Show that the function $f(x) = (x^2 + a)/2$ is a contraction on $[0,a]$ when $0<a<1$

Question

Let $0 < a < 1$ and $f(x) = (x^2 + a)/2$. Show that $f : [0, a] \to [0, a]$ and that $f$ is a contraction. Find the fixed point of $f$."

For this problem, since there isn't any metric $d(x,y)$ mentioned, can I just assume that this $d(x,y)=|x-y|$?

If so, then let any $x$, $y$ from $[0, a]$, then $d(f(x),f(y))=(|x+y|\cdot|x-y|)/2$ and $d(x,y)=|x-y|$. Then since $x$, $y$ belongs to $[0, a]$, I have $|x+y|/2$ belongs to $[0, a]$ as well, then the contraction coefficient is $a$. Is this the right way to do it?

Answer 1

You are right. And that about fixed point was a mistake.

Answer 2

It seems you are right. Moreover, $d(x,y)=|x-y|$ is the standard metric on $\mathbb R$. But you also should show that $f(x)\in [0;a]$ for each $x\in [0,a]$.