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$\sqrt{x^2+1}$ uniformly continuous on (0, 1)?

How to deal with such problems? Please help. I know the definition of U C. but unable to handle the problem.

rama_ran
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    The function $\sqrt{x^2+1}$ is differentiable on the interval $(0,1)$ and it's derivative is bounded by $\frac{1}{2}$. – Tornado Apr 16 '17 at 04:04
  • https://math.stackexchange.com/questions/569928/sqrt-x-is-uniformly-continuous – dantopa Apr 16 '17 at 04:05
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    The function extends fine to $[0,1]$, where it must be uniformly continuous, by compactness of the domain. – Lubin Apr 16 '17 at 04:06
  • In fact it's uniformly continuous on all of $\mathbb R.$ – zhw. Apr 16 '17 at 05:57
  • @zhw. then for $x,y\in \mathbb{R}$ how to show $\frac{|x+y|}{\sqrt{x^2+1}+\sqrt{y^2+1}}|x-y| \leq\ \frac{|x|+|y|}{2}|x-y| \leq |x-y|$ – rama_ran Apr 16 '17 at 06:54
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    Why prove that when you can show $|f'(x)|\le 1$ everywhere very simply, so by the MVT $|f(y)-f(x)| \le |y-x|$ everywhere. – zhw. Apr 16 '17 at 19:11

2 Answers2

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Hint: Observe \begin{align} \left|\sqrt{x^2+1}-\sqrt{y^2+1}\right| =& \frac{|x^2-y^2|}{\sqrt{x^2+1}+\sqrt{y^2+1}} \\ =&\ \frac{|x+y|}{\sqrt{x^2+1}+\sqrt{y^2+1}}|x-y|\\ \leq&\ \frac{|x|+|y|}{2}|x-y| \leq |x-y| \end{align} when $x, y \in (0, 1)$.

Edit: This is more or less the solution. However, you should try to use the definition of uniform continuity to fill in the details.

More Edit: To show $f(x)= \sqrt{x^2+1}$ is uniformly continuous on all of $\mathbb{R}$ observe \begin{align} \frac{|x|+|y|}{\sqrt{x^2+1}+\sqrt{y^2+1}} \leq \frac{|x|}{\sqrt{x^2+1}}+\frac{|y|}{\sqrt{y^2+1}}\leq 2. \end{align}

Jacky Chong
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  • The interval in the question is $(0,1)$ and you have taken $[0, 1]$. Have you taken it intensionally? If yes, please mention the cause behind it. – rama_ran Apr 16 '17 at 05:36
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    @rama_ran It really doesn't matter for this problem whether we consider $(0, 1)$ or $[0, 1]$. – Jacky Chong Apr 16 '17 at 05:42
  • @rama_ran The only time we used the fact that we have $(0, 1)$ is in the last and second to last steps in the derivation of the above inequality. For both those inequalities, we don't really care whether we can plug 1 or 0 in. – Jacky Chong Apr 16 '17 at 05:55
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$f$ is continuous on $[0,1]$ and differentiable in $(0,1)$ so if $0\le x\le y\le 1$ an application of the mean value theorem gives $|f(x)-f(y)|=|f'(c)(x-y)|$ for some $c\in (x,y).$ The result now follows because $|f'|\le 1$ on $(0,1)$.

If you don't want to use MVT you can observe that $f$ extends to a continuous function on the compact set $[0,1]$.

Matematleta
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  • Please be kind enough to elaborate what you are suggesting to extend f to a continuous function on the compact set [0,1]. – rama_ran Apr 16 '17 at 05:24
  • $f(x)=\sqrt{x^2+1}$ is even continuous on the $closed$ interval $[0,1]$. Since this is a closed interval it is compact, which makes $f$ uniformly continuous there, and in particular on $(0,1)$. – Matematleta Apr 16 '17 at 14:21