Proving that $x-x^2$ is a contraction without mean value theorem

Solution 1:

Sure, it's fairly straightforward.

Consider that $f(x) - f(y) = (x - x^2) - (y - y^2) = (x - y) - (x^2 - y^2) = (x - y) (1 - x - y)$.

Then in particular, we have $d(f(x), f(y)) = |x - y| |1 - x - y| = d(x, y) |1 - x - y|$.

Now we see that $-1 \leq 1 - x - y \leq 1$. In particular, the extremes are only achieved when $x = y \in \{0, 1\}$, so when $x \neq y$, we have $-1 < 1 - x - y < 1$. In this case, we see that $|1 - x - y| < 1$. Therefore, we see that when $x \neq y$, we have $d(f(x), f(y)) = d(x, y) |1 - x - y| < d(x, y)$.

Now suppose the map were a strict contraction. Then take some $c \in (0, 1)$ such that for all $x \neq y$, we have $d(f(x), f(y)) < c d(x, y)$.

Then let $k = \frac{1 + c}{2}$; then $0 < c < k < 1$. And we see that $d(f(1), f(k)) = d(1, k) |1 - 1 - k| = d(1, k) |k| = d(1, k) k > d(1, k) c > d(f(1), f(k))$. This is a contradiction.

Solution 2:

Hint: $$f(x)-f(y) = (x-y)(1-(x+y)), $$ $$0 \le x+y \le 2.$$