Given a real-valued function $f$ which is continuous on $[0,1]$. Assume that $0 \le f(x) \le 1$ for each $x \in [0,1]$. Prove that there is at least one point $c \in [0,1]$ for which $f(c) = c$.
Define $g(x) = f(x) - x$. Then $g(0) = f(0)$ and $g(1) = f(1) - 1$. If $g(0) = 0,$ then $f(0) = 0$. If $g(1) = 0$, then $f(1) = 1$ and in either case this is satisfied. We can suppose then that $0 < c < 1$, and so $g(c) = f(c) - c$. Since $0 < f(c) < 1$, we have that $-c < g(c) < 1 - c$. It's clear that $-c$ and $1-c$ have opposiet signs, so by Bolzano's theorem, there exists a $k \in (c,1-c)$ such…