IMC 2000 · Day 1 · P1

a) Yes.

Proof: Let $A = \{ x \in [0,1] : f(x) > x \}$. If $f(0) = 0$ we are done, if not then $A$ is non-empty ($0$ is in $A$) bounded, so it has supremum, say $a$. Let $b = f(a)$.

I. case: $a < b$. Then, using that $f$ is monotone and $a$ was the sup, we get $b = f(a) \le f((a+b)/2) \le (a+b)/2$, which contradicts $a < b$.

II. case: $a > b$. Then we get $b = f(a) \ge f((a+b)/2) > (a+b)/2$ contradiction.

Therefore we must have $a = b$.

b) No. Let, for example, $$f(x) = 1 - x/2 \quad \text{if } x \le 1/2$$ and $$f(x) = 1/2 - x/2 \quad \text{if } x > 1/2$$ This is clearly a good counter-example.