# Prove that function defines a contraction

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?

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

So

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

By rationalization we get:

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

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 Thank you for fast answer! I still wonder where that substitution comes from, but at least I can understand the proof now. – lpaul7 Jan 2 at 14:45 @lpaul7 Added something to explain the substitution. Of course using the rationalisation, you don;t even need to do the substitution.... – N. S. Jan 2 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 $\,\sqrt c>1\,$ .

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