon$ is $\epsilon=\sqrt{\delta}$. By example we can see that $\epsilon=\delta$ does not work because consider for example $x=0.01$ and $x=0.11$. Then $\delta=.1$ but $|\sqrt{0.01}-\sqrt{0.11}|=0.231...>\delta$.

• Your concavity relation seems off. Take (x, y) as (5,4) – markovchain Jun 25 '18 at 4:44
• @markovchain Fixed, I was wrong there. – Will Fisher Jun 25 '18 at 4:50

Continuity on a compact interval implies uniform continuity, so you need continuity of $\sqrt{x}$ as well as compactness for your argument.

Let $x \in [a,b]$, $a>0, b>a.$

$|√x-√y| =$

$\dfrac{|x-y|}{√x+√y} \le (2√a)^{-1}|x-y|$ .

Let $\epsilon >0$ be given.

Choose $\delta = (2√a)\epsilon$ then

$|x-y| \lt \delta$ implies

$|√x-√y| \le (2√a)^{-1}|x-y| \lt \epsilon.$

If the MVT can be used, then note that for $x\in [.01,100],$

$$f'(x) = \frac{1}{2\sqrt x} \le \frac{1}{2\sqrt {.01}}.$$

Let $C$ be the number on the right. Then by the MVT, $|f(y)-f(x)| \le C|y-x|,$ and this allows you to choose $\delta = \epsilon/C.$