Showing that a function $f$ on some interval $I$ is a contractive function The sequence $(u_{n})_{n\geq1}$ is recursively defined by :
$u_{1} = α$  $   ϵ  $  $ [0, 16]$
$u_{n+1} = \sqrt{20 - u_{n}} $    ,for every natural number $n$
Show that the function $f(u) = \sqrt{20 - u} $ on the interval $I = [0, 16]$ is a contraction of $I$.
What I have done so far is :
To prove that $f$ is a contraction on $I$ , we need to show $2$ things :
$1)$ for $x$ ϵ $I$, $f(x)$ ϵ $I$ as well.
$2)$ for $u, v$ ϵ $I$ we have $|F(u) - F(v)| \leq c| u - v |$
I first drew the curve with equation $y = \sqrt{20 - u}$ in the $u,y$-plane, and the line $y=u$.  

Then I found the intersection point of the line and the curve, which is 4. I am not sure where to go from this.  In fact, I am not sure if the graph I drew was correct.
Meanwhile in the solution, it says :
If $ 0 \leq u \leq 16$, then $2 \leq $ $\sqrt{20 - u}$ $ \leq$ $ \sqrt{20} $ $\leq 5$.
I have no idea where did it come from.
Can anyone please give me a clue on how to solve this? and probably some explanations on how the solution is related to the intersection point that I found. Thanks!
 A: $$|f(u)-f(v)|=|\sqrt{20 - u}-\sqrt{20-v}|=\frac{|(\sqrt{20 - u}-\sqrt{20-v})(\sqrt{20 - u}+\sqrt{20-v})|}{\sqrt{20 - u}+\sqrt{20-v}}\\ = \frac{|u-v|}{\sqrt{20 - u}+\sqrt{20-v}}\le \frac{|u-v|}{\sqrt{20 - 16}+\sqrt{20-16}} =\frac14 |u-v|, \forall u,v \in[0,16].$$
That is,
$$|f(u)-f(v)|\le \frac14 |u-v|, \forall u,v \in[0,16],$$ from where it follows that is contractive.
A: The graph you drew is fine (though the dependent variable should be $u$ and not $x$), and shows that the function has exactly one fixed point: $u=4.$ This is not, however, what was to be shown.
The kicker here is that $0\le u\le 16.$ Consequently, $0\ge-u\ge-16,$ so $20\ge20-u\ge4=2^2.$ Taking the square root shows us that $2\le f(u)\le \sqrt{20}.$ Since $20\le 25=5^2,$ then we also know that $\sqrt{20}\le 5.$
Thus, if $0\le u\le 16,$ then $2\le f(u)\le 5,$ so $0\le f(u)\le 16,$ meaning that condition $1$ holds, as desired.

Now, what you have shown is that $f$ is a contraction map on a closed, bounded interval. It turns out that every such map has a unique fixed point, so in showing that it is a contraction map, we can conclude that it has a fixed point, though we can't necessarily say what it is. You've shown in your graph that said fixed point is $u=4,$ as I mentioned above, so it's just a nice confirmation that the general result does apply to this specific case.
