Prove that function defines a contraction

Question

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.

Answer 1

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

So

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

Added

By rationalization we get:

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

Answer 2

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