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Prove that $f(x) = \sqrt{x^2 + x}$ where $x \in [0, +\infty)$ is uniformly continuous.

So lets take:

${\mid \sqrt{x^2+x} - \sqrt{y^2+y} \mid}^2 \, \leqslant \,\, {\mid \sqrt{x^2+x} - \sqrt{y^2+y} \mid}{\mid \sqrt{x^2+x} + \sqrt{y^2+y} \mid} \,\, = \,\, {\mid x^2+x-y^2-y \mid} \,\, = \,\, {\mid (x+y)(x-y)+(x-y) \mid} \,\, = \,\, {\mid (x-y)(x+y+1) \mid } < {\epsilon}^2 \Rightarrow \,\, {\mid \sqrt{x^2+x} - \sqrt{y^2+y} \mid} < \epsilon$

So now we know that if we take $\delta = {\epsilon}^2$ the condition for uniform continuity of this function will be met because ${\mid x - y \mid} < (\delta = {\epsilon}^2) \Rightarrow {\mid f(x)-f(y) \mid} < \epsilon$

Is this proof valid? Or I miss something?

VirrageS
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3 Answers3

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For $\;x\in [1,\infty)\;$:

$$\left(\sqrt{x^2+x}\right)'=\frac{2x+1}{2\sqrt{x^2+x}}\le\frac{2x+1}{2x}=1+\frac1{2x}\stackrel{\text{Why?}}\le\frac32$$

Thus, having a bounded derivative makes $\;\sqrt{x^2+x}\;$ uniformly continuous in $\;[1,\infty]\;$, and being continuous in the bounded, closed interval $\;[0,1]\;$ it is unif. continuous there as well. Thus, it is unif. continuous in the whole non-negative interval.

Timbuc
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  • Why is it true that if is uniformly continuous in $[1, \infty]$ and $[0, 1]$ then is uniformly continuous on whole interval $[0, \infty]$. Is there any explanation for that? – VirrageS Mar 20 '15 at 16:30
  • Yes, there is @VirrageS. For a proof, see: http://math.stackexchange.com/questions/217683/if-f-is-unformly-continuous-on-two-sets-show-that-f-is-also-uniformly-continuou –  Mar 20 '15 at 16:38
  • It's a fairly easy exercise, @VirrageS . Try it. Your "problem" may only be when you take $;x\in[0,1);,;;y\in(1,\infty);$ and they are $;\delta$-close, but... – Timbuc Mar 20 '15 at 16:38
  • Thanks, I think I get it. I will try to do it. – VirrageS Mar 20 '15 at 16:43
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Let's work with $f$ on an interval $[a,\infty[$, with $a>0$.

$$f'(x) = \frac{2x + 1}{2\sqrt{x^2 + x}} \le \frac{2x + 1}{2x} = 1 + \frac{1}{2x} \le 1 + \frac{1}{2a} $$

Now using MVT, $f$ becomes $(1 + \frac{1}{2a})$-Lipschitz on $[a,\infty[$. But Lipschitz $\implies$ Uniformly continuous.

For $[0,a]$, use the fact that this is a closed interval in $\mathbb R$, and that $f$ is continuous on this interval.

Thus $f$ is uniformly continuous on $[0,a]$ and $[a,\infty[$ each, hence on their union as well.

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Hint: Multiply by conjugate: $$\left|\sqrt{x^2+x}-\sqrt{y^2+y}\right|=\left|\sqrt{x^2+x}-\sqrt{y^2+y}\right|\frac{|\sqrt{x^2+x}+\sqrt{y^2+y}|}{|\sqrt{x^2+x}+\sqrt{y^2+y}|}$$ Note that this part of the proof only works with $[A,\infty)$, where $A > 0$. For the interval $[0,A]$, observe that $[0,A]$ is closed, and $f$ is continuous over that closed interval. There should be a theorem that says if $f$ is continuous over the closed interval then $f$ is uniformly continuous over that interval.

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