I'm reading A First Course In Dynamics, chapter about Lipschitz continuity.

There is an example I can't understand.

The function $ f(x) = \sqrt x $ defines a contraction on $ [1, \infty) $. To prove this, we show that for $ x \geq 1 $ and $ t \geq 0 $ we have $ \sqrt {x + t} \leq \sqrt x + t/2$.

Why is this enough to prove it?

Thank you for any advice.


Let $x,y \in [1, \infty)$, with $x <y$. Let $t= y-x$. Then $y=x+t$.


$$|f(y)-f(x)|=\sqrt{x+t}-\sqrt{x} \leq \frac{t}{2}=\frac{y-x}{2}$$


By rationalization we get:

$$\sqrt{x+t}-\sqrt{x}=\frac{x+t-x}{\sqrt{x+t}+\sqrt{x}}=\frac{t}{\sqrt{x+t}+\sqrt{x}}\leq \frac{t}{2}$$

  • $\begingroup$ Thank you for fast answer! I still wonder where that substitution comes from, but at least I can understand the proof now. $\endgroup$ – lpaul7 Jan 2 '13 at 14:45
  • $\begingroup$ @lpaul7 Added something to explain the substitution. Of course using the rationalisation, you don;t even need to do the substitution.... $\endgroup$ – N. S. Jan 2 '13 at 14:47

Another approach, using the fact that $\,\sqrt x\,$ is derivable in $\,[1,\infty)\,$ , for $\,x,y,\in [1,\infty)\,\,,\,x<y\,$ :

$$\exists\,c\in (x,y)\,\,\,s.t.\,\,\,\sqrt y-\sqrt x=\frac{y-x}{2\sqrt c}<\frac{y-x}{2}$$

since $\,\sqrt c>1\,$ .

  • $\begingroup$ Thank you, I like your approach more than one in the book. $\endgroup$ – lpaul7 Jan 2 '13 at 15:30
  • $\begingroup$ Anytime. Perhaps the book has some educative purpose trying that approach, but as for simplicity and clearity the mean value theorem takes it easily. $\endgroup$ – DonAntonio Jan 2 '13 at 15:41
  • $\begingroup$ Sorry, if I understand correctly, square root of x is greater or equal than 1, isn't it? $\endgroup$ – lpaul7 Jan 2 '13 at 15:53
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    $\begingroup$ I'm talking about "since $ \sqrt x>1 $ ." in your answer, but thanks for explanation! $\endgroup$ – lpaul7 Jan 2 '13 at 16:03
  • 1
    $\begingroup$ Makes much more sense now :) $\endgroup$ – lpaul7 Jan 2 '13 at 16:05

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