If $0$ $\leq$ $f(t)$ $\leq$ $1$, and $0$ $\leq$ $\frac{\partial f(t)}{\partial t}$ $\leq$ $\frac 12$. Does $f(t)$ have any fixed points?

Question

Let $f(t)$ be a differentiable function for $t$ $\in$ $[0,1]$ satisfying the above,

Does $f(t)$ have any fixed points?

I can easily prove there always exists fixed points without the second condition using $MVT$,

does $0$ $\leq$ $\frac{\partial f(t)}{\partial t}$ $\leq$ $\frac 12$ change anything?

I am very curious to know the answer of this problem, and note fixed points are when; $f(x)=x$

Answer 1

Here is a hint. Let g(t)=f(t)-t. Now use the intermediate value theorem to show that there are fixed points. For bonus, show that the fixed point is unique! Good luck!

Answer 2

You definitely have fixed points, by an Intermediate Value theorem argument: let $g(t):= f(t)-t$. Then $g(0) \ge 0$ and $g(1)< 0$ so using only continuity one necessarily obtains a fixed point, which solves $g(t^*)=0$.

But, $f'(t) \ge 0$ implies that your function is always increasing, and $f' \le \frac{1}{2}$ implies that $g'(t) < 0$ for all $t$, in particular at $t^*$.

So what if, for sake of contradiction, $g$ had two zeroes: $t^*$ and $t^{**}$? Without loss $t^* < t^{**}$, hence $g(t^{**})<g(t^*)$, but by hypothesis $g(t^*) = g(t^{**})$! Thus your fixed point is unique.